লগ ইননিবন্ধন
বিষয়সমূহ

PrepBank · বিষয়ভিত্তিক প্রশ্ন

Probability, Permutation and Combination

মোট প্রশ্ন৯৬৯এই পাতা১০০প্রতি পাতা১০০
প্রশ্ন দেখুনঅনুশীলন
ঘনত্ব
উত্তর
উত্তরিতবর্তমানপুনরায় দেখুনঅসম্পূর্ণ

Probability, Permutation and Combination

PrepBank · পাতা ২ / ১০ · ১০১–২০০ / ৯৬৯

১০১.
In a lottery, there are 8 prizes and 12 blanks. A lottery is drawn at random. What is the probability of getting a prize?
  1. ক
    1/2
  2. খ
    1/5
  3. গ
    2/5
  4. ঘ
    4/5
ব্যাখ্যা

Question: In a lottery, there are 8 prizes and 12 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Solution: 
Here, 
Number of prizes = 8
Number of blanks = 12
∴ Total outcome = 8 + 12
= 20

∴ Probability of getting a prize = Number of prizes/Total outcome 
= 8/20
= 2/5 

১০২.
Find the value of (x + 1)!/(x - 2)! 
  1. ক
    x
  2. খ
    x2 - x
  3. গ
    x2 - 1
  4. ঘ
    x(x2 - 1)
ব্যাখ্যা

Question: Find the value of (x + 1)!/(x - 2)! 

Solution:
Here,
(x + 1)!/(x - 2)!
= [x(x + 1)(x - 1)(x - 2)!]/(x - 2)! 
= x(x + 1)(x - 1)
= x(x2 - 1) 

১০৩.
What is the probability of rolling an odd number on a standard six-sided die?
  1. ক
    2/3
  2. খ
    1/3
  3. গ
    1/2
  4. ঘ
    5/6
ব্যাখ্যা

Question: What is the probability of rolling an odd number on a standard six-sided die?
(Officer Cash 2022 অনুযায়ী)

Solution:
A standard six-sided die has the numbers,
1, 2, 3, 4, 5, 6
Odd numbers between 1 and 6 are = 1, 3, 5
So, favorable outcomes = 3
And total possible outcomes = 6

∴ Probability = Favorable outcomes/Total outcomes ​
= 3/6 ​
= 1/​2

১০৪.
In how many different ways can the letters of the word 'WATER' be arranged?
  1. ক
    ক) 100 ways
  2. খ
    খ) 110 ways
  3. গ
    গ) 120 ways
  4. ঘ
    ঘ) 130 ways
ব্যাখ্যা
Question: In how many different ways can the letters of the word 'WATER' be arranged?

Solution:
the given words contain 5 diffrerent letters.

∴ they can be arranged in = 5! ways
= 120 ways 
১০৫.
How many possible two-digit numbers can be formed by using the digits 3, 5 and 7 (repetition of digits is allowed)?
  1. ক
    12
  2. খ
    27
  3. গ
    18
  4. ঘ
    9
ব্যাখ্যা

Question: How many possible two-digit numbers can be formed by using the digits 3, 5 and 7 (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 

১০৬.
In a bag, there are 5 red, 3 green, and 2 blue balls. If two balls are drawn one after the other without replacement, what is the probability that the first one is green and the second one is blue?
  1. ক
    1/15
  2. খ
    3/5
  3. গ
    2/3
  4. ঘ
    1/10
ব্যাখ্যা

Question: In a bag, there are 5 red, 3 green, and 2 blue balls. If two balls are drawn one after the other without replacement, what is the probability that the first one is green and the second one is blue?

Solution:
মোট বলের সংখ্যা = 5 (লাল) + 3 (সবুজ) + 2 (নীল) = 10টি
প্রথম বলটি সবুজ হওয়ার সম্ভাবনা = 3/10
প্রথম বল তোলার পর বাক্সে বলের সংখ্যা = 10 - 1 = 9
দ্বিতীয় বলটি নীল হওয়ার সম্ভাবনা = 2/9

সুতরাং, প্রথমটি সবুজ ও দ্বিতীয়টি নীল হওয়ার সম্ভাবনা = (3/10) × (2/9)
= 6/90 
= 1/15

১০৭.
A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?
  1. ক
    3/10
  2. খ
    23/60
  3. গ
    7/12
  4. ঘ
    41/60
  5. ঙ
    5/6
ব্যাখ্যা
Question: A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

Solution:
Let's calculate the opposite probability of NOT getting a matched set and subtract this value from 1.

This could happen only if we pick all three same hand BLUE gloves; two same hand BLUE gloves and any green glove; or two same hand GREEN gloves and any BLUE glove

BBB: (6/10) × (2/9) × (1/8) =1/60 (after we pick a blue glove, 6/10, then there are 2 same hand gloves left out of the total 9 gloves : 2/9, and so on);

BBG: (6/10) × (2/9) × (4/8) × 3 = 12/60, [multiplying by 3 as this scenario can occur in 3 different ways: BBG, BGB, GBB;]

GGB: (4/10) × (1/9) × (6/8) × 3 = 6/60

P = 1 - (1/60 + 12/60 + 6/60) = 41/60
১০৮.
One card is drawn from a pack of 52 cards. What is the probability that the card drawn is either a red card or a king?
  1. ক
    ক) 1/2
  2. খ
    খ) 6/13
  3. গ
    গ) 7/13
  4. ঘ
    ঘ) 27/52
ব্যাখ্যা

Here, n(S) = 52
There are 26 red cards (including 2 kings) and there are 2 more kings.
Let E = event of getting a red card or a king.
Then, n(E) = 28
∴P(E)=n(E)/n(S)
=28/52
=7/13

১০৯.
Two dice are thrown simultaneously. What is the probability of getting the sum of the face number is odd?
  1. ক
    ক) 1/2
  2. খ
    খ) 2/3
  3. গ
    গ) 3/4
  4. ঘ
    ঘ) 1/3
ব্যাখ্যা
প্রশ্ন: Two dice are thrown simultaneously. What is the probability of getting the sum of the face number is odd?

সমাধান: 
The odd sum is 3, 5, 7, 9, 11
The number of ways getting 3 = (1, 2), (2, 1) = 2
The number of ways getting 5 = (1, 4), (2, 3), (3, 2), (4, 1) = 4
The number of ways getting 7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6
The number of ways getting 9 = (3, 6), (4, 5), (5, 4), (6, 3) = 4
The number of ways getting 11 = (5, 6), (6,5) = 2
The number of ways = 2 + 4 + 6 + 4 + 2 = 18
The total number of ways = 6 × 6 = 36

Probability = 18/36 = 1/2
∴ The probability of getting an odd sum is 1/2
১১০.
A committee of 3 members is to be selected out of 3 men and 2 women. What is the probability that the committee has at least one woman?
  1. ক
    11/20
  2. খ
    2/3
  3. গ
    3/8
  4. ঘ
    9/10
ব্যাখ্যা

Question: A committee of 3 members is to be selected out of 3 men and 2 women. What is the probability that the committee has at least one woman?

Solution:
Total member = 3 + 2 = 5

Committee can be form with at least 1 one woman:
1 woman, 2 men : 2C1 ×  3C2 = 2 × 3 = 6
2 women, 1 man: 2C2 × 3C1 = 1 × 3 = 3
∴ The total number of ways to make committe with at least 1 one woman: 6 + 3 = 9

The total number of ways to make committe with all members = 5C3 = 10

∴ The probability that the committee has at least woman = 9/10

১১১.
How many arrangements can be made out of the letters of the word DRAUGHT, the vowels never being separated?
  1. ক
    ক) 1440
  2. খ
    খ) 720
  3. গ
    গ) 360
  4. ঘ
    ঘ) 640
  5. ঙ
    ঙ) None of these
ব্যাখ্যা

There are 7 letters in the word DRAUGHT, the two vowels are A and U.
Since, the vowels are not to be separated; AU can be considered as one entity.
Therefore, the number of letters will be 6 instead of 7.
The permutations will be P(6,6) = 6! ways.
But the two vowels A and U can be arranged in two ways, i.e. AU and UA.
The required number of arrangements = 2!.6! = 1440 ways.

১১২.
The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?
  1. ক
    320
  2. খ
    520
  3. গ
    242
  4. ঘ
    342
ব্যাখ্যা

Question: The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?

Solution:
Here,
The order of each letter in the dictionary is ABLORU.

Now,
with A in the beginning, the remaining letters can be permuted = 5! ways.
= 120 ways

Similarly,
with B in the beginning, the remaining letters can be permuted = 5! ways.
= 120 ways

With L in the beginning,
the first word will be LABORU, the second will be LABOUR.

Hence, the rank of the word LABOUR = 5! + 5! + 2
= 120 + 120 + 2
= 242

১১৩.
Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 4/7. What is the probability that both of them get selected?
  1. ক
    8/35
  2. খ
    34/35
  3. গ
    27/35
  4. ঘ
    None of these
ব্যাখ্যা

P(A) = 2/5
P(B) = 4/7
E = {A and B both get selected}

P(E) = P(A) × P(B)
= (2/5) × (4/7)
= 8/35

১১৪.
In a simultaneous throw of two dice, what is the probability of getting a total of 7?
  1. ক
    ক) 1/6
  2. খ
    খ) 1/18
  3. গ
    গ) 1/9
  4. ঘ
    ঘ) 1/36
ব্যাখ্যা
Question: In a simultaneous throw of two dice, what is the probability of getting a total of 7?

Solution:
We know that in a simultaneous throw of two dice,
n(S) = 6 × 6 = 36
Let E = event of getting a total of 7
= {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

We know,
P(E) = n(E)/n(S)
= 6/36
= 1/6
১১৫.
A box contains 5 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one white ball is to be included in the draw?
  1. ক
    ক) 125 ways
  2. খ
    খ) 169 ways
  3. গ
    গ) 185 ways
  4. ঘ
    ঘ) 220 ways
ব্যাখ্যা
Question: A box contains 5 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one white ball is to be included in the draw?

Solution:
Total number of balls = 5 + 3 + 4 
= 12 balls

choosing 3 balls from 12 balls = 12C3
= 220 ways

choosing three balls from black and red balls = 7C3
= 35 ways 

∴ ways can 3 balls be drawn from the box, if at least one white ball is to be included in the draw is = 220 - 35 ways
= 185 ways
১১৬.
A box contains 20 electric bulbs, out of which 4 are defective. Two balls are chosen at random from this box. The probability that at least one of them is defective, is:
  1. ক
    3/8
  2. খ
    13/19
  3. গ
    7/19
  4. ঘ
    11/25
ব্যাখ্যা

Question: A box contains 20 electric bulbs, out of which 4 are defective. Two balls are chosen at random from this box. The probability that at least one of them is defective, is:

Solution:
Given that,
Total bulbs = 20
Defective bulbs = 4
Non-defective bulbs = 20 - 4 = 16
Two bulbs are chosen at random (without replacement)

Now,
P(both non-defective) = (16/20) × (15/19) = 240/380 = 12/19

And,
∴ P(at least one defective) = 1 - P(both non-defective)
= 1 - (12/19)
= (19 - 12)/19
= 7/19
∴ The probability that at least one of them is defective is 7/19

১১৭.
Five persons, A, M, J, R, and P, sit randomly in five chairs in a row. What is the probability that R and M sit next to each other?
  1. ক
    2/5
  2. খ
    2/3
  3. গ
    1/5
  4. ঘ
    1/6
ব্যাখ্যা
Question: Five persons, A, M, J, R, and P, sit randomly in five chairs in a row. What is the probability that R and M sit next to each other?

Solution:
Total possibilities = 5! = 120
favorabole events = (4! × 2!)
= (24 × 2)
= 48

∴ probability = 48/120
= 2/5
১১৮.
The number of straight lines that can be drawn out of 12 points of which 8 are collinear is-
  1. ক
    39
  2. খ
    29
  3. গ
    49
  4. ঘ
    59
  5. ঙ
    69
ব্যাখ্যা
Question: The number of straight lines that can be drawn out of 12 points of which 8 are collinear is-

Solution:
The required number of lines = 12C2 - 8C2 + 1 = 66 - 28 + 1 = 39
১১৯.
How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can take more than 10 letters?
  1. ক
    510
  2. খ
    105
  3. গ
    10C5
  4. ঘ
    10P5
ব্যাখ্যা
Question: How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can take more than 10 letters?

Solution: 
Each letter has 5 independent choices (any of the 5 boxes).
So for:
Letter 1 → 5 choices, 
Letter 2 → 5 choices, 
...
Letter 10 → 5 choices. 

Since all letters are independent, 
Total number of ways = 510
১২০.
In how many ways can a group of 5 members be formed by selecting 3 boys out of 6 boys and 2 girls out of 5 girls?
  1. ক
    200
  2. খ
    350
  3. গ
    462
  4. ঘ
    30
  5. ঙ
    None of these
ব্যাখ্যা
Question: In how many ways can a group of 5 members be formed by selecting 3 boys out of 6 boys and 2 girls out of 5 girls?

Solution:
Number of ways 3 boys can be selected out of 6 = 6C3 = 6!/[(3!) × (3!)] = (6 × 5 × 4) / (3 × 2 × 1) = 20

Number of ways 2 girls can be selected out of 5 = 5C2 = 5!/[(2 !) × (3 !)] = (5 × 4)/(2 × 1) = 10

Therefore, total number of ways of forming the group = 20 × 10 = 200 
১২১.
nC1 + nC2 + nC3 + .. .. nCn = ?
  1. ক
    ক) 2n
  2. খ
    খ) 2n-1
  3. গ
    গ) {n(n-1)(n2+1)}/2
  4. ঘ
    ঘ) 2n - 1
ব্যাখ্যা
We know that, (1+x)n = nC0 + nC1x + nC2x2 + nC3x3 + ..... + nCnxn

Let us consider, x=1

Then,
(1+1)n = nC0 + nC1×1 + nC2×12 + nC3×13..... + nCn×1n
2n = 1 + nC1 + nC2 + nC3 + .... + nCn
nC1 + nC2 + nC3 + .... + nCn = 2n − 1
১২২.
There are 8 black, 5 red and 7 green marbles in a jar. If a marble is picked at random, what is the probability of having either a black or a green marble?
  1. ক
    5/12
  2. খ
    3/4
  3. গ
    2/3
  4. ঘ
    7/12
ব্যাখ্যা

Question: There are 8 black, 5 red and 7 green marbles in a jar. If a marble is picked at random, what is the probability of having either a black or a green marble?

Solution:
মোট মার্বেলের সংখ্যা = 8 + 5 + 7 = 20

কালো মার্বেল পাওয়ার সম্ভাবনা, P(Black) = 8/20

সবুজ মার্বেল পাওয়ার সম্ভাবনা P(Green) = 7/20

যেহেতু কালো এবং সবুজ মার্বেল পাওয়া দুটি বিচ্ছিন্ন (mutually exclusive) ঘটনা,
∴ কালো অথবা সবুজ মার্বেল পাওয়ার সম্ভাবনা (P(Black or Green) = P(Black) + P(Green)
= 8/20 + 7/20
= (8 + 7) / 20 = 15/20
= 3/4

অতএব, কালো অথবা সবুজ মার্বেল পাওয়ার সম্ভাবনা হলো 3/4।

• Shortcut:
মোট মার্বেলের সংখ্যা = 8 + 5 + 7 = 20
অনুকূল ঘটনা = কালো + সবুজ = 8 + 7 = 15টি
∴ সম্ভাবনা = অনুকূল ঘটনা/মোট ঘটনা = 15/20 = 3/4

১২৩.
In how many ways can 5 letters be posted in 4 letter boxes?
  1. ক
    ক) 512
  2. খ
    খ) 1024
  3. গ
    গ) 625
  4. ঘ
    ঘ) 20
  5. ঙ
    ঙ) None of these
ব্যাখ্যা

First letter can be posted in 4 letter boxes in 4 ways.
Similarly the second letter can be posted in 4 letter boxes in 4 ways and so on.
Hence all the 5 letters can be posted in = 4 x 4 x 4 x 4 x 4 = 1024

১২৪.
How many 3-digit numbers can be formed from the digits 4, 6, 5, 9 and 2, which are divisible by 5 and none of the digits is repeated?
  1. ক
    12 ways
  2. খ
    8 ways
  3. গ
    20 ways
  4. ঘ
    36 ways
ব্যাখ্যা
Question: How many 3-digit numbers can be formed from the digits 4, 6, 5, 9 and 2, which are divisible by 5 and none of the digits is repeated?

Solution:
We know,
the number will be divisible by 5 if the last number is 5.
So, first number can be chosen in = 4C1 ways 
= 4 ways

As the digit is not repeated
the second number can be chosen in = 3C1 
= 3 ways

∴ Total ways = 4 × 3 ways 
= 12 ways
১২৫.
In how many different ways can the letter of the word JUDGE be arranged in such a ways that the vowels always come together?
  1. ক
    ক) 48
  2. খ
    খ) 96
  3. গ
    গ) 120
  4. ঘ
    ঘ) 140
ব্যাখ্যা
The word 'JUDGE' has 5 letters in which 'JDG' are consonants and 'UE' are vowels.
On keeping vowels together, we get JDG(UE)
∴ Number of arrangements = 4! × 2 !
                                           =24 × 2
                                           = 48
১২৬.
In how many ways can the letters of the word 'ENGINEERING' be arranged such that the first letter is always 'R'?
  1. ক
    22,700
  2. খ
    25,200
  3. গ
    27000
  4. ঘ
    30,240
ব্যাখ্যা

Question: In how many ways can the letters of the word 'ENGINEERING' be arranged such that the first letter is always 'R'?

Solution:
'ENGINEERING' শব্দটিতে মোট 11টি বর্ণ রয়েছে।

শর্ত: প্রথম স্থানে 'R' স্থির।
এখন বাকি 11 - 1 = 10টি স্থানে বাকি বর্ণগুলোকে সাজাতে হবে।
বাকি 10টি বর্ণের মধ্যে পুনরাবৃত্ত অক্ষর:
E (3 বার), N (3 বার), G (2 বার), I (2 বার)।

∴ বাকি ১০টি বর্ণের বিন্যাস সংখ্যা = 10!/(3! × 3! × 2! × 2!)
= 3,628,800/(6 × 6 × 2 × 2)
= 3,628,800/144
= 25,200

∴ প্রথম অক্ষর 'R' রেখে 'ENGINEERING' শব্দটির বিন্যাস সংখ্যা হলো 25,200.

১২৭.
One-fourth of the boys and three-eight of the girls in a school participated in the annual ports. What proportional part of the total student population of the school participated in the annual sports?
  1. ক
    ক) 4/12
  2. খ
    খ) 5/8
  3. গ
    গ) 8/12
  4. ঘ
    ঘ) None of these
ব্যাখ্যা
মনেকরি
ছাত্রসংখ্যা 4জন এবং  প্রতিযোগিতায় অংশ গ্রহণ করেছে 1 জন 
ছাত্রী সংখ্যা 8 জন এবং প্রতিযোগিতায় অংশ গ্রহণ করেছে 3 জন 
মোট ছাত্রছাত্রী = 4 + 8  
                        =12 জন 
মোট প্রতিযোগিতায় অংশ গ্রহণ করেছে= 1 + 3    
                                                            = 4

প্রতিযোগিতায় ছাত্রছাত্রীদের 4/12 অংশ গ্রহণ করেছে। 
১২৮.
If a committee of 3 people is to be selected from among 6 married couples such that the committee does not include two people who are married to each other, how many such committees are possible?
  1. ক
    160
  2. খ
    220
  3. গ
    960
  4. ঘ
    480
ব্যাখ্যা
Question: If a committee of 3 people is to be selected from among 6 married couples such that the committee does not include two people who are married to each other, how many such committees are possible?

Solution:
Number of ways 3 people can be chosen from 12 = 12C3 = 220

Number of ways a couple can be chosen = 6
Number of ways the third person can be chosen = 10C1 = 10

Hence total number of ways a couple with another person can be chosen = 6 × 10 = 60

Therefore total number of ways a couple cannot be chosen = 220 - 60 = 160
১২৯.
A box contains 4 red, 5 green and 6 black balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or black?
  1. ক
    ক) 1/3
  2. খ
    খ) 2/15
  3. গ
    গ) 2/3
  4. ঘ
    ঘ) 1/9
ব্যাখ্যা
Question: A box contains 4 red, 5 green and 6 black balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or black?

Solution: 
Total number of balls = (4 + 5 + 6) = 15
P(drawing a red ball or a black ball) = P(red) + P(black)
                                                          = (4/15) + (6/15)
                                                          = (4 + 6)/15
                                                          = 10/15
                                                          = 2/3
∴ The probability is 2/3.
১৩০.
There are three prizes to be distributed among five students. If no students gets more than one prize, then this can be done in:
  1. ক
    ক) 10
  2. খ
    খ) 12
  3. গ
    গ) 16
  4. ঘ
    ঘ) 24
ব্যাখ্যা
3 prize among 5 students can be distributed in 5C3 ways = 10 ways.
১৩১.
What is the probability of getting a sum 9 from two throws of a dice?
  1. ক
    ক) 1/12
  2. খ
    খ) 1/8
  3. গ
    গ) 1/6
  4. ঘ
    ঘ) 1/9
ব্যাখ্যা
Question: What is the probability of getting a sum 9 from two throws of a dice?

Solution:
Total events = 6 × 6 = 36
Set of expected events = {(3, 6), (4, 5), (5, 4), (6, 3)}

∴ Probability = 4/36 = 1/9
১৩২.
The income of a broker remains unchanged through the rate of commission is increased from 4% to 5%. the percentage of slump in business is -
  1. ক
    ক) 10%
  2. খ
    খ) 20%
  3. গ
    গ) 60%
  4. ঘ
    ঘ) 80%
ব্যাখ্যা
Question: The income of a broker remains unchanged through the rate of commission is increased from 4% to 5%. the percentage of slump in business is -

Solution:
দালালের কমিশন ৪% থেকে ৫% পেলেও আয় বৃদ্ধি পায় নি, অর্থাৎ মূল ব্যবসায় মন্দা চলছে।

∴ দালালের কমিশন পূর্বের ৪% = বর্তমানের ৫%

মনে করি, মূল ব্যবসা ১০০ টাকার
তাহলে, দালালের পূর্বের আয় = ১০০ × ৪% = ৪ টাকা

∴ দালালের বর্তমান আয়, ৫% = ৪ টাকা
∴ মূল ব্যবসা, ১০০% = (৪ × ১০০)/৫ = ৮০ টাকা

তাহলে মূল ব্যবসা কমেছে = ১০০ - ৮০ = ২০ টাকা
১৩৩.
A fair coin is flipped three times. What is the probability that coin lands head each time?
  1. ক
    ক) 1/2
  2. খ
    খ) 1/4
  3. গ
    গ) 1/6
  4. ঘ
    ঘ) 1/8
ব্যাখ্যা
Question: A fair coin is flipped three times. What is the probability that coin lands head each time? 

Solution: 
একটি মুদ্রাকে তিনবার নিক্ষেপ করলে নমুনা বিন্দু হবে = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
= 8 টি
নমুনা ক্ষেত্র থেকে দেখা যাচ্ছে, প্রতিবারে H অর্থাৎ তিনটিই H আসার অনুকূল ফলাফল 1টি 

নির্ণেয় সম্ভাবনা = 1/8
১৩৪.
In a simultaneous throw of two coins, the probability of getting at least one head is: 
  1. ক
    ক) 1/2
  2. খ
    খ) 1/4
  3. গ
    গ) 3/​4
  4. ঘ
    ঘ) 1/3
ব্যাখ্যা
Question: In a simultaneous throw of two coins, the probability of getting at least one head is: 

Solution: 
Here S= {HH,HT,TH,TT}
Let E= event of getting at least on head ={HH,HT,TH}

∴P(E)= n(E)/n(S)​ = 3/​4
১৩৫.
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
  1. ক
    120
  2. খ
    240
  3. গ
    360
  4. ঘ
    720
  5. ঙ
    None
ব্যাখ্যা
Question: In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Solution:
The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in = 3! = 6 ways.

 Required number of ways = (120 × 6) = 720.
১৩৬.
There are 5 red and 3 black balls in a bag. Probability of drawing a black ball is -
  1. ক
    ক) 5/8
  2. খ
    খ) 1/2
  3. গ
    গ) 3/8
  4. ঘ
    ঘ) 1/4
ব্যাখ্যা

মোট বলের সংখ্যা = (5 + 3) = 8 টি
কালো বলের সংখ্যা = 3 টি
একটি কালো বল উঠার সম্ভাব্যতা = 3/8 টি

১৩৭.
In how many ways can the letters of the word "AUTHOR" be arranged such that the vowels are only in the odd positions?
  1. ক
    36
  2. খ
    72
  3. গ
    18
  4. ঘ
    108
ব্যাখ্যা

Question: In how many ways can the letters of the word "AUTHOR" be arranged such that the vowels are only in the odd positions?

Solution:
এখানে
মোট বর্ণ আছে 6টি
স্বরবর্ণ অর্থাৎ Vowel আছে (A, O, U) 3টি
ব্যঞ্জনবর্ণ অর্থাৎ Consonant আছে (T, H, R) 3টি

স্বরবর্ণ 3টি বিজোড় স্থানে রেখে বিন্যাস সংখ্যা = 3! = 6
বাকি 3টি ব্যঞ্জনবর্ণ 3টি জোড় স্থানে রেখে বিন্যাস সংখ্যা = 3! = 6

∴ স্বরবর্ণগুলোকে কেবল বিজোড় স্থানে রেখে মোট বিন্যাস সংখ্যা = 6 × 6
= 36

অতএব, AUTHOR শব্দটিকে স্বরবর্ণগুলোকে কেবল বিজোড় স্থানে রেখে মোট 36 উপায়ে সাজানো যাবে।

১৩৮.
How many 4-digit numbers can be formed from the digits 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
  1. ক
    ক) 12 ways
  2. খ
    খ) 18 ways
  3. গ
    গ) 24 ways
  4. ঘ
    ঘ) 36 ways
ব্যাখ্যা
Question: How many 4-digit numbers can be formed from the digits 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Solution:
Number will be divisible by 5 if the last number is 5.
So, first number can be chosen in 4C1 ways 
= 4 ways
As the digit is not repeated
second number can be chosen in 3C1 
= 3 ways

As the digit is not repeated
third number can be chosen in 2C1 
= 2 ways

∴ Total ways = 4 × 3 × 2 ways
= 24 ways
১৩৯.
A bag contains 6 blue marbles, 4 green marbles, and 5 black marbles. If one marble is drawn at random from the bag, what is the probability that the marble is green?
  1. ক
    2/5
  2. খ
    1/3
  3. গ
    2/15
  4. ঘ
    4/15
ব্যাখ্যা

Question: A bag contains 6 blue marbles, 4 green marbles, and 5 black marbles. If one marble is drawn at random from the bag, what is the probability that the marble is green?

Solution:
Total number of marbles = 6 + 4 + 5 = 15
Number of green marbles = 4

Therefore, P(green) = Number of green marbles/Total marbles
= 4/15

So the probability is 4/15.

১৪০.
A box contains 2 red, 3 green and 2 white balls. Two balls are drawn at random. What is the probability that none of the balls drawn is white?
  1. ক
    10/21
  2. খ
    11/21
  3. গ
    15/21
  4. ঘ
    21/10
  5. ঙ
    None
ব্যাখ্যা
Total number of balls = 2 + 3 + 2 = 7
If n(S) be the number of ways of drawing 2 balls out of 7
then n(S) = 7C2 = 21
If n(E) be the number of ways of drawing 2 balls out of (2 + 3) balls
then n(E) = 5C2 = 10
The probability that none of the balls drawn is white = 10/21
১৪১.
If two coins are tossed, what is the probability of getting at most one head?
  1. ক
    1/4
  2. খ
    1/2
  3. গ
    3/4
  4. ঘ
    1
ব্যাখ্যা

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

Solution:
দুটি মুদ্রা নিক্ষেপ করলে সম্ভাব্য ফলাফলগুলো হলো:
S = {HH, HT, TH, TT}
এখানে, মোট ফলাফল সংখ্যা = 4

"at most one head" মানে সর্বোচ্চ একটি head অর্থাৎ 0টি head অথবা 1টি head.
0টি head এর ক্ষেত্রে: TT (1টি)
1টি head এর ক্ষেত্রে: HT, TH (2টি)
∴ অনুকূল ফলাফল n(E) = 1 + 2 = 3টি

সুতরাং, সর্বোচ্চ একটি head পাওয়ার সম্ভাবনা = n(E)/n(S)
= 3/4

১৪২.
In a box, there are 5 black pens, 3 white pens and 4 red pens. In how many ways can 2 black pens, 2 white pens and 2 red pens can be chosen?
  1. ক
    ক) 180
  2. খ
    খ) 220
  3. গ
    গ) 240
  4. ঘ
    ঘ) 160
  5. ঙ
    ঙ) None of these
ব্যাখ্যা

Number of ways of choosing 2 black pens from 5 black pens in 5C2 ways.
Number of ways of choosing 2 white pens from 3 white pens in 3C2 ways.
Number of ways of choosing 2 red pens from 4 red pens in 4C2 ways.
By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 = 180 ways

১৪৩.
Six friends Rita, Anika, Zara, Lima, Arif, and Sayeed sit randomly in a row of six chairs. What is the probability that Rita and Anika do not sit next to each other?
  1. ক
    2/3
  2. খ
    1/2
  3. গ
    3/4
  4. ঘ
    3/5
ব্যাখ্যা

Question: Six friends Rita, Anika, Zara, Lima, Arif, and Sayeed sit randomly in a row of six chairs. What is the probability that Rita and Anika do not sit next to each other?

Solution:
Total number of possibilities = 6! = 720

Number of possibilities where Rita and Anika sit together = 5! × 2! 
​= 120 × 2
= 240

So the possibilities where Rita and Anika do not sit together = 720 - 240
= 480

∴Probability that Rita and Anika do not sit next to each other = 480/720
= 2/3

১৪৪.
A bag contains 6 red, 4 green, and 5 blue balls. Two balls are drawn without replacement. What is the probability that the first ball is red and the second ball is green or blue?
  1. ক
    13/15
  2. খ
    9/35
  3. গ
    21/65
  4. ঘ
    12/35
ব্যাখ্যা

Question: A bag contains 6 red, 4 green, and 5 blue balls. Two balls are drawn without replacement. What is the probability that the first ball is red and the second ball is green or blue?
​
​Solution:
​Given that,
​Red balls = 6
​Green balls = 4
​Blue balls = 5
​Total balls = 6 + 4 + 5 = 15

​We know 
​Probability = Favorable outcomes/Total outcomes

​Now,
​First ball is Red = 6/15 = 2/5
​If first is red, then remaining balls = 14
​Second ball is Green or Blue = (4 + 5)/14 = 9/14
​
​∴ Required probability = (2/5) × (9/14) 
​= 9/35
​
​∴ The required probability is 9/35.

১৪৫.
Two cards are drawn from a pack of 52 cards. What is the probability that both are black cards?
  1. ক
    ক) 29/34
  2. খ
    খ) 1/2
  3. গ
    গ) 13/102
  4. ঘ
    ঘ) 25/102
ব্যাখ্যা
Question: Two cards are drawn from a pack of 52 cards. What is the probability that both are black cards?

Solution:
number of black cards = 26

probability both card black = 26C2/52C2
= 325/1326
= 25/102
১৪৬.
A certain jar contains 60 jellybeans 22 white, 18 green, 11 yellow, 5 red and 4 purple. If a jellybeans is to be chosen at random, what is the probability that the jellybean will be neither red nor purple ?
  1. ক
    ক) 0.09
  2. খ
    খ) 0.85
  3. গ
    গ) 0.54
  4. ঘ
    ঘ) 0.91
ব্যাখ্যা
Question: A certain jar contains 60 jellybeans 22 white, 18 green, 11 yellow, 5 red and 4 purple. If a jellybeans is to be chosen at random, what is the probability that the jellybean will be neither red nor purple ?

Solution: 
Total = 60
White = 22
Green = 18
Yellow = 11
Red = 5
Purple = 4

লাল ও বেগুনি জেলি সংখ্যা = 5 + 4 = 9
লাল ও বেগুনি জেলি উঠার সম্ভাবনা = 9/60
লাল ও বেগুনি জেলি না উঠার সম্ভাবনা = 1 - 9/60
= (60 - 9)/60
= 51/60
= 0.85
১৪৭.
A school has 8 basketball players. A 5-member team and a captain (selected from the remaining players) will be chosen out of these 8 players. How many different selections can be made?
  1. ক
    120 ways
  2. খ
    156 ways
  3. গ
    168 ways
  4. ঘ
    210 ways
ব্যাখ্যা

Question: A school has 8 basketball players. A 5-member team and a captain (selected from the remaining players) will be chosen out of these 8 players. How many different selections can be made?

Solution:
We can select the 5 member team out of the 8 in = 8C5 ways
= 8!/(5! × 3!)
= 56 ways

The captain can be selected from amongst the remaining 3 players in = 3C1
= 3 ways.

∴ The total ways the selection of 5 players and a captain can be made = 56 × 3 ways
= 168 ways

১৪৮.
Find the number of ways the letters of the word ‘RUBBER can be arranged?
  1. ক
    ক) 450
  2. খ
    খ) 362
  3. গ
    গ) 250
  4. ঘ
    ঘ) 180
ব্যাখ্যা
Question: Find the number of ways the letters of the word ‘RUBBER can be arranged?

Solution: 
The word ‘RUBBER’ contains 6 letters: 2R, 2B, 1 U, 1 E
Therefore,
The required Number of ways: 
6!/{(2!) × (2!)}
= 180
১৪৯.
A license plate begins with three letters. If the possible letters are A, B, C, D , how many different permutations of these letters can be made if no letter is used more than once?
  1. ক
    ক) 12
  2. খ
    খ) 18
  3. গ
    গ) 20
  4. ঘ
    ঘ) 24
ব্যাখ্যা
Question: A license plate begins with three letters. If the possible letters are A, B, C, D , how many different permutations of these letters can be made if no letter is used more than once?

Solution: 
For the first letter, there are 5 possible choices. After that letter is chosen, there are 4 possible choices. Finally, there are 3 possible choices.
4 × 3 × 2
= 24
১৫০.
There are 10 points in a plane out of which 4 are collinear. Find the number of triangles formed by the points as vertices.
  1. ক
    ক) 116
  2. খ
    খ) 120
  3. গ
    গ) 124
  4. ঘ
    ঘ) 132
ব্যাখ্যা
The number of triangle can be formed by 10 points = 10C3
Similarly, the number of triangle can be formed by 4 points when no one is collinear = 4C3
In the question, given 4 points are collinear, Thus, required number of triangle can be formed,
= 10C3 - 4C3
= 120 - 4
= 116
১৫১.
In how many ways the letters of the word 'DIFFERENT' can be arranged?
  1. ক
    720
  2. খ
    5040
  3. গ
    40080
  4. ঘ
    90720
ব্যাখ্যা
Question: In how many ways the letters of the word 'DIFFERENT' can be arranged?

Solution:
Total nnumber of letters in the word 'DIFFERENT' = 9
Repeating letters:
F = 2 times
E = 2 times
∴ Required no. of ways = 9!/(2! × 2!)
= 90720
১৫২.
What is the probability of getting a sum of 6 if two dice are thrown?
  1. ক
    1/9
  2. খ
    5/36
  3. গ
    2/3
  4. ঘ
    None of these
ব্যাখ্যা
Question: What is the probability of getting a sum of 6 if two dice are thrown?

Solution:
In two throws a dice, n(S) = 6 × 6 = 36
Let E is the event of getting a sum of 6.
E = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
So, n(E) = 5

∴ P(E) = n(E)/n(S)
= 5/36
১৫৩.
A college has 8 basketball players. A 4 member's team and a captain will be selected out of these 8 players. How many different selections can be made?
  1. ক
    280
  2. খ
    520
  3. গ
    480
  4. ঘ
    120
ব্যাখ্যা

Question: A college has 8 basketball players. A 4 member's team and a captain will be selected out of these 8 players. How many different selections can be made?

Solution: 
We can select the 4 member team out of the 8 in =  8C4 ways
= 8!/4!(8 - 4)! = 8!/(4! × 4!)
= 70 ways

The captain can be selected from amongst the remaining 4 players in 4 ways.

∴ The total ways the selection of 4 players and a captain can be made = 70 × 4 ways
= 280 ways

So the total number of different selections that can be made is 280.

১৫৪.
Two groups, A and B wrote an exam. The probability of A's pass is 2/7 and the probability of B's pass is 2/5. What is the probability that only one of them is passed out?
  1. ক
    ক) 5/6
  2. খ
    খ) 1/3
  3. গ
    গ) 18/35
  4. ঘ
    ঘ) 16/35
ব্যাখ্যা

Let, A be the event of the group A pass
Let, B be the event of the group B pass

Then,
A'= Event of the group A's fail and B'= event of the group B's fail.
Therefore, p(A) = 2/7 and p(B) = 2/5,
P(A') = 1 - P(A) = 1- 2/7 = 5/7 and P(B') = 1- P(B) = 1- 2/5 = 3/5

Required probability = P[( A And B') Or (B And A')]
= P[( A And B') Or (B And A')]
= P[( A And B') + (B And A')]
= P[( A And B')] + P[(B And A')]
= p(A) x P(B') + P(A') x P(B)
= (2/7 x 3/5) + (2/5 x 5/7)
= (6/35 + 10/35)
= 16/35

১৫৫.
In how many ways, a committee consisting of 5 men and 6 women can be formed from 8 men and 10 women?
  1. ক
    1760
  2. খ
    5040
  3. গ
    11760
  4. ঘ
    86400
ব্যাখ্যা

Required number of ways = (8C5 × 10C6)
= (8C3 × 10C4)
= (8 × 7 × 6)/3! × (10 × 9 × 8 × 7)/4!
= (8 × 7 × 6)/6 ×(10 × 9 × 8 × 7)/(4 × 3 × 2 × 1)
= 11760.

১৫৬.
In how many ways can 3 people be seated in a row containing 6 seats?
  1. ক
    110
  2. খ
    120
  3. গ
    130
  4. ঘ
    140
ব্যাখ্যা
Question: In how many ways can 3 people be seated in a row containing 6 seats?

Solution:
First person can be seated in 6 ways, the second person in 5 ways and the third person in 4 ways.
Then, by fundamental principle, total number of ways in which three persons can be seated in 6 seats in a row is (6 × 5 × 4) ways
= 120 ways
১৫৭.
Five persons, Shahin, Masud, Jayed, Raj, and Alom, sit randomly in five chairs in a row. What is the probability that Raj and Masud sit next to each other?
  1. ক
    ক) 1/5
  2. খ
    খ) 2/5
  3. গ
    গ) 1/3
  4. ঘ
    ঘ) 1/60
ব্যাখ্যা
Question: Five persons, Shahin, Masud, Jayed, Raj, and Alom, sit randomly in five chairs in a row. What is the probability that Raj and Masud sit next to each other?

Solution:
Total possibilities = 5! = 120
favorabole events = 4! × 2! 
= 24 × 2
= 48

probability = 48/120
= 2/5
১৫৮.
How many different six digit numbers can be formed using all of the following digits 2, 2, 5, 5, 5, 4?
  1. ক
    30
  2. খ
    60
  3. গ
    90
  4. ঘ
    120
ব্যাখ্যা
Question: How many different six digit numbers can be formed using all of the following digits 2, 2, 5, 5, 5, 4?

Solution:
প্রদত্ত অঙ্ক মোট 6টি যার মধ্যে 2টি 2 এবং 3টি 5 আছে।

∴ নির্ণেয় ছয় অঙ্কবিশিষ্ট মোট গঠিত সংখ্যা = 6!/(2! × 3!) টি
= 60 টি 
১৫৯.
In a box of 5 eggs, 2 are rotten. What is the probability that two eggs chosen at random from the box are rotten?
  1. ক
    2/5
  2. খ
    1/16
  3. গ
    1/10
  4. ঘ
    1/5
  5. ঙ
    13/20
ব্যাখ্যা

Question: In a box of 5 eggs, 2 are rotten. What is the probability that two eggs chosen at random from the box are rotten?

Solution: 
Given,
Total number of eggs = 5
Number of rotten eggs = 2
Number of non-rotten (good) eggs = 3
Number of eggs to be chosen = 2

We can think of this as picking one egg and then picking another without replacing the first one.
The probability that the first egg picked is rotten = 2/5
  
Now, if the first egg was rotten, there is now 1 rotten egg left and 4 total eggs left.
The probability that the second egg picked is also rotten = 1/4
  
Hence, the probability of selecting 2 rotten eggs in a row = (2/5) × (1/4)
= 1/10

১৬০.
Two unbiased coins are tossed. What is the probability of getting at least 1 tail?
  1. ক
    1/2
  2. খ
    1/3
  3. গ
    2/3
  4. ঘ
    3/4
  5. ঙ
    None of the above
ব্যাখ্যা

Question: Two unbiased coins are tossed. What is the probability of getting at least 1 tail?

Solution:
Total outcomes = {TT, TH, HT, HH} = 4
Favorable outcomes = {TT, TH, HT} = 3

So, the probability of getting at least 1 tail = Favorable outcomes/Total outcomes
= 3/4

১৬১.
One card is drawn from a deck of 52 cards, well-shuffled. Calculate the probability that the card will not be an ace.
  1. ক
    ক) 12/13
  2. খ
    খ) 51/52
  3. গ
    গ) 3/26
  4. ঘ
    ঘ) 1/13
ব্যাখ্যা
Question: One card is drawn from a deck of 52 cards, well-shuffled. Calculate the probability that the card will not be an ace.

Solution:
Total number of Ace is 4.
∴ The probability of a card will be 4/53 = 1/13

∴ The probability that the card will not be an ace = 1 - (1/13)
= (13 - 1)/13
= 12/13 
১৬২.
Five children, Shihan, Masum, Jayed, Rana, and khairul, sit randomly in five chairs in a row. What is the probability that khairul and Masum don't sit next to each other?
  1. ক
    1/5
  2. খ
    2/5
  3. গ
    3/5
  4. ঘ
    1/2
ব্যাখ্যা
Question: Five children, Shihan, Masum, Jayed, Rana, and khairul, sit randomly in five chairs in a row. What is the probability that khairul and Masum don't sit next to each other?

Solution:
Total possibilities = 5! = 120
 events where  khairul and Masum sit next to each other= 4! × 2! 
= 24 × 2
= 48

probability that they sit next to each other= 48/120
= 2/5

∴ probability that khairul and Masum don't sit next to each other = 1 - 2/5
= (5 - 2)/5
= 3/5
১৬৩.
At the end of a banquet 10 people shake hands with each other. How many handshakes will there be in total?
  1. ক
    100
  2. খ
    20
  3. গ
    45
  4. ঘ
    90
ব্যাখ্যা
Question: At the end of a banquet 10 people shake hands with each other. How many handshakes will there be in total?

Solution:
Two people can make 1 handshake
∴ No. of handshakes = 10C2 = 45
১৬৪.
In how many different ways can the letters of the word MAGIC can be formed?
  1. ক
    24 ways
  2. খ
    120 ways
  3. গ
    240 ways
  4. ঘ
    720 ways
  5. ঙ
    None of these
ব্যাখ্যা
Question: In how many different ways can the letters of the word MAGIC can be formed?

Solution:
he word "MAGIC" consists of 5 distinct letters: M, A, G, I, C.
The formula for finding the number of permutations of n distinct items is given by n!
Here,
n = 5.

So, we need to calculate 5!
5! = 5 × 4 × 3 × 2 × 1 = 120

Therefore, the number of different ways the letters of the word 'MAGIC' can be formed is 120.
১৬৫.
A letter is taken out at random from 'ASSISTANT'  and another is taken out from 'STATISTICS'. The probability that they are the same letter is :
  1. ক
    19/90 
  2. খ
    35/96
  3. গ
    35/73
  4. ঘ
    1/5
ব্যাখ্যা
Question: A letter is taken out at random from 'ASSISTANT'  and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

Solution: 
For S = (3/9) × (3/10) = 1/10
For A = (2/9) × (1/10) = 1/45
For I = (1/9) × (2/10) = 1/45
For T = (2/9) × (3/10) = 1/15

Total probability = (1/10) + (1/45) + (1/45) + (1/15)
= 19/90 
১৬৬.
A dice is cast twice, and the sum of the appearing numbers is 10. The probability that the number 5 has appeared at least once is-
  1. ক
    2/3
  2. খ
    1/2
  3. গ
    1/5
  4. ঘ
    1/3
ব্যাখ্যা

Question: A dice is cast twice, and the sum of the appearing numbers is 10. The probability that the number 5 has appeared at least once is-

Solution:
A dice is cast twice
Sum of the numbers = 10
Find probability that number 5 appears at least once

Now,
All possible pairs (first die, second die) whose sum = 10 is,
(4, 6), (5, 5), (6, 4)
∴ Total outcomes = 3

And,
Outcomes with at least one 5
From the above pairs we get only 1 outcome has at least one 5.

 ∴ Probability = Favorable Outcome/Total outcomes = 1/3

১৬৭.
An integer n between 1 and 100, inclusive, is to be chosen at random. What is the probability that n(n + 1) will be divisible by 5?
  1. ক
    1/5
  2. খ
    2/5
  3. গ
    1/2
  4. ঘ
    1/3
ব্যাখ্যা
Question: An integer n between 1 and 100, inclusive, is to be chosen at random. What is the probability that n(n + 1) will be divisible by 5?

Solution: 
total number = 100
n(n+1) will be divisible by 5 if n or n + 1 is divisible by 5

when n is divisible by 5, there are 20 such numbers (5, 10, 15, 20, 25,.....,100)
when n + 1 is divisible by 5, there are 20 such numbers (4, 9, 14,.....,99)

∴ probability = (20 + 20)/100
= 40/100
= 2/5
১৬৮.
A group has 8 men and 7 women. In how many ways can a committee of 5 people be formed if the number of women is at least 3?
  1. ক
    785
  2. খ
    988
  3. গ
    1281
  4. ঘ
    1125
ব্যাখ্যা

Question:  A group has 8 men and 7 women. In how many ways can a committee of 5 people be formed if the number of women is at least 3?
​
​Solution:
​Given that,
​Total committee size = 5

And for women ≥ 3, possible distributions:
Three ways to formed the committee
1. 3 women + 2 men
2. 4 women + 1 man
3. 5 women + 0 men
​
​Now, 1st case- 3 women + 2 men
​Choose 3 women from 7, (7C3) = 35
​Choose 2 men from 8,  (8C2) = 28
​∴ Total ways = 35 × 28 = 980
​
​2nd case- 4 women + 1 man
​Choose 4 women from 7, (7C4) = 35
Choose 1 man from 8, (8C1) = 8
 ​∴ Total ways = 35 × 8 = 280
​
​And 3rd case-​5 women + 0 men
Choose 5 women from 7,  (7C5) = 21
No men to choose
​∴ Total ways = 21
​
​∴ ​Total ways = 980 + 280 + 21= 1281

১৬৯.
In how many ways can 5 people be seated around a circular table?
  1. ক
    20
  2. খ
    24
  3. গ
    60
  4. ঘ
    120
ব্যাখ্যা
Question: In how many ways can 5 people be seated around a circular table?

Solution: 
For n people around a circle, the number of distinct arrangements is = (n -1)!
So, for 5 people = (5 − 1)! = 4! = 24
১৭০.
Out of six men and seven women, a five-member committee is to be chosen. What’s the likelihood that it will have exactly two men and three women?
  1. ক
    75/429
  2. খ
    177/429
  3. গ
    49/327
  4. ঘ
    175/429
  5. ঙ
    None of the above
ব্যাখ্যা
Question: Out of six men and seven women, a five-member committee is to be chosen. What’s the likelihood that it will have exactly two men and three women?
(ছয়জন পুরুষ ও সাতজন নারী থেকে পাঁচ সদস্যের একটি কমিটি গঠন করতে হবে। এর মধ্যে ঠিক দুইজন পুরুষ ও তিনজন নারী থাকার সম্ভাবনা কত?)

Solution:
Total member = 6 + 7 = 13
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

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
১৭১.
A bag contains 3 red balls and 2 blue balls. If two balls are drawn without replacement, what is the probability that both are red?
  1. ক
    3/10
  2. খ
    3/5
  3. গ
    6/25
  4. ঘ
    9/25
ব্যাখ্যা
Question: A bag contains 3 red balls and 2 blue balls. If two balls are drawn without replacement, what is the probability that both are red?

Solution: 
Total balls = 3 red + 2 blue = 5 balls. 
Probability that the first ball is red = 3/5

After removing 1 red ball, we have:
Remaining red balls = 2
Total remaining balls = 4
So, the probability that the second ball is red = 2/4 = 1/2 

∴ Total probability (both red) = 3/5 × 1/2
= 3/10 
১৭২.
Tickets numbered 1 to 24 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 3 or 4?
  1. ক
    11/24
  2. খ
    1/2
  3. গ
    5/12
  4. ঘ
    3/8
ব্যাখ্যা
Question: Tickets numbered 1 to 24 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 3 or 4?

Solution:
Here, S = {1, 2, 3, 4, ...., 23, 24}
Let E = event of getting a multiple of 3 or 4
= {3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21, 24}

∴P(E) = n(E)/n(S)
= 12/24
= 1/2
১৭৩.
How many different ways can the letters in the word ATTEND be arranged?
  1. ক
    260
  2. খ
    180
  3. গ
    360
  4. ঘ
    420
ব্যাখ্যা
Question: How many different ways can the letters in the word ATTEND be arranged?

Solution:
There are 6 letter in the word 'ATTEND' whereas, T comes two times.
So, required number of ways = 6!/2!
= 360
১৭৪.
A basket contains three blue and four red balls. If three balls are drawn at random from the basket, what is the probability that all the three are either blue or red?
  1. ক
    2/5
  2. খ
    1/7
  3. গ
    3/8
  4. ঘ
    None of these
ব্যাখ্যা
Question: A basket contains three blue and four red balls. If three balls are drawn at random from the basket, what is the probability that all the three are either blue or red?

Solution:
Probability to be a Blue = 3C3/7C3 = 1/35
Probability to be a Red = 4C3/7C3 = 4/35

∴ Required probability = (1/35) + (4/35)
= (1 + 4)/35
= 5/35
= 1/7
১৭৫.
A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of them is defective is-
  1. ক
    ক) 7/19
  2. খ
    খ) 4/19
  3. গ
    গ) 12/19
  4. ঘ
    ঘ) 21/95
ব্যাখ্যা
Question: A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of them is defective is-

Solution:
Total bulbs 20
Number of defective bulbs 4
∴ Number of non-defective bulbs (20 - 4) = 16.

The probability of non-defective bulbs is 16C2/20C2 = 120/190 = 12/19

∴ The probability of at least 1 bulb is defective = 1 - (12/19) = 7/19 
১৭৬.
In a meeting, every person shakes hands with every other person exactly once. If the total number of handshakes was 28, how many people were in the meeting?
  1. ক
    7
  2. খ
    8
  3. গ
    9
  4. ঘ
    10
  5. ঙ
    None of the above
ব্যাখ্যা

Question: In a meeting, every person shakes hands with every other person exactly once. If the total number of handshakes was 28, how many people were in the meeting?

Solution:
Let,
the number of people be n.

ATQ,
number of total handshakes,
n(n - 1)/2 = 28
⇒ n(n - 1) = 28 × 2
⇒ n2 - n = 56
⇒ n2 - n - 56 = 0
⇒ n2 - 8n + 7n - 56 = 0
⇒ n(n - 8) + 7(n - 8) = 0
⇒ (n - 8)(n + 7) = 0

∴ n = 8, - 7

So, the number of people be 8.

১৭৭.
A bag contains 8 red and 5 green balls. Two balls are drawn randomly. Find the probability that one ball is red and other is green.
  1. ক
    ক) 5/13
  2. খ
    খ) 8/13
  3. গ
    গ) 2/13
  4. ঘ
    ঘ) 20/39
ব্যাখ্যা
Question: A bag contains 8 red and 5 green balls. Two balls are drawn randomly. Find the probability that one ball is red and other is green.

Solution: 
A bag contains 8 red and 5 green balls. 
Total balls = 8 + 5 = 13
Two balls drawn 13C2 = 78
One ball is red and other is green = 8C1 × (5C1) = 8 × 5 = 40

Required probability = 40/78 = 20/39
১৭৮.
A problem is assigned to three students. The chances of solving it for the first, second, and third students are 1/3, 1/4, and 1/6, respectively. What is the probability that the problem will be solved by at least one of them? 
  1. ক
    5/12
  2. খ
    7/12
  3. গ
    7/10
  4. ঘ
    4/9
  5. ঙ
    None
ব্যাখ্যা

Question: A problem is assigned to three students. The chances of solving it for the first, second, and third students are 1/3, 1/4, and 1/6, respectively. What is the probability that the problem will be solved by at least one of them?

Solution:
Probability of 1st student solving the problem = 1/3
Probability of 1st student not solving the problem = 1 - 1/3 = 2/3

Probability of 2nd student solving the problem = 1/4
Probability of 2nd student not solving the problem = 1 - 1/4 = 3/4

Probability of 3rd student solving the problem = 1/6
Probability of 3rd student not solving the problem = 1 - 1/6 = 5/6

Probability that none of the students solve the problem = (2/3) × (3/4) × (5/6)
= 5/12

∴ Probability that the problem will be solved = 1 - 5/12
= 7/12

∴ The probability that the problem will be solved is 7/12.

১৭৯.
What is the probability that a randomly chosen 4-digit number has all distinct digits?
  1. ক
    63/125
  2. খ
    423/500
  3. গ
    18/25
  4. ঘ
    89/100
ব্যাখ্যা
Question: What is the probability that a randomly chosen 4-digit number has all distinct digits?

Solution: 
Total 4-digit numbers = 9999 − 1000 + 1 = 9000

A 4-digit number has digits in the form: ABCD, where:
A (thousands place): Can be 1-9 (9 options, cannot be 0).
B (hundreds place): Can be 0-9, except A (9 options).
C (tens place): Can be 0-9, except A and B (8 options).
D (units place): Can be 0-9, except A, B, and C (7 options).
 
Total numbers with distinct digits = 9 × 9 × 8 × 7 = 4536

∴ Probability = 4536/9000 = 63/125 
১৮০.
Out of 5 men and 3 women, a committee of 3 members is to be formed so that it has 1 woman and 2 men. In how many different ways can it be done?
  1. ক
    10
  2. খ
    20
  3. গ
    30
  4. ঘ
    23
  5. ঙ
    None of these
ব্যাখ্যা
Question: Out of 5 men and 3 women, a committee of 3 members is to be formed so that it has 1 woman and 2 men. In how many different ways can it be done?

Solution:
Number of selections = Number of ways of selecting 2 men out of 5 men × Number of ways of selecting 1 woman out of 3 women
= 5C2× 3C1
= 10 × 3
= 30
১৮১.
A coin is tossed four times. What is the probability of getting head on all tosses?
  1. ক
    ক) 1/4
  2. খ
    খ) 1/8
  3. গ
    গ) 1/16
  4. ঘ
    ঘ) 3/16
ব্যাখ্যা

Probability of getting head on all tosses = 1/2 × 1/2 × 1/2 × 1/2 = 1/16

১৮২.
The King and Queen of black color are taken out from a deck of 52 playing cards. A card is drawn from the remaining well-shuffled cards. The probability of getting a spade card is-
  1. ক
    11/48
  2. খ
    13/48
  3. গ
    11/52
  4. ঘ
    13/52
ব্যাখ্যা

Question: The King and Queen of black color are taken out from a deck of 52 playing cards. A card is drawn from the remaining well-shuffled cards. The probability of getting a spade card is-

Solution:
Number of Kings and Queens of black color = 4
∴ Remaining cards = 52 - 4
= 48

Spade card is black, hence 1 queen and 1 king of spade is removed, remaining spade card = 13 - (1 + 1) 
= 11

∴ Probability = 11/48

১৮৩.
In how many ways 5 students can be chosen from the class of 10 students?
  1. ক
    ক) 152
  2. খ
    খ) 252
  3. গ
    গ) 352
  4. ঘ
    ঘ) 452
ব্যাখ্যা
Question: In how many ways 5 students can be chosen from the class of 10 students?

Solution:
ways 5 students can be chosen from the class of 10 students is = 10C5
= 10!/(5! 5!)
= 252
১৮৪.
A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 5 from part A and 8 from part B, in how many ways can he choose the questions?
  1. ক
    10340
  2. খ
    11240
  3. গ
    11340
  4. ঘ
    11360
ব্যাখ্যা
Question: A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 5 from part A and 8 from part B, in how many ways can he choose the questions?

Solution:
ways to choose 5 from part A = 10C5
ways to choose 8 from part B = 10C8

choose 5 from part A and 8 from part B = 10C5 × 10C8
= {10!/(5! 5!)} × {10!/(2! 8!)}
= 11340
১৮৫.
In how many ways can 5 people from a group of 8 people be seated around a circular table?
  1. ক
    1200
  2. খ
    560
  3. গ
    2520
  4. ঘ
    1344
ব্যাখ্যা

Question: In how many ways can 5 people from a group of 8 people be seated around a circular table?

Solution:
5 people out of 8 = 8C5
= 8!/5!(8 - 5)!
= 8!/(3! × 5!)
​= (8 × 7 × 6 × 5!)/(6  × 5!)
= 56

And 5 people around a circular table = (5 - 1)! = 4! = 24

∴ Total ways = 24 × 56 = 1344

১৮৬.
A bag contains 4 white, 5 red, and 6 blue balls. One ball is drawn at random. What is the probability that the ball drawn is neither white nor blue?
  1. ক
    4/15
  2. খ
    2/5
  3. গ
    1/3
  4. ঘ
    1/2
ব্যাখ্যা

Question: A bag contains 4 white, 5 red, and 6 blue balls. One ball is drawn at random. What is the probability that the ball drawn is neither white nor blue?

Solution:
Total number of balls, n(S) = 4 + 5 + 6 = 15

Let E = event that the ball is neither white nor blue (which means the ball is red)

Number of red balls, n(E) = 5

∴ Probability, P(E) = n(E)/n(S)
= 5/15
 = 1/3

১৮৭.
In a simultaneous throw of a pair of dice, what is the probability of getting a total more than 8?
  1. ক
    5/18
  2. খ
    13/18
  3. গ
    1/4
  4. ঘ
    7/18
ব্যাখ্যা

Question: In a simultaneous throw of a pair of dice, what is the probability of getting a total more than 8?

Solution:
When two fair six-sided dice are thrown together. Then we get
Total outcomes = 6 × 6 = 36

And, Count outcomes with sum > 8

We want sum > 8 ⇒ sums = 9, 10, 11, 12
Sum 9 = (3, 6), (4, 5), (5, 4), (6, 3) ⇒ 4 outcomes
Sum 10 = (4, 6), (5, 5), (6, 4) ⇒ 3 outcomes
Sum 11 = (5, 6), (6, 5) ⇒ 2 outcomes
Sum 12 = (6, 6) ⇒ 1 outcome

∴ Total favorable outcomes =  4 + 3 + 2 + 1 = 10

∴ Probability(sum > 8) = favorable outcomes/total outcomes
= 10/36
= 5/18

So the probability of getting a total more than 8 is 5/18.

১৮৮.
If a person purchases 15 of the 3000 tickets sold in a raffle that awards one prize, what is the probability that this person will not win?
  1. ক
    0
  2. খ
    1/200
  3. গ
    1/2
  4. ঘ
    199/200
  5. ঙ
    1
ব্যাখ্যা
Question: If a person purchases 15 of the 3000 tickets sold in a raffle that awards one prize, what is the probability that this person will not win?

Solution:
Probability for him to win = 15/3000 = 1/200

Probability for him to NOT win = 1 - 1/200 = 199/200
১৮৯.
In what ways the letters of the word "RUMOUR" can be arranged?
  1. ক
    180
  2. খ
    150
  3. গ
    200
  4. ঘ
    230
ব্যাখ্যা
Question: In what ways the letters of the word "RUMOUR" can be arranged?

Solution:
The word RUMOUR consists of 6 words in which R and U are repeated twice.
Therefore, the required number of permutations = 6!/(2! × 2!)
= 180
১৯০.
What is probability of getting a sum 9 from two throws of a dice?
  1. ক
    ক) 1/9
  2. খ
    খ) 1/4
  3. গ
    গ) 1/12
  4. ঘ
    ঘ) 1/18
ব্যাখ্যা
Question: What is probability of getting a sum 9 from two throws of a dice?

Solution: 
Two dice were thrown.
Total outcomes: 36
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.
P(E) = n(E)/n(S)
= 4/36
= 1/9
১৯১.
A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
  1. ক
    3/4
  2. খ
    4/7
  3. গ
    1/8
  4. ঘ
    3/7
ব্যাখ্যা
Question: A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

Solution:
Let number of balls = (6 + 8) = 14.
Number of white balls = 8.
P (drawing a white ball) = 8/14 = 4/7.
১৯২.
What is the probability of getting 53 Mondays in a leap year?
  1. ক
    3/8
  2. খ
    5/9
  3. গ
    2/5
  4. ঘ
    2/7
  5. ঙ
    3/13
ব্যাখ্যা

Question: What is the probability of getting 53 Mondays in a leap year?

Solution:
1 year = 365 days . A leap year has 366 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks = 52 × 7 = 364days
366 - 364 = 2 days

In a leap year there will be 52 Sundays and 2 days will be left.
These 2 days can be:
1. Sunday, Monday
2. Monday, Tuesday
3. Tuesday, Wednesday
4. Wednesday, Thursday
5. Thursday, Friday
6. Friday, Saturday
7. Saturday, Sunday
Of these total 7 outcomes, the favourable outcomes are 2.

Hence the probability of getting 53 days = 2/7

১৯৩.
In how many different ways can the letters of the word "DESIGN" be arranged so that the vowels are at the two ends?
  1. ক
    24
  2. খ
    32
  3. গ
    48
  4. ঘ
    60
ব্যাখ্যা
Question: In how many different ways can the letters of the word "DESIGN" be arranged so that the vowels are at the two ends?

Solution:
The given word "DESIGN" contains 4 consonants and 2 vowel. 
At the two ends the two vowels can be arranged in 2! = 2 ways.
Remaining 4 letters can be arranged in = 4!
= 24 

∴ Total number of ways = (24 × 2) = 48
১৯৪.
In how many ways can 12 identical balls be distributed among 4 distinct boxes such that each box contains at least one ball?
  1. ক
    165
  2. খ
    220
  3. গ
    286
  4. ঘ
    330
ব্যাখ্যা
Question: In how many ways can 12 identical balls be distributed among 4 distinct boxes such that each box contains at least one ball?

Solution: 
Since each box must get at least one ball, give 1 ball to each box first.
Balls used = 4
Remaining balls = 12 - 4 = 8

Now, distributing 8 identical balls among 4 boxes with no restrictions. 
Formula = n + k - 1Ck - 1 ;where n = 8, k = 4
= 11C3 
= 11!/(3! × 8!)
= (11 × 10 × 9)/(3 × 2 × 1)
= 990/6
= 165
১৯৫.
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
  1. ক
    32
  2. খ
    48
  3. গ
    64
  4. ঘ
    96
  5. ঙ
    69
ব্যাখ্যা

We may have (1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
= 3 x (6 x 5)/(1 x 2) + (3 x 2)/(1 x 2) x 6 + 1
= (45 + 18 + 1)
= 64.

১৯৬.
There is a group of 5 men, 6 women and 8 children. 1 man, 1 woman and one child are going to be selected to play a game. In how many ways can the selection be done?
  1. ক
    ক) 240 ways
  2. খ
    খ) 480 ways
  3. গ
    গ) 120 ways
  4. ঘ
    ঘ) None of these
ব্যাখ্যা
5 জন পুরুষ থেকে 1 জন পুরুষ বাছাই করার উপায় = 5c1 = 5
6 জন মহিলা  থেকে 1 জন মহিলা বাছাই করার উপায় = 6c1 = 6
8 শিশু থেকে 1 জন শিশু বাছাই করার উপায় = 8c1 = 8
মোট বাছাই করার উপায় =  5 × 6 × 8 = 240
১৯৭.
A box contains 9 blue balls, 6 green balls, and 10 yellow balls. One ball is drawn at random. What is the probability that the ball drawn is neither blue nor yellow?
  1. ক
    1/4
  2. খ
    6/25
  3. গ
    3/5
  4. ঘ
    2/25
ব্যাখ্যা

Question: A box contains 9 blue balls, 6 green balls, and 10 yellow balls. One ball is drawn at random. What is the probability that the ball drawn is neither blue nor yellow?

Solution:
Number of blue balls = 9
Number of green balls = 6
Number of yellow balls = 10

∴ Total balls = 9 + 6 + 10 = 25

Let, event E = The ball drawn is neither blue nor yellow, so it must be green.

∴ Number of favorable outcomes = 6

∴ P(E) = 6/25

১৯৮.
In how many different ways can the letters of the AUCTION be arranged in such a way that the vowels always come together?
  1. ক
    ক) 30
  2. খ
    খ) 48
  3. গ
    গ) 144
  4. ঘ
    ঘ) 576
ব্যাখ্যা

The given word contains 7 different letters.
Keeping the vowels (AUIO) together, we take them as 1 letter.
Then,
we have to arrange the letters CTN(AUIO).
Now, 4 letters can be arranged in 4! = 24 ways.
The vowels (AUIO) can be arranged themselves in 4! = 24 ways.
∴ Required number of ways = (24 × 24)
= 576.

১৯৯.
There are 20 stops for the local trains running between Churchgate and Virar. A passenger travelling between any two stops needs to buy a ticket. How many types of tickets are required to be made to meet all the possibilities?
  1. ক
    ক) 72
  2. খ
    খ) 190
  3. গ
    গ) 380
  4. ঘ
    ঘ) 760
ব্যাখ্যা

We need to SELECT people.
[SELECT = Combination = nCr = n!/r!(n-r)!

There are 20 stations. A ticket is needed between 2 stops.
That means, we simply need to select 2 stops from possible 20 stops.

That can be done by 20C2 = 20!/2!(20 - 2)! = 20!/2!18! = 190 ways.

This is when we start from one side.
When we travel from the other side we will need a separate ticket.
That means while going from A to B and B to A, we will need separate tickets.
So again on another side, we need 190 tickets.

Total tickets = 190 + 190 = 380 tickets.

২০০.
A committee of 4 members is to be formed from 6 men and 5 women. In how many different ways can the committee be formed if it must contain at least 2 men?
  1. ক
    265
  2. খ
    360
  3. গ
    420
  4. ঘ
    540
ব্যাখ্যা

Question: A committee of 4 members is to be formed from 6 men and 5 women. In how many different ways can the committee be formed if it must contain at least 2 men?

Solution:
To form a committee of 4 members with at least 2 men, we consider the following possible cases:

Case 1: 2 Men and 2 Women
Ways = 6C2 × 5C2
= 15 × 10
= 150

Case 2: 3 Men and 1 Woman
Ways = 6C3 × 5C1
= 20 × 5
= 100

Case 3: 4 Men and 0 Women
Ways = 6C4 × 5C0
= 15 × 1
= 15

Total number of ways = 150 + 100 + 15
= 265