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PrepBank · বিষয়ভিত্তিক প্রশ্ন

Probability, Permutation and Combination

মোট প্রশ্ন৯৬৯এই পাতা১০০প্রতি পাতা১০০
প্রশ্ন দেখুনঅনুশীলন
ঘনত্ব
উত্তর
উত্তরিতবর্তমানপুনরায় দেখুনঅসম্পূর্ণ

Probability, Permutation and Combination

PrepBank · পাতা ৬ / ১০ · ৫০১–৬০০ / ৯৬৯

৫০১.
How many two-digit numbers can be formed using the digits 3, 5, and 7 if repetition of digits is allowed?
  1. ক
    9 possible two-digit numbers 
  2. খ
    10 possible two-digit numbers 
  3. গ
    8 possible two-digit numbers 
  4. ঘ
    7 possible two-digit numbers 
ব্যাখ্যা

Question: How many two-digit numbers can be formed using the digits 3, 5, and 7 if repetition of digits is allowed?

Solution:
Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3.

∴ 9 possible two-digit numbers can be formed.

The 9 possible two-digit numbers are-
33, 35, 37, 53, 55, 57, 73, 75, 77 

৫০২.
In how many different orders can 8 different colors of flowers be arranged in a straight line?
  1. ক
    ক) 8
  2. খ
    খ) 64
  3. গ
    গ) 40,320
  4. ঘ
    ঘ) 80,640
ব্যাখ্যা
Different orders of colors can be = 8! = 40,320
৫০৩.
In how many ways can 3 boys and 3 girls be arranged in a line if all boys must stand together?
  1. ক
    72
  2. খ
    144
  3. গ
    240
  4. ঘ
    360
ব্যাখ্যা
Question: In how many ways can 3 boys and 3 girls be arranged in a line if all boys must stand together?

Solution: 
Since all 3 boys must stand together, we can consider them as one combined unit.

Number of ways to arrange these 4 units (3 girls + 1 group of boys) = 4! = 24
Number of arrangements among the 3 boys = 3! = 6

∴ Total arrangements = 24 × 6 = 144
৫০৪.
Three unbiased coins are tossed. What is the probability of getting at least two tails?
  1. ক
    ক) 1/4
  2. খ
    খ) 1/2
  3. গ
    গ) 1/8
  4. ঘ
    ঘ) 2/3
ব্যাখ্যা
Question: Three unbiased coins are tossed. What is the probability of getting at least two tails?

Solution:
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at least two tails.
Then E = {TTT, TTH, THT, HTT}

∴ P(E) = n(E)/n(S)
= 4/8
= 1/2
৫০৫.
Five students are to be arranged on five chairs for a photograph. Three of these are girls and the rest are boys. Find out the number of ways in which all three girls do not occupy consecutive seats.
  1. ক
    ক) 120
  2. খ
    খ) 36
  3. গ
    গ) 84
  4. ঘ
    ঘ) 76
ব্যাখ্যা

As per the question, three girls can’t occupy consecutive seats but two can.

Therefore, if we find the number of ways in which all three girls occupy consecutive seats and subtract this number from the total number of ways in which the five people can be arranged among themselves, we will get the required answer.

5 students can be arranged among themselves in 5p5 ways = 120 ways.

Assume that the 3 girls are one entity. The total number of ways in which they can be arranged among themselves = 3! = 6
Also, the set of three girls and the other students can be arranged among themselves in 3! = 6 ways.
Thus, the total number of ways in which three girls are together = 6 × 6 = 36

Thus, a number of ways in which all 3 girls will not occupy consecutive seats = 120 – 36 = 84.

৫০৬.
How many 4 letter code can be formed using the first 9 letters of the English alphabets, if no letter can be repeated?
  1. ক
    3024
  2. খ
    3036
  3. গ
    3021
  4. ঘ
    3034
ব্যাখ্যা
Question: How many 4 letter code can be formed using the first 9 letters of the English alphabets, if no letter can be repeated?

Solution:
By fundamental principle, it is (9 × 8 × 7 × 6) ways
= 3024 ways

Alternative:
9P4 = 9!/(9 - 4)!
= 9!/5!
= (9 × 8 × 7 × 6 × 5!)/5!
= 3024 ways
৫০৭.
There are 5 red, 4 black and 3 white color bulbs. If a bulb is picked at random, what is the probability of having either a red or a black bulb?
  1. ক
    ক) 2/3
  2. খ
    খ) 5/12
  3. গ
    গ) 3/4
  4. ঘ
    ঘ) none of the above
ব্যাখ্যা
Question: There are 5 red, 4 black and 3 white color bulbs. If a bulb is picked at random, what is the probability of having either a red or a black bulb? 

Solution: 
Red color bulb = 5
Black color bulb = 4
White color bulb = 5

Total bulb = 5 + 4 + 3 = 12

Total red and black bulb = (5 + 4) = 9
The probability of having either a red or a black bulb = 9/12 = 3/4
৫০৮.
In how many different ways can be letters of the word 'CYCLE' be arranged?
  1. ক
    32 ways
  2. খ
    60 ways
  3. গ
    80 ways
  4. ঘ
    110 ways
ব্যাখ্যা

Question: In how many different ways can be letters of the word 'CYCLE' be arranged?

Solution:
CYCLE whereas total 5 letters and C comes two times.

So, arrangements are = 5!/2! 
= 60 ways

৫০৯.
In how many ways can you rearrange the word JUMBLE such that the rearranged word starts with a vowel?
  1. ক
    120
  2. খ
    230
  3. গ
    250
  4. ঘ
    360
  5. ঙ
    None of these
ব্যাখ্যা
Question: In how many ways can you rearrange the word JUMBLE such that the rearranged word starts with a vowel?

Solution:
JUMBLE is a six-lettered word.
Since the rearranged word has to start with a vowel, the first letter can be either U or E.
The balance 5 letters can be arranged in 5P5 or 5! ways.

∴  Total number of words = 2 × 5! = 240.
৫১০.
A box contains 4 red, 5 green and 6 white balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or green?
  1. ক
    ক) 2/5
  2. খ
    খ) 3/5
  3. গ
    গ) 1/7
  4. ঘ
    ঘ) 7/15
ব্যাখ্যা

Total number of balls = (4 + 5 + 6)
= 15.
P(drawing a red ball or a green ball)
= P(red) + P(green)
= (4/15 + 5/15)
= 9/15
= 3/5.

৫১১.
If there are 12 persons in a party and if each of them shake hands with each other, then number of hand shakes in party are:
  1. ক
    22
  2. খ
    33
  3. গ
    66
  4. ঘ
    88
ব্যাখ্যা
Question: If there are 12 persons in a party and if each of them shake hands with each other, then number of hand shakes in party are:

Solution:

Two people can make 1 handshake.

∴ Number of handshakes
= 12C2
= (12 × 11)/(2 × 1)
= 66

∴ The number of hand shakes in party are: 66
৫১২.
From 6 men and 4 women, a committee of 3 people is to be formed including at least 1 woman. In how many ways can this be done?
  1. ক
    80
  2. খ
    120
  3. গ
    100
  4. ঘ
    90
  5. ঙ
    None of these
ব্যাখ্যা
Question: From 6 men and 4 women, a committee of 3 people is to be formed including at least 1 woman. In how many ways can this be done?

Solution:
Total people = 6 men + 4 women = 10 people.
Total ways to choose 3 people,
10C3 = 10!/3!(10 - 3)!
= (10 × 9 × 8 × 7!)/(3! × 7!)
= 120

Again,
Ways to choose 3 people with no women (all men),
6C3 = 6!/3!(6 - 3)!
= 20

∴ Ways to choose 3 people with at least 1 woman = 120 - 20 = 100.
৫১৩.
Committee X has 4 members, committee Y has 5 members, and these committees have no members in common. If a task force is to be formed consisting of one member of X and one member of Y, how many different task forces are possible?
  1. ক
    30
  2. খ
    24
  3. গ
    20
  4. ঘ
    62
ব্যাখ্যা
Question: Committee X has 4 members, committee Y has 5 members, and these committees have no members in common. If a task force is to be formed consisting of one member of X and one member of Y, how many different task forces are possible?

Solution:
Committee X has 4 members: Choosing one out of 4 = 4C1 =4
Committee Y has 5 members: Choosing one out of 5 = 5C1 =5

total = 4 × 5 = 20
৫১৪.
In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?
  1. ক
    360
  2. খ
    36
  3. গ
    72
  4. ঘ
    60
ব্যাখ্যা
Question: In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?

Solution:
ABACUS is a 6 letter word with 3 of the letters being vowels.
If the 3 vowels have to appear together as stated in the question, then there will 3 consonants and a set of 3 vowels grouped together.
One group of 3 vowels and 3 consonants are essentially 4 elements to be rearranged.
The number of possible rearrangements is 4! = 24

The group of 3 vowels contains two a s and one u
The 3 vowels can rearrange amongst themselves in 3!/2! = 3

Hence, the total number of rearrangements in which the vowels appear together are = 24 × 3 = 72
৫১৫.
In how many different ways can five friends sit for a photograph of five chairs in a row?
  1. ক
    120 ways
  2. খ
    24 ways
  3. গ
    240 ways
  4. ঘ
    720 ways
ব্যাখ্যা
Question: In how many different ways can five friends sit for a photograph of five chairs in a row?

Solution:
We have to find total number of arrangements of 5 persons seated in a row.
We know that arrangement of n different things can be done in n! ways.

So, arrangements of 5 persons can be done in 5! = 120 ways.
৫১৬.
Ramesh has a garment shop. He currently has 6 black, 4 red, 2 white and 3 blue shirts of the same size in the stock. He picks 2 shirts randomly for the display. What is the probability that either both shirts are white or blue?
  1. ক
    1/105
  2. খ
    1/35
  3. গ
    4/105
  4. ঘ
    1/15
ব্যাখ্যা

Probability = What we want/Total
Or = Add
And = Multiply

Both white OR both blue
Total shirts = 15
There are 2 white and 3 blue shirts

Probability for 2 white shirts = (2/15) × (1/14) = 1/105
Probability for 2 blue shirts = (3/15) × (2/14) = 1/35

∴ Total probability = (1/105) + (1/35) = 4/105.

৫১৭.
There are 8 multiple-choice questions in an examination. Each question has 4 options. In how many different ways can these questions be answered?
  1. ক
    24 ways
  2. খ
    84 ways
  3. গ
    48 ways
  4. ঘ
    28 ways
ব্যাখ্যা

Question: There are 8 multiple-choice questions in an examination. Each question has 4 options. In how many different ways can these questions be answered?

Solution:
For each multiple-choice question, there are 4 possible answers (options A, B, C, or D).

Since there are 8 questions, and each question can be answered independently:

Total number of ways = 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4
= 48 ways
= 65536 ways

৫১৮.
In how many different ways can the letters of the word 'BANANA' be arranged?
  1. ক
    ক) 30
  2. খ
    খ) 120
  3. গ
    গ) 60
  4. ঘ
    ঘ) 720
ব্যাখ্যা
Question: In how many different ways can the letters of the word 'BANANA' be arranged?

Solution:
The word BANANA
Here total letters = 6
No of A = 3
No of N = 2

∴ Arrangement = 6!/(3! × 2!)
= 60
৫১৯.
A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is-
  1. ক
    2/91
  2. খ
    1/22
  3. গ
    3/22
  4. ঘ
    2/77
  5. ঙ
    None of these
ব্যাখ্যা
Question: A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is-

Solution:
Let S be the sample space.
Then,
n(S) = number of ways of drawing 3 balls out of 15 = 15C3 = 455. 

Let E = event of getting all the 3 red balls.
n(E) = 5C3 = 10

∴ P(E) = n(E)/n(S) = 10/455 = 2/91
৫২০.
A student is required to solve 6 out of the 10 questions in a test. The questions are divided into two sections of 5 questions each. In how many ways can the student select the questions to solve if not more than 4 questions can be chosen from either section?
  1. ক
    150
  2. খ
    180
  3. গ
    200
  4. ঘ
    220
ব্যাখ্যা
Question: A student is required to solve 6 out of the 10 questions in a test. The questions are divided into two sections of 5 questions each. In how many ways can the student select the questions to solve if not more than 4 questions can be chosen from either section?

Solution: 
Number of ways = (5C4 × 5C2) + (5C3 × 5C3) + (5C2 × 5C4) 
= 5 × 10 + 10 × 10 + 10 × 5 
= 200
৫২১.
A volunteer group is organizing a charity event. The group consists of 3 male volunteers and 5 female volunteers. If a team of 4 volunteers is to be chosen at random to lead the event, what is the probability that the team will include at least 2 women?
  1. ক
    5/7
  2. খ
    3/7
  3. গ
    3/10
  4. ঘ
    13/14
  5. ঙ
    None
ব্যাখ্যা
Question: A volunteer group is organizing a charity event. The group consists of 3 male volunteers and 5 female volunteers. If a team of 4 volunteers is to be chosen at random to lead the event, what is the probability that the team will include at least 2 women?

Solution:
Given,
Total people = 8
∴ Ways of selecting 4 people from 8 = 8C4
= 70

We want at least 2 women, so there are 3 possible combinations:
1st combination: 2 women and 2 men = 5C2 × 3C2 = 10 × 3 = 30
2nd combination: 3 women and 1 man = 5C3 × 3C1 = 10 × 3 = 30
3rd combination: 4 women 0 man = 5C4 = 5

∴ Total outcomes = 30 + 30 + 5
= 65

∴ Probability = 65/70
= 13/14
৫২২.
How many ways the letters of the word 'DEPOSIT' can be arranged?
  1. ক
    5040
  2. খ
    2520
  3. গ
    1008
  4. ঘ
    49
ব্যাখ্যা
Question: How many ways the letters of the word 'DEPOSIT' can be arranged?

Solution:
the given words contain 7 diffrerent letters.

∴ they can be arranged in = 7! ways
= 5040 ways
৫২৩.
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 4 times, what is the probability that it will land heads up on the first 3 flips and not on the last flip?
  1. ক
    1/8
  2. খ
    1/12
  3. গ
    1/16
  4. ঘ
    1/18
ব্যাখ্যা
Question: If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 4 times, what is the probability that it will land heads up on the first 3 flips and not on the last flip?

Solution: 
The probability of landing heads and not landing on heads is same = 1/2
The probability of first three heads =(1/2) × (1/2) × (1/2)
The probability of last  landing not on heads = 1/2
The total probability =(1/2) × (1/2) × (1/2) × (1/2)
= 1/ 24
= 1/16
৫২৪.
In how many different ways can the letters of the word 'CORRUPTION' be arranged so that the vowels always come together?
  1. ক
    ক) 30240
  2. খ
    খ) 30200
  3. গ
    গ) 30420
  4. ঘ
    ঘ) None of the above
ব্যাখ্যা
Question: In how many different ways can the letters of the word 'CORRUPTION' be arranged so that the vowels always come together?

Solution: 
In the word 'CORRUPTION', we treat the vowels OUIO as one letter.

Thus, we have CRRPTN (OUIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =    7!/2!   = 2520.

Now, 4 vowels in which O occurs 2 times and the rest are different, can be arranged in    4!/2!   = 12 ways.

Required number of ways = (2520 x 12) = 30240.
৫২৫.
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
  1. ক
    128
  2. খ
    196
  3. গ
    346
  4. ঘ
    280
ব্যাখ্যা
Question: A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

Solution:
The student can choose 4 questions from first 5 questions or he can also choose 5 questions from the first five questions.

∴ No. of choices available to the student = 5C4 × 8C6 + 5C5 × 8C5
= 5 × 28 + 1 × 56
= 196
৫২৬.
All possible three digit numbers are formed by 1, 2, 3. If one number is chosen randomly; the probability that it would be divisible by 111 is -
  1. ক
    ক) 0
  2. খ
    খ) 2/9
  3. গ
    গ) 1/3
  4. ঘ
    ঘ) 1/4
ব্যাখ্যা

1,2,3 অংকগুলো দ্বারা গঠিত সংখ্যাগুলো হলোঃ 123, 132, 213, 231, 312, 321 (6টি)
এখন এর মধ্যে 111 দ্বারা বিভাজ্য সংখ্যা একটিও নেই।
Probability = 0/6 = 0

৫২৭.
In how many ways can 4 people from a group of 6 people be seated around a circle table?
  1. ক
    65
  2. খ
    90
  3. গ
    120
  4. ঘ
    144
ব্যাখ্যা

Question: In how many ways can 4 people from a group of 6 people be seated around a circle table?

Solution:
এখানে 6 জন মানুষের মধ্য থেকে 4 জনকে নিয়ে একটি বৃত্তাকার টেবিলে বসাতে হবে।

প্রথমে 6 জন থেকে 4 জনকে বাছাই করার উপায় = 6C4
= 6!/(4! × 2!) 
= (6 × 5)/(2 × 1)
= 15


আমরা জানি, n জন ব্যক্তিকে একটি বৃত্তাকার টেবিলে বসানোর উপায় = (n - 1)!
সুতরাং, বাছাইকৃত 4 জন ব্যক্তিকে বৃত্তাকার টেবিলে বসানোর উপায় = (4 - 1)! = 3!
= 3 × 2 × 1 = 6 উপায়।

∴ মোট বিন্যাস সংখ্যা = (বাছাই করার উপায়) × (বৃত্তাকারে বসানোর উপায়)
= 15 × 6
= 90 ways

৫২৮.
There are five comics numbered from 1 to 5. In how many ways can they be arranged, so that part-1 and part-4 are never together?
  1. ক
    ক) 48
  2. খ
    খ) 72
  3. গ
    গ) 96
  4. ঘ
    ঘ) 120
ব্যাখ্যা
Question: There are five comics numbered from 1 to 5. In how many ways can they be arranged, so that part-1 and part-4 are never together?

Solution:
The total number of ways in which 5 part can be arranged = 5!
= 120.
The total number of ways in which part-1 and part-4 are always together:
= 4! × 2!
= 48

∴ Therefore, the total number of arrangements, in which they are not together is:
= 120 - 48
= 72
৫২৯.
How many distinct permutations can be formed with the letters of “BALLOON”?
  1. ক
    720
  2. খ
    840
  3. গ
    1260
  4. ঘ
    2520
ব্যাখ্যা
Question: How many distinct permutations can be formed with the letters of “BALLOON”?

Solution: 
BALLOON has 7 letters.
B – 1 time
A – 1 time
L – 2 times
O – 2 times
N – 1 time

∴ Total permutations = 7!/(2! × 2!) = 1260
৫৩০.
A bag contains 7 red and 9 green marbles. One marble is drawn at random. What is the probability that the marble drawn is not red?
  1. ক
    1/7
  2. খ
    1/9
  3. গ
    7/16
  4. ঘ
    9/16
ব্যাখ্যা

Question: A bag contains 7 red and 9 green marbles. One marble is drawn at random. What is the probability that the marble drawn is not red?

Solution:
Number of red marbles = 7
Number of green marbles = 9
Total number of marbles = 7 + 9 = 16

P (red marble) = 7/16

∴ P (not red marble) = 1 - (7/16)
= 9/16

৫৩১.
A student is required to solve 6 out of the 10 questions in a test. The questions are divided into two sections of 5 questions each. In how many ways can the student select the questions to solve if not more than 4 questions can be chosen from either section?
  1. ক
    250 ways
  2. খ
    100 ways
  3. গ
    150 ways
  4. ঘ
    200 ways
  5. ঙ
    300 ways
ব্যাখ্যা
Question: A student is required to solve 6 out of the 10 questions in a test. The questions are divided into two sections of 5 questions each. In how many ways can the student select the questions to solve if not more than 4 questions can be chosen from either section?

Solution:
Possibility 1: This can be done in 5C4 × 5C2 = 5 × 10 = 50 ways
Possibility 2: This can be done in 5C3 × 5C3 = 10 × 10 = 100 ways
Possibility 3:  This can be done in 5C2 × 5C4 = 10 × 5 = 50 ways

Total number of ways = 50 + 100 + 50 = 200 ways
৫৩২.
A five-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following pattern. The first digit must be 2 or 4, 3rd digit must be a multiple of 3 and no two consecutive digits should be same. How many different codes are possible?
  1. ক
    2,016
  2. খ
    3,888
  3. গ
    4,608
  4. ঘ
    5,184
  5. ঙ
    6,236
ব্যাখ্যা
Question: A five-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following pattern. The first digit must be 2 or 4, 3rd digit must be a multiple of 3 and no two consecutive digits should be same. How many different codes are possible?

Solution:
• A five-digit code uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
• The first digit must be 2 or 4
• 3rd digit must be a multiple of 3
• No two consecutive digits should be same

Let the 5-digit number is abcde.
• Ways to fill ‘a’ = 2(2 or 4)
• Ways to fill ‘c’ = 4 (0,3,6,9)
• Hence, we filled two different digits in ‘a’ and ‘c’.
    o So, out of 10 digits, b cannot take 2 digits.
    o Thus, ‘b’ can be filled in 8 ways.
• Now, ‘d’ can be filled in 9 ways as similar digit cannot be in ‘c’ and ‘d’.
• Similarly, ‘e’ can be filled in 9 ways.

Thus, total ways to fill= 2 × 8 × 4 × 9 × 9 = 5184
৫৩৩.
How many 4 letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', where repetition of letters is not allowed?
  1. ক
    2080
  2. খ
    3650
  3. গ
    4040
  4. ঘ
    5040
  5. ঙ
    None
ব্যাখ্যা
Question: How many 4 letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', where repetition of letters is not allowed?

Solution:
Here,
'LOGARITHMS' contains 10 different letters.

So the number of word = Number of arrangements of 10 letters, taking 4 at a time
= 10P4
= (10 × 9 × 8 × 7)
= 5040
৫৩৪.
Find the number of triangles that can be formed by joining the angular points of a polygon of 5 sides as vertices.
  1. ক
    8
  2. খ
    10
  3. গ
    15
  4. ঘ
    20
ব্যাখ্যা
Question: Find the number of triangles that can be formed by joining the angular points of a polygon of 5 sides as vertices.

Solution:
 the number of triangles which can be formed by joining the angular points of a polygon of 5 sides as vertices
= 5C3
= 5!/(3! × 2!)
= 10 ways
৫৩৫.
In a party 15 people shake their hands with each other. How many times did the hand-shakes take place?
  1. ক
    105
  2. খ
    120
  3. গ
    135
  4. ঘ
    165
ব্যাখ্যা
Question: In a party 15 people shake their hands with each other. How many times did the hand-shakes take place?

Solution:
Number of hand-shakes = 15C2 = 105
৫৩৬.
Three gentlemen and three ladies are candidates for two vacancies. A voter has to vote for two candidates. In how many ways can one cast his vote?
  1. ক
    15
  2. খ
    12
  3. গ
    10
  4. ঘ
    18
ব্যাখ্যা
Total number of vacancies = 2
Number of candidates = 6
There are 6 candidates and a voter has to vote for any two of them.
So, the required number of ways = 6C2 = 15
৫৩৭.
What is the probability of getting an even number when a dice is rolled?
  1. ক
    1/5
  2. খ
    1/3
  3. গ
    1/2
  4. ঘ
    1/4
  5. ঙ
    None of these
ব্যাখ্যা
Question: What is the probability of getting an even number when a dice is rolled?

Solution:
The sample space when a dice is rolled, S = (1, 2, 3, 4, 5 and 6)
So, n(S) = 6

E is the event of getting an even number.
So, n(E) = 3

∴ Probability = n(E)/n(S) = 3/6 =1/2
৫৩৮.
There are 5 green 7 red balls. Two balls are selected one by one without replacement. Find the probability that first is green and second is red.
  1. ক
    35/132
  2. খ
    35/144
  3. গ
    139/144
  4. ঘ
    12/23
  5. ঙ
    None of these
ব্যাখ্যা
Question: There are 5 green 7 red balls. Two balls are selected one by one without replacement. Find the probability that first is green and second is red.

Solution:
P (G) × P (R)
= (5/12) × (7/11)
= 35/132
৫৩৯.
When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?
  1. ক
    25
  2. খ
    41
  3. গ
    42
  4. ঘ
    22
ব্যাখ্যা

The question requires you to find a number of the outcomes in which at most 3 coins turn up as heads.
i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.

The number of outcomes in which 0 coins turn heads is,
6C0 = 1 outcome.

The number of outcomes in which 1 coin turns head is,
6C1 = 6 outcomes.

The number of outcomes in which 2 coins turn heads is,
6C2 = 15 outcomes.

The number of outcomes in which 3 coins turn heads is,
6C3

Therefore, total number of outcomes
= 1 + 6 + 15 + 20
= 42 outcomes.

৫৪০.
A box contains 5 detective and 15 non-detective bulbs. Two bulbs are chosen at random. Find the probability that both the bulbs are non-defective.
  1. ক
    ক) 5/19
  2. খ
    খ) 3/20
  3. গ
    গ) 21/38
  4. ঘ
    ঘ) 25/38
  5. ঙ
    ঙ) None of these
ব্যাখ্যা

n(S) = 20C2 = 190
n(E) = 15C2 = 105
Therefore,
P(E) = 105/190
= 21/38

৫৪১.
Tickets numbered 1 to 30 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 2 and 3?
  1. ক
    ক) 2/3
  2. খ
    খ) 1/6
  3. গ
    গ) 1/2
  4. ঘ
    ঘ) none of these
ব্যাখ্যা
Question: Tickets numbered 1 to 30 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 2 and 3?

Solution: 
Number which are multiple of 2 and 3 between 1 to 30  = {6, 12, 18, 24, 30} 

∴ The probability is =  5/30 = 1/6
৫৪২.
If two coins are tossed, what is the probability of getting at least one head?
  1. ক
    1/2
  2. খ
    1/4
  3. গ
    2/3
  4. ঘ
    3/4
ব্যাখ্যা

Question: If two coins are tossed, what is the probability of getting at least one head?

Solution:
দুটি মুদ্রা নিক্ষেপ করলে নমুনা ক্ষেত্রটি হলো,
S = {HH, HT, TH, TT}
এখানে মোট ফলাফল সংখ্যা, n(S) = 4

"at least one head" বলতে বোঝায় কমপক্ষে 1টি head অর্থাৎ 1 টি অথবা 2 টিও হতে পারে।

1টি head আছে এমন ফলাফল: {HT, TH}
2টি head আছে এমন ফলাফল: {HH}

∴ অনুকূল ফলাফল, n(E) = {HT, TH, HH}
∴ n(E) = 3

∴ সম্ভাবনা P(E) = n(E)/n(S)
∴ P(E) = 3/4

৫৪৩.
If there are 15 dots on a circle,how many triangles can be formed?
  1. ক
    ক) 455
  2. খ
    খ) 450
  3. গ
    গ) 469
  4. ঘ
    ঘ) 500
  5. ঙ
    ঙ) None of these
ব্যাখ্যা

There are 15 dots in total,and to make a triangle we need to select any three of those dots.
So, 15C3 = 455

৫৪৪.
Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
  1. ক
    6/5
  2. খ
    3/4
  3. গ
    1/2
  4. ঘ
    2/3
ব্যাখ্যা

Question: Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

Solution:
In a simultaneous throw of two dice,
we have n(S) = (6 × 6) = 36

Now, we find the odd product,
E = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5 ,5)}

∴ n(E) = 9

∴ P(odd product)= 9/36 = 1/4

​​Now, we find the even product,
​probability(product even) = 1 - (1/4) = 3/4

৫৪৫.
A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?
  1. ক
    1/7
  2. খ
    3/5
  3. গ
    3/7
  4. ঘ
    3/8
ব্যাখ্যা
Question:  A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

Solution:
the probability that the team will have exactly 2 women is = (5C2 × 3C2)/8C4
= 30/70
= 3/7
৫৪৬.
An examination paper consists of 8 questions divided into two parts 'A' and 'B'. Part A consists of 4 questions, and Part B consists of 4 questions. A candidate is required to attempt 5 questions selecting at least 2 questions from each part. In how many ways can the candidate select the question?
  1. ক
    56 ways
  2. খ
    96 ways
  3. গ
    72 ways
  4. ঘ
    48 ways
  5. ঙ
    120 ways
ব্যাখ্যা
Question: An examination paper consists of 8 questions divided into two parts 'A' and 'B'. Part A consists of 4 questions, and Part B consists of 4 questions. A candidate is required to attempt 5 questions selecting at least 2 questions from each part. In how many ways can the candidate select the question?

Solution:
Possibility 1: This can be done in 4C2 × 4C3 = 6 × 4 = 24 ways
Possibility 2: This can be done in 4C3 × 4C2 = 4 × 6 = 24 ways

∴ Total number of ways = 24 + 24 = 48 ways
৫৪৭.
5P2 - 5C2 =?
  1. ক
    ক) 0
  2. খ
    খ) 5
  3. গ
    গ) 10
  4. ঘ
    ঘ) 15
ব্যাখ্যা
প্রশ্ন: 5P2 - 5C2 =?

সমাধান: 
5P2
= 5!/(5 - 2)!
= 5!/3!
= 20

5C2
= 5!/2!(5 - 2)!
= 5!/2! 3!
= 10

∴ 5P2 - 5C2 = 20 - 10
= 10
৫৪৮.
In how many ways can 8 football players be divided into two teams with an equal number of players?
  1. ক
    70
  2. খ
    60
  3. গ
    50
  4. ঘ
    35
ব্যাখ্যা

Question: In how many ways can 8 football players be divided into two teams with an equal number of players?

Solution:
8 জন খেলোয়ার কে সমান দুটি ভাগে ভাগ করলে 8/2 = 4 জন করে দল গঠন করা যাবে।
 
তাহলে, 8 জন থেকে 4 জন করে নিয়ে পাই,  8C4 = 8!/[4!(8 - 4)!] 
= (8 × 7 × 6 × 5 × 4!)/(4! × 4!)
= (8 × 7 × 6 × 5)/(4 × 3 × 2)
= 70 

∴ দল গঠন করার উপায় = 70/2 = 35

৫৪৯.
In how many ways can a cricket eleven be chosen out of 13 players? 
  1. ক
    75
  2. খ
    78
  3. গ
    82
  4. ঘ
    85
ব্যাখ্যা
Question: In how many ways can a cricket eleven be chosen out of 13 players? 

Solution:
১৩ জনের মধ্য থেকে ক্রিকেট খেলার জন্য ১১ জন বাছাই করার উপায় = ১৩C১১
= ১৩!/১১!(১৩ - ১১)!
= ১৩!/১১! ২!
= ৭৮
৫৫০.
A committee is composed of w women and m men. If 3 women and 2 men are added to the committee, and if one person is selected at random from the enlarged committee, then the probability that a woman is selected can be represented by-
  1. ক
    w/m
  2. খ
    w/(w + m)
  3. গ
    (w + 3)/(m + 2)
  4. ঘ
    (w + 3)/(w + m + 3)
  5. ঙ
    (w + 3)/(w + m + 5)
ব্যাখ্যা
Question: A committee is composed of w women and m men. If 3 women and 2 men are added to the committee, and if one person is selected at random from the enlarged committee, then the probability that a woman is selected can be represented by-

Solution:
A committee is composed of w women and m men.

The total number of members on the enlarged committee is (w + 3) + (m + 2)
= w + 3 + m + 2
= w + m + 5;

The total number of women on the enlarged committee is = w + 3

∴ The probability that a woman is selected is P = favorable/total = (w + 3)/(w + m + 5).
৫৫১.
A bag contains 5 red, 3 green, and 4 blue balls. Two balls are drawn without replacement. What is the probability that the first ball is green and the second ball is red or blue?
  1. ক
    7/46
  2. খ
    11/36
  3. গ
    5/16
  4. ঘ
    9/44
ব্যাখ্যা

Question: A bag contains 5 red, 3 green, and 4 blue balls. Two balls are drawn without replacement. What is the probability that the first ball is green and the second ball is red or blue?

Solution:
Given that,
Red balls = 5
Green balls = 3
Blue balls = 4
∴ Total balls = 5 + 3 + 4 = 12

We know Probability = Favorable outcomes/Total outcomes

Now, First ball is Green = 3/12 = 1/4

If first is green, then the remaining balls = 11
Second ball is Red or Blue = (5 + 4)/11 = 9/11

∴ Required probability = (1/4) × (9/11)
= 9/44

৫৫২.
What is the probability of having 53 Sundays in a leap year?
  1. ক
    2/53
  2. খ
    7/53
  3. গ
    1/7
  4. ঘ
    2/7
ব্যাখ্যা

Question: What is the probability of having 53 Sundays in a leap year?

Solution:
Leap year = 366 Days
= {(52 × 7) + 2} Days
= 52 Weeks + 2 Days

∴ Probability of having 53 Sundays = 2/7

৫৫৩.
If two letters are taken at random from the word HOME, what is the probability that none of the letters would be vowels?
  1. ক
    ক) 1/6
  2. খ
    খ) 1/2
  3. গ
    গ) 1/3
  4. ঘ
    ঘ) 1/4
  5. ঙ
    ঙ) None of the above
ব্যাখ্যা

P(first letter is not vowel) = 2/4
P(second letter is not vowel) = 1/3
So, probability that none of letters would be vowels is = 2/4 × 1/3
= 1/6

৫৫৪.
A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 3 women?
  1. ক
    1/7
  2. খ
    2/7
  3. গ
    3/7
  4. ঘ
    5/7
ব্যাখ্যা
Question: A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 3 women?

Solution:
A small company employs 3 men and 5 women.
Total people = 8

ways of selecting 4 people from 8 = 8C4
= 70

ways of selecting 3 women from 5 = 5C3
ways of selecting 1 men from 3 = 3C1

∴ probability = (5C3 × 3C1)/ 70
= (10 × 3)/70
= 3/7
৫৫৫.
A fair coin is tossed 4 times. What is the probability of getting exactly 3 heads?
  1. ক
    1/2
  2. খ
    1/3
  3. গ
    1/4
  4. ঘ
    1/5
ব্যাখ্যা

Question: A fair coin is tossed 4 times. What is the probability of getting exactly 3 heads?

Solution:
For 4 tosses, the total number of possible outcomes is = 24 = 16
The number of ways to choose 3 Heads out of 4 tosses is = 4C3 = 4

∴ Probability of getting exactly 3 heads = 4/16 = 1/4

৫৫৬.
A bag contains a certain number of bolts out of which some are 'defective' while the remaining are 'non-defective'. The probability of picking a 'non-defective' bolt is 200% more than picking a 'defective' bolt from the bag. If 40 more 'non- defective' bolts are added to the bag then the probability of picking a 'defective' bolt becomes 20%. Now two bolts are picked from the bag then what is the probability that at most one of the two bolts is 'defective'?
  1. ক
    179/212
  2. খ
    185/212
  3. গ
    199/212
  4. ঘ
    195/212
ব্যাখ্যা
Question: A bag contains a certain number of bolts out of which some are 'defective' while the remaining are 'non-defective'. The probability of picking a 'non-defective' bolt is 200% more than picking a 'defective' bolt from the bag. If 40 more 'non- defective' bolts are added to the bag then the probability of picking a 'defective' bolt becomes 20%. Now two bolts are picked from the bag then what is the probability that at most one of the two bolts is 'defective'?

Solution:
Let the number of non-defective bolts in the bag = N
The number of defective bolts in the bag = D
∴ The total number of bolts in the bag = N + D

The probability of picking a 'non-defective' bolt is 200% more than picking a 'defective' bolt from the bag.
ATQ,
N/(N + D) = 300% × D/(N + D)
∴ N = 3D ..............(i)

If 40 more 'non-defective' bolts are added to the bag then the probability of picking a 'defective' bolt becomes 20%.
D/(N + D + 40) = 20/100
⇒ D/(3D+ D+ 40) = 1/5 [Using (i)]
⇒ 5D = 4D + 40
∴ D = 40
∴ N = 3D = 120

The total number of bolts in the bag = 40 + 120 = 160

Two bolts are picked from the bag then the probability that at most one of the two bolts is 'defective' = 1 - P (Both the selected bolts are defective)

∴ Required probability = 1 - (40C2/ 160C2)
= 1 - (780/12720)
= 1 - (78/1272)
= 1 - (13/212)
= (212 - 13)/212
= 199/212
Hence, the correct answer is 199/212.
৫৫৭.
According to meteorological records, it rained on 21 days in the month of June last year. What is the probability that it will rain on fourth of June this year?
  1. ক
    1/10
  2. খ
    3/10
  3. গ
    7/10
  4. ঘ
    9/10
ব্যাখ্যা
Question: According to meteorological records, it rained on 21 days in the month of June last year. What is the probability that it will rain on fourth of June this year?

Solution:
June month has 30 days
favorable events = 21 days

∴ the probability that it will rain on fourth of June this year = 21/30
= 7/10
৫৫৮.
In how many different ways can the letters of the word 'WEDDING' be arranged?
  1. ক
    ক) 3520
  2. খ
    খ) 2920
  3. গ
    গ) 2520
  4. ঘ
    ঘ) 1520
ব্যাখ্যা
Question: In how many different ways can the letters of the word 'WEDDING' be arranged?

Solution:
D is taken two times.

the word 'WEDDING' be arranged in = 7!/2!
= 5040/2
= 2520 
৫৫৯.
If x and y are positive integers such that x + y = 6, what is the probability that x ≥ 3?
  1. ক
    1/4
  2. খ
    2/5
  3. গ
    3/4
  4. ঘ
    3/5
ব্যাখ্যা
Question: If x and y are positive integers such that x + y = 6, what is the probability that x ≥ 3?

Solution:
Since both x and y must be positive integers,
total possible ways = (1, 5), (2 ,4), (3, 3), (4, 2), (5,1) = 5
Number of favorable outcomes = (3, 3), (4, 2), (5, 1) = 3

∴ So the probability that (x ≥ 3) = 3/5
৫৬০.
A committee of 5 members is to be formed by selecting out of 6 man and 7 women. In how many different ways the committee can be formed if it should have 2 men and 3 women?
  1. ক
    ক) 450
  2. খ
    খ) 525
  3. গ
    গ) 575
  4. ঘ
    ঘ) 615
ব্যাখ্যা
Question: A committee of 5 members is to be formed by selecting out of 6 man and 7 women. In how many different ways the committee can be formed if it should have 2 men and 3 women?

Solution:
2 men can be selected out of 6 men in  6C2 ways
3 women can be selected out of 7 women in 7C3 ways

Required number of ways = 6C2 × 7C3 = 15 × 35 = 525
৫৬১.
In a lottery, there are 16 prizes and 32 blanks. A lottery is drawn at random. What is the probability of getting a prize?
  1. ক
    ক) 1/16
  2. খ
    খ) 1/32
  3. গ
    গ) 1/3
  4. ঘ
    ঘ) None of the above
ব্যাখ্যা
Question: In a lottery, there are 16 prizes and 32 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Solution: 
Total outcome = 16 + 32 = 48 
Favorable outcome = 16
P (getting a prize) =16/48
                              = 1/3
৫৬২.
How many different registration numbers can be formed using two distinct letters followed by two distinct digits?
  1. ক
    42,500
  2. খ
    50,500
  3. গ
    58,500
  4. ঘ
    72,500
  5. ঙ
    None of the above
ব্যাখ্যা

Question: How many different registration numbers can be formed using two distinct letters followed by two distinct digits?

Solution:
Here,
Number of ways to choose and arrange two distinct letters out of 26 alphabets = 26P2  
= 26 × 25  
= 650

Number of ways to choose and arrange two distinct digits out of 10 digits (0 - 9) = 10P2  
= 10 × 9  
= 90

Total number of registration numbers = 650 × 90  
= 58,500

৫৬৩.
A bag contains 6 red balls, 9 black balls, and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is neither red nor white?
  1. ক
    3/7
  2. খ
    9/20
  3. গ
    11/20
  4. ঘ
    4/9
ব্যাখ্যা

Question: A bag contains 6 red balls, 9 black balls, and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is neither red nor white?

Solution:

Total balls = 6 + 9 + 5 = 20

Favorable outcomes = balls that are neither red nor white, that is, black balls = 9

∴ P(black) = Favorable outcomes/Total outcomes
= 9/20

Therefore, the probability is 9/20.

৫৬৪.
A committee of 3 men and 2 women is to be formed from 6 men and 4 women. How many ways can this be done?
  1. ক
    60
  2. খ
    120
  3. গ
    240
  4. ঘ
    360
ব্যাখ্যা
Question: A committee of 3 men and 2 women is to be formed from 6 men and 4 women. How many ways can this be done?

Solution:
Ways to choose 3 men out of 6 = 6C3 = 20
Ways to choose 2 women out of 4 = 4C2 = 6

Total number of ways = 20 × 6 = 120
৫৬৫.
How many ways can five different rings be worn on four fingers of one hand?
  1. ক
    2056
  2. খ
    625
  3. গ
    512
  4. ঘ
    1024
ব্যাখ্যা
Question: How many ways can five different rings be worn on four fingers of one hand?

Solution: 
Each ring may be worn on any of the 4 fingers.
So, each ring may be worn in 4 different ways.
∴ 5 rings may be worm in (4×4×4×4×4) = 45 = 1024 ways.
৫৬৬.
There are 10 oranges in a basket. Find the no. of ways in which 3 oranges are chosen from the basket?
  1. ক
    120
  2. খ
    240
  3. গ
    180
  4. ঘ
    210
ব্যাখ্যা
Question: There are 10 oranges in a basket. Find the no. of ways in which 3 oranges are chosen from the basket?

Solution:
Required number of ways = 10C3
= 10!/{3! (10 - 3)!}
= 10!/(3! 7!)
= (10 × 9 × 8)/(3 × 2)
= 120
৫৬৭.
In how many ways can 3 boys and 3 girls be selected from 12 boys and 9 girls?
  1. ক
    ক) 304
  2. খ
    খ) 9240
  3. গ
    গ) 14880
  4. ঘ
    ঘ) 18480
ব্যাখ্যা
Question: In how many ways can 3 boys and 3 girls be selected from 12 boys and 9 girls?

Solution:
12 জন বালক হতে প্রতিবার 3 জন বালক বেছে নেয়া যায় = 12C3 = 220 উপায়ে
9 জন বালিকা হতে প্রতিবার 3 জন বালিকা বেছে নেয়া যায় = 9C3 = 84 উপায়ে

∴ মোট বেছে নেয়া যায় = 220 × 84 = 18480 উপায়ে
৫৬৮.
Find the probability of selecting a prime number from a set of numbers 1 to 15 (both inclusive).
  1. ক
    ক) 3/5
  2. খ
    খ) 1/15
  3. গ
    গ) 7/15
  4. ঘ
    ঘ) 2/5
ব্যাখ্যা
Question: Find the probability of selecting a prime number from a set of numbers 1 to 15 (both inclusive).

Solution:
1 থেকে 15 পর্যন্ত মোট সংখ্যা = 15
1 থেকে 15 পর্যন্ত মৌলিক সংখ্যা 6টি যথাক্রমে 2, 3, 5, 7, 11, 13.

∴ নির্ণেয় সম্ভাব্যতা = 6/15 = 2/5
৫৬৯.
A box contains 6 bottles of variety 1 drink, 3 bottles of variety 2 drink, and 4 bottles of variety 3 drink. Three bottles of them are drawn at random, what is the probability that the three are not of the same variety?
  1. ক
    ক) 833/858
  2. খ
    খ) 752/833
  3. গ
    গ) 632/713
  4. ঘ
    ঘ) none of these
ব্যাখ্যা

Total number of drink bottles = 6 + 3 + 4 = 13.
Let S be the sample space
Then, n(S) = number of ways of taking 3 drink bottles out of 13.
Therefore, n(S) = 13C3
= (13 x 12 x 11)/(1 x 2 x 3)
= 66 x 13
= 858.
Let E be the event of taking 3 bottles of the same variety
Then, E = event of taking (3 bottles out of 6) or (3 bottles out of 3) or (3 bottles out of 4)
n(E) = 6C3 + 3C3 + 4C3
= (6 x 5 x 4 )/ (1 x 2 x 3) + 1 + (4 x 3 x 2) / (1 x 2 x 3)
= 20 + 1 + 4
= 25
The probability of taking 3 bottles of the same variety = n(E)/n(S)
= 25/858.
Then, the probability of taking 3 bottles are not of the same variety = 1 - 25/858
= 833/858.

৫৭০.
How many diagonals can be drawn in a pentagon?
  1. ক
    4
  2. খ
    5
  3. গ
    6
  4. ঘ
    3
ব্যাখ্যা
Question: How many diagonals can be drawn in a pentagon?

Solution: 
A pentagon has 5 sides. We obtain the diagonals by joining the vertices in pairs.
Total number of sides and diagonals,
= 5C2
= 10
This includes its 5 sides also.

∴ Diagonals = 10 – 5 = 5
৫৭১.
12 people at a party shake hands once with everyone else in the room. How many handshakes took place?
  1. ক
    72
  2. খ
    66
  3. গ
    76
  4. ঘ
    64
  5. ঙ
    None of these
ব্যাখ্যা

There are 12 people, so this is our n value.
So, 12C2= 66

৫৭২.
In an examination paper, there are two groups each containing 5 questions. A candidate is required to attempt 5 questions but not more than 3 questions from any group. In how many ways can 5 questions be selected?
  1. ক
    ক) 60
  2. খ
    খ) 100
  3. গ
    গ) 200
  4. ঘ
    ঘ) 288
ব্যাখ্যা
Question: In an examination paper, there are two groups each containing 5 questions. A candidate is required to attempt 5 questions but not more than 3 questions from any group. In how many ways can 5 questions be selected?

Solution: 
5 questions can be selected in the following ways,
2 question from first group and 3 question from second group Or 3 question from first group and 2 question from second group.
= (5C2 × 5C3) + (5C3 × 5C2)
= 100 + 100
= 200
৫৭৩.
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?
  1. ক
    60
  2. খ
    70
  3. গ
    90
  4. ঘ
    210
ব্যাখ্যা
Question: If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

Solution: 
There are two types of triangles possible:
With two vertices on the line with 4 points and the third vertex on the line with 5 points = 4C2 × 5C1 = 30
With two vertices on the line with 5 points and the third vertex on the line with 4 points = 4C1 × 5C2 = 40

total waya = 30 + 40 = 70
৫৭৪.
Two students are selected from a class of 5 girls and 12 boys. Find the probability that a particular pair of girl and boy is selected.
  1. ক
    1/72
  2. খ
    1/136
  3. গ
    5/136
  4. ঘ
    1/17
ব্যাখ্যা
Question: Two students are selected from a class of 5 girls and 12 boys. Find the probability that a particular pair of girl and boy is selected.

Solution: 
total number of students is = 12 + 5 = 17
a pair of students can be chosen from 17 students in = 17C2 ways
= (17!)/(2!15!)
= 136 ways

there is one probability that a particular pair of girl and boy is selected.

total probability = 1/136
৫৭৫.
7Pr = 210 and 7Cr = 35 then what is the value of r?
  1. ক
    3
  2. খ
    6
  3. গ
    4
  4. ঘ
    5
ব্যাখ্যা

Question: 7Pr = 210 and 7Cr = 35 then what is the value of r?
​
​​​​Solution:
​Given that,
​7Pr = 210 and 7Cr = 35
​
​We know that,
​nPr​  = r! × nCr​
​⇒ 210 = r! × 35
 ​⇒ ​r! = 210/35
​ ​⇒ r! = 6
 ​⇒ ​r! = 3!
∴ ​r = 3

৫৭৬.
In how many ways can a cricket eleven be chosen out of 15 players?
  1. ক
    780
  2. খ
    1365
  3. গ
    940
  4. ঘ
    1445
ব্যাখ্যা
Quiestion: In how many ways can a cricket eleven be chosen out of 15 players?

Solution:
Required number of ways = 15C11
= 15C(15 - 11)
= 15C4
= (14 × 13 × 12 ×11) /(4 × 3 × 2)
= 1365
৫৭৭.
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
  1. ক
    266
  2. খ
    5040
  3. গ
    11760
  4. ঘ
    86400
  5. ঙ
    None of these
ব্যাখ্যা
Question: In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

Solution:
Required number of ways = (8C5 × 10C6)
= 56 × 210
= 11760
৫৭৮.
In How many different ways can the letters of the word 'PRESENT' be arranged?
  1. ক
    5040
  2. খ
    2520
  3. গ
    1260
  4. ঘ
    630
ব্যাখ্যা
Question: In How many different ways can the letters of the word 'PRESENT' be arranged?

Solution:
The word 'PRESENT' contains 7 letters, with 2 E and all other 5 different.

Ways = 7!/2!
= 2520
৫৭৯.
If A and B events with P(A) = 2/5, P(B) = 3/10 and P(A ∩ B) = 1/10. Find P(A̅ ∩ B̅) = ?
  1. ক
    5/7
  2. খ
    2/5
  3. গ
    4/3
  4. ঘ
    1/2
  5. ঙ
    3/5
ব্যাখ্যা
Question: If A and B events with P(A) = 2/5, P(B) = 3/10 and P(A ∩ B) = 1/10. Find P(A̅ ∩ B̅) = ?

Solution:
Given that,
P(A) = 2/5
P(B) = 3/10
P(A ∩ B) = 1/10

Now,
P(A̅ ∩ B̅) = 1 - P(A ∪ B) .......(1)

Now,
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
= (2/5) + (3/10) - (1/10)
= (4 + 3 - 1)/10
= 6/10
= 3/5

From (1) We get,
P(A̅ ∩ B̅) = 1 - P(A ∪ B) = 1 - (3/5) = 2/5
৫৮০.
A meeting is attended by 750 students. 450 of the students are females. Half the female students are less than thirty years old, and one-fourth of the male students are less than thirty years old. If one of the students of the meeting is selected at random to receive a prize, what is the probability that the person selected is not less than thirty years old?
  1. ক
    ক) 2/5
  2. খ
    খ) 3/5
  3. গ
    গ) 1/5
  4. ঘ
    ঘ) 7/13
ব্যাখ্যা
Question: A meeting is attended by 750 students. 450 of the students are females. Half the female students are less than thirty years old, and one-fourth of the male students are less than thirty years old. If one of the students of the meeting is selected at random to receive a prize, what is the probability that the person selected is not less than thirty years old?

Solution:
 A meeting is attended by 750 students. 450 of the students are females. 

Half the female students are less than thirty years old
number of females less than thirty years old = 450/2
= 225

male students = 750 - 450
= 300
one-fourth of the male students are less than thirty years old.
number of males less than thirty years old = 300/4
= 75

total number of students less than thirty age = 225 + 75
= 300

the probability that the person selected is less than thirty years old = 300/750
= 2/5

∴  the probability that the person selected is not less than thirty years old = 1 - (2/5)
= (5 - 2)/5
= 3/5
৫৮১.
In how many different way can the letters of the word "ORANGE" be arranged?
  1. ক
    120
  2. খ
    320
  3. গ
    360
  4. ঘ
    720
ব্যাখ্যা
Question: In how many different way can the letters of the word "ORANGE" be arranged?

Solution:
the given words contain 6 diffrerent letters.

∴ they can be arranged in = 6! ways
= 720 ways
৫৮২.
If two fair coins are flipped, what is the probability that one will come up heads and the other tails?
  1. ক
    ক) 1/4
  2. খ
    খ) 1/3
  3. গ
    গ) 1/2
  4. ঘ
    ঘ) 3/4
ব্যাখ্যা

দুটি একইরকম মুদ্রা ছুড়ে মারা হলে নমুনাক্ষেত্রটি হবে = HH, HT, TH, TT
হেড এবং টেল আসার সম্ভাবনা = 2/4 = 1/2

৫৮৩.
In a class, 30% of the students offered English, 20% offered Bengali and 10% offered both. If a student is selected at random, What is the probability that he has offered English or Bengali ?
  1. ক
    ক) 1/5
  2. খ
    খ) 2/3
  3. গ
    গ) 3/5
  4. ঘ
    ঘ) 2/5
ব্যাখ্যা
P(E) = 30/100 = 3/10;
P(B) = 20/100 = 1/5
and P(E∩B) = 10/100 =1/10
P (E or B)
= P(E∪B)
= P(E) + P(B) - P(E∪B)
= (3/10+1/5−1/10)
= 4/10
= 2/5
৫৮৪.
In a survey, it was found that 70% of people read Ittefaq, 60% read Sangbad, and 40% read both newspapers. If a person is chosen at random, find the probability that they read either Ittefaq or Sangbad.
  1. ক
    1/10
  2. খ
    3/10
  3. গ
    7/10
  4. ঘ
    9/10
ব্যাখ্যা

Question: In a survey, it was found that 70% of people read Ittefaq, 60% read Sangbad, and 40% read both newspapers. If a person is chosen at random, find the probability that they read either Ittefaq or Sangbad.

Solution:
ধরি, ইত্তেফাক পড়ার ঘটনা A 
এবং সংবাদ পড়ার ঘটনা B
P(A) = 70/100 = 7/10 
P(B) = 60/100 = 6/10
P(A ∩ B)= 40/100 = 4/10

নিরপেক্ষভাবে বাছাই করলে একজন লোকের ইত্তেফাক বা সংবাদ পড়ার সম্ভাব্যতা P( A ∪ B)

∴ P(A ∪ B) = P(A) + P(B) - P(A ∩ B) 
= (7/10) + (6/10) - (4/10)
= (7 + 6 - 4)/10
= 9/10

৫৮৫.
The probability that a card drawn from a pack of 52 cards will be a diamond or a king is:
  1. ক
    3/52
  2. খ
    4/13
  3. গ
    1/26
  4. ঘ
    1/52
ব্যাখ্যা

Question: The probability that a card drawn from a pack of 52 cards will be a diamond or a king is: 

Solution: 
Here, n(S) = 52
There are 13 cards of diamonds (including one king), and there are three more kings.

Let E = the event of getting a diamond or a king
Then, n(E) = (13 + 3) = 16

∴ P(E) = n(E)/n(S)
= 16/52
= 4/13

৫৮৬.
If the probability that Mokhles will miss at least one of the ten jobs assigned to him is 0.55, then what is the probability that he will do all ten jobs?
  1. ক
    ক) 1
  2. খ
    খ) 0.1
  3. গ
    গ) 0.45
  4. ঘ
    ঘ) 0.85
ব্যাখ্যা
Question: If the probability that Mokhles will miss at least one of the ten jobs assigned to him is 0.55, then what is the probability that he will do all ten jobs?

Solution:
There are only two cases:
1) Mokhles will miss at least one of the ten jobs.
2) Mokhles will not miss any of the ten jobs.

Hence, (The probability that Mokhles will miss at least one of the ten jobs) + (The probability that he will not miss any job) = 1. Since the probability that Mokhles will miss at least one of the ten jobs is 0.55, this equation becomes
⇒ 0.55+ (The probability that he will not miss any job) = 1
⇒ (The probability that he will not miss any job) = 1 - 0.55
⇒ (The probability that he will not miss any job) = 0.45
৫৮৭.
A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?
  1. ক
    1/4
  2. খ
    1/2
  3. গ
    3/4
  4. ঘ
    7/12
ব্যাখ্যা
Question: A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

Solution:
১ম ছাত্রের সমস্যাটি সমাধান করার সম্ভাবনা ১/২
১ম ছাত্রের সমস্যাটি সমাধান না করার সম্ভাবনা ১ - ১/২ = ১/২

২য় ছাত্রের সমস্যাটি সমাধান করার সম্ভাবনা ১/৩
২য় ছাত্রের সমস্যাটি সমাধান না করার সম্ভাবনা ১ - ১/৩ = ২/৩

৩য় ছাত্রের সমস্যাটি সমাধান করার সম্ভাবনা ১/৪
৩য় ছাত্রের সমস্যাটি সমাধান না করার সম্ভাবনা ১ - ১/৪ = ৩/৪

তিন ছাত্রেরই সমস্যা সমাধান না করার সম্ভাবনা = (১/২) × (২/৩) × (৩/৪)
= ১/৪

∴ সমস্যাটি সমাধানের সম্ভাবনা = ১ - ১/৪
= ৩/৪ 
৫৮৮.
In a club, 60% members are male and 70% members are graduates. Also 50% of the graduate members are male. What percentage of the club members are female and non-graduate?
  1. ক
    ক) 5%
  2. খ
    খ) 10%
  3. গ
    গ) 15%
  4. ঘ
    ঘ) 20%
ব্যাখ্যা
Question: In a club, 60% members are male and 70% members are graduates. Also 50% of the graduate members are male. What percentage of the club members are female and non-graduate?

Solution: 
ধরি 
x জন সদস্য নারী ও নন-গ্র্যাজুয়েট 

প্রশ্নমতে,
100 = 70 + (60 - 35) + x
x + 25 = 30
x = 30 - 25
x = 5
৫৮৯.
A committee of 5 members is to be formed by selecting out of 6 men and 7 women. What is the probability that the committee has exactly 2 men and 3 women?
  1. ক
    105/354
  2. খ
    96/363
  3. গ
    102/455
  4. ঘ
    175/429
ব্যাখ্যা
Question: A committee of 5 members is to be formed by selecting out of 6 men and 7 women. What is the probability that the committee has exactly 2 men and 3 women?

Solution:
Total member = 3 + 2 = 5

2 men can be selected out of 6 men = 6C2 ways
3 women can be selected out of 7 women in = 7C3 ways
Required number of ways = 6C2 × 7C3 = 15 × 35 = 525

The total number of ways to make committee with all members = 13C5 = 1287

∴ The probability that the committee has exactly 2 men and 3 women = 525/1287
= 175/429
৫৯০.
How many numbers are there between 100 and 1000 inclusive, having at least one of their digits 7?
  1. ক
    250
  2. খ
    252
  3. গ
    255
  4. ঘ
    260
ব্যাখ্যা
Question: How many numbers are there between 100 and 1000 inclusive, having at least one of their digits 7?

Solution:
total numbers = 901

3 digit number without 7 = (8 × 9 × 9)
= 648
so, numbers are there between 100 and 1000 inclusive without 7 is = 648 + 1 = 649

∴ numbers are there between 100 and 1000 inclusive, having at least one of their digits 7 = 901 - 649
= 252
৫৯১.
A committee of 3 members is to be formed by selecting out of 5 men and 4 women. In how many different ways the committee can be formed if it should have 1 man and 2 women?
  1. ক
    ক) 10
  2. খ
    খ) 11
  3. গ
    গ) 25
  4. ঘ
    ঘ) 30
ব্যাখ্যা
Question: A committee of 3 members is to be formed by selecting out of 5 men and 4 women. In how many different ways the committee can be formed if it should have 1 man and 2 women?

Solution:
We can select 1 men from 5 men in = 5C1 ways = 5
2 women from 4 women in = 4C2 = 6
Committee can be formed = 5 × 6 = 30
৫৯২.
In how many different ways can the letters of the word "JASHORE" be arranged so that the vowels always come together?
  1. ক
    120
  2. খ
    720
  3. গ
    114
  4. ঘ
    126
ব্যাখ্যা

Question: In how many different ways can the letters of the word "JASHORE" be arranged so that the vowels always come together?

Solution:
The word J A S H O R E = 7 letters
There, Vowels = A, O, E (3 vowels)
Consonants = J, S, H, R (4 consonants)

Now,
Treat the 3 vowels (A, O, E) as a single unit.
So now we have, [A O E]  J  S  H  R
Total units to arrange = 5 (1 vowel block + 4 consonants)

Number of ways to arrange these 5 units = 5! = 120 ways

And
The 3 vowels (A, O, E) can be arranged among themselves in = 3! = 6 ways

Total number of arrangements = (ways to arrange the 5 units) × (ways to arrange vowels inside the block)
= 120 × 6
= 720

So there are 720 different ways to arrange the letters of 'JASHORE' such that the vowels always come together.

৫৯৩.
Out of 5 men and 6 women, how many ways can you form a 4-member committee with at least 2 women?
  1. ক
    252
  2. খ
    265
  3. গ
    240
  4. ঘ
    150
ব্যাখ্যা
Question: Out of 5 men and 6 women, how many ways can you form a 4-member committee with at least 2 women?

Solution:
6 জন মহিলা থেকে 2 জন  ও 5 জন পুরুষ থেকে 2 জন নিয়ে 4 সদস্যের কমিটি গঠন করা যায়,
= 6C2 × 5C2
= {6! / (2! × 4! )} × {5! / (2! × 3!)}
= {(6 × 5) / 2}  × {(5 × 4) / 2}
= 15 × 10 
= 150 উপায়ে 

6 জন মহিলা থেকে 3 জন ও 5 জন পুরুষ থেকে 1 জন নিয়ে 4 সদস্যের কমিটি গঠন করা যায়,
= 6C3 × 5C1
= {6! / (3! × 3!)} × {5! / (1! × 4!)}
= {(6 × 5 × 4) / 6}  ×  5
= 20 × 5
= 100 উপায়ে 

আবার, 
6 জন মহিলা থেকে 4জন নিয়ে ( পুরুষ সদস্য ব্যতীত ) 4 সদস্যের কমিটি গঠন করা যায়,
= 6C4
= 6! / (4! × 2!)
= (6 × 5) /2
= 15 উপায়ে

মোট উপায় = 150 + 100 + 15 = 265 টি 
৫৯৪.
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, and 7 if no digit is repeated?
  1. ক
    360
  2. খ
    180
  3. গ
    120
  4. ঘ
    60
ব্যাখ্যা
Question: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, and 7 if no digit is repeated?

Solution:
Given digits: 1, 2, 3, 4, 6, 7
Number of digits = 6

Number of possible digits at unit’s place = 3 (2, 4 and 6)
⇒ Number of permutations = 3P1 = 3

When one of the digits is taken in units’ place, then the number of possible digits available = 5
⇒ Number of permutations = 5P2 = 5!/(5 - 2)!
= 5!/3!
= 120/6
= 20

∴ The total number of permutations = 3 × 20 = 60.
৫৯৫.
A bag contains 4 red, 5 yellow and 6 pink balls. Two balls are drawn at random. What is the probability that none of the balls drawn is yellow in colour?
  1. ক
    ক) 1/7
  2. খ
    খ) 3/7
  3. গ
    গ) 2/7
  4. ঘ
    ঘ) 5/14
ব্যাখ্যা

Number of red balls = 4
Number of yellow ball = 5
Number of pink ball = 6
Total number of balls = 4 + 5 + 6 = 15
Total possible outcomes = selection of 2 balls out of 15 balls = 15C2
= 15!/2!(15 - 2)!
= 15!/(2! × 13!)
= (15 × 14)/(1 × 2)
= 105.
Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls.
10C2
= 10!/2!(10 - 2)!
= 10!/2!8!
= (10 × 9)/(1 × 2)
= 45.
∴ Required Probability = 45/105
= 3/7

৫৯৬.
If nC2 = 105, then n =?
  1. ক
    14
  2. খ
    15
  3. গ
    16
  4. ঘ
    17
  5. ঙ
    18
ব্যাখ্যা
Question: If nC2 = 105, then n =?

Solution:
nC2 = 105
⇒ n!/{2! ×(n - 2)!} = 105
⇒ n(n - 1) = 210
⇒ n2 - n - 210 = 0
⇒ n2 - 15n + 14n - 210 = 0
⇒ n(n - 15) + 14(n - 15) = 0
⇒ (n - 15)(n + 14) = 0
∴ n = 15, - 14 [- 14 is not acceptable]
৫৯৭.
There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. The number of ways in which the votes can be given is
  1. ক
    1024
  2. খ
    560
  3. গ
    462
  4. ঘ
    None
ব্যাখ্যা
Question: There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. The number of ways in which the votes can be given is

Solution:
Since there are 4 candidates for the post of a lecturer and 5 men voting, each voter can choose one candidate from the 4. The number of ways in which the votes can be given can be calculated by considering that each of the 5 men can choose one candidate.

For each man, there are 4 possible choices of candidates. Therefore, for 5 men, the total number of ways the votes can be given = 45 = 1024
৫৯৮.
In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?
  1. ক
    ক) 40
  2. খ
    খ) 216
  3. গ
    গ) 20
  4. ঘ
    ঘ) 720
ব্যাখ্যা
Question: In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?
Solution:
6টি শিশু ৩টি সবুজ শার্ট গ্রহণ করার উপায় = 6c3 = (6 × 5 × 4)/(3 × 2× 1) = 20
বাকি ৩টি শিশু ৩টি লাল শার্ট গ্রহণ করার উপায় = 3C3 = (3 × 2 × 1)/(3 × 2 × 1) = 1
সুতরাং, মোট উপায় সংখ্যা = 20 ×1 = 20টি 
৫৯৯.
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done ?
  1. ক
    ক) 857
  2. খ
    খ) 896
  3. গ
    গ) 756
  4. ঘ
    ঘ) 675
ব্যাখ্যা
Ways in which at least 3 men are selected;
⇒ 3 men + 2 women
⇒ 4 men + 1 woman
⇒ 5 men + 0 woman

Number of ways = 7C3 × 6C2 + 7C4 × 6C1 + 7C5 × 6C0
                           = 35 × 15 + 35  × 6 + 21
                           = 756
                     
∴ The required no of ways = 756
৬০০.
In your bookshelf, you have five favorite books. If you decide to arrange these five books in every possible combination and moved just one book in every half a minute. How much time it will take you to arrange?
  1. ক
    ক) 3 hours
  2. খ
    খ) 1 hour
  3. গ
    গ) 2 hours
  4. ঘ
    ঘ) 30 hours
ব্যাখ্যা
Question: In your bookshelf, you have five favorite books. If you decide to arrange these five books in every possible combination and moved just one book in every half a minute. How much time it will take you to arrange?

Solution:
5টি বইকে সাজানো যায় মোট = 5! উপায়ে 
= 120 উপায়ে 

বই 1 বার সরাতে সময় লাগে 1/2 মিনিট 
বই 120 বার সরাতে সময় লাগে 120/2 মিনিট 
= 60 মিনিট
= 1 ঘণ্টা