পেট্রোবাংলা প্রিলি ও লিখিত সমন্বিত প্রস্তুতি [Archived]

Question: A coin is thrown 3 times. What is the probability that atleast one head is obtained?

Solution:
Sample space = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Total number of ways = 2³ = 8.
Fav. Cases = 7

∴ Probability that atleast one head is obtained = 7/8

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

Question: A box contains three white balls, four black balls and three red balls. The number of ways in which three balls can be drawn from the box so that at least one of the balls is black is-

Solution:
The required number of ways
(a) 1 black and 2 others = ⁴C₁ × ⁶C₂ = 4 × 15 = 60

(b) 2 black and 1 other = ⁴C₂ × ⁶C₁ = 6 × 6 = 36

(c) All the three black = ⁴C₃ = 4

∴ Total = 60 + 36 + 4 = 100

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

Question: In a cricket championship, there are 21 matches. If each team plays one match with every other team, the number of teams is-

Solution:
Let n be the number of teams.
ⁿC₂ = 21
⇒ n!/{2! × (n - 2)!} = 21
⇒ {n(n - 1)(n - 2)!}/{2! × (n - 2)!} = 21
⇒ n(n - 1)/2 = 21
⇒ n(n - 1) = 42
⇒ n² - n - 42 = 0
⇒ n - 7n + 6n - 42 = 0
⇒ n(n - 7) + 6(n - 7) = 0
⇒ (n - 7)(n + 6)
∴ n = 7 [Negative value is not acceptable]

Question: What is the probability of getting a sum of 7 when two dice are thrown?

Solution:
Total number of ways = 6 × 6 = 36 ways.
Favorable cases = (1, 6) (6, 1) (2, 5) (5, 2) (3, 4) (4, 3) = 6 ways.

∴ Probability = 6/36 = 1/6

Question: The number of four-digit telephone numbers having at least one of their digits repeated is-

Solution:
The number of four-digit telephone numbers which can be formed using the digits of 0, 1, 2,...., 9 is 10⁴ = 10000
The number of four digit telephone numbers which have none of their digits repeated is ¹⁰P₄ = 5040

Hence the required number =10000 - 5040 = 4960

Question: 1 card is drawn at random from the pack of 52 cards. Find the Probability that it is an honor card.

Solution:
honor cards = (A, J, Q, K) 4 cards from each suits = 4 × 4 = 16

∴ P(honor card) = 16/52 = 4/13

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.

Question: Three dice are rolled together. What is the probability as getting at least one 4?

Solution:
Total number of ways = 6 × 6 × 6 = 216
Probability of getting number 4 at least one time
= 1 - (Probability of getting no number 4)
= 1 - (5/6) × (5/6) × (5/6)
= 1 - 125/216
= (216 - 125)/216
= 91/216

Question: In an examination, a candidate is required to pass all five different subjects. The number of ways he can fail is-

Solution:
The candidate will fail if he fails either in 1 or 2 or 3 or 4 or 5 subjects,
∴ Required number of ways ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅
= 5 + 10 + 10 + 5 + 1
= 31

Question: Two cards are drawn from the pack of 52 cards. Find the probability that both are diamonds or both are kings.

Solution:
Total no. of ways = ⁵²C₂
Case I: Both are diamonds = ¹³C₂
Case II: Both are kings = ⁴C₂

∴ P(both are diamonds or both are kings) = (¹³C₂ + ⁴C₂)/⁵²C₂
= (78 + 6)/1326
= 84/1326
= 42/663
= 14/221

Question: Nine chairs are numbered 1 to 9. Three women and four men wish to occupy one chair each. First the women chose the chairs from amongst the chair marked 1 to 5; and then the men select the chairs from amongst the remaining. The number of possible arrangements is-

Solution:
Women can select 3 chairs from chairs numbered 1 to 5 in ⁵C₃ ways
and remaining 6 chairs can be selected by 4 men in ⁶C₄ ways.

Hence the required number of ways = ⁵C₃ × ⁶C₄

Question: A problem is given to three persons P, Q, R whose respective chances of solving it are 2/7, 4/7, 4/9 respectively. What is the probability that the problem is solved?

Solution:
P এর সমাধান করার সম্ভাবনা = 2/7
∴ P এর সমাধান না করার সম্ভাবনা = 1 - 2/7 = 5/7

Q এর সমাধান করার সম্ভাবনা = 4/7
∴ Q এর সমাধান না করার সম্ভাবনা = 1 - 4/7 = 3/7

R এর সমাধান করার সম্ভাবনা = 4/9
∴ R এর সমাধান না করার সম্ভাবনা = 1 - 4/9 = 5/9

∴ তিনজন এর সমাধান না করার সম্ভাবনা = (5/7) × (3/7) × (5/9)
= 75/441
= 25/147

∴ তিনজনের সমাধান করার সম্ভাবনা = 1 - 25/147
= 122/147

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.

Question: In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

Solution:
Required number of ways = (⁷C₅ × ³C₂)
= 21 × 3
= 63

Question: Find the probability that a leap year has 52 Sundays.

Solution:
A leap year can have 52 Sundays or 53 Sundays.
In a leap year, there are 366 days out of which there are 52 complete weeks & remaining 2 days.

Now, these two days can be (Sat, Sun) (Sun, Mon) (Mon, Tue) (Tue, Wed) (Wed, Thur) (Thur, Friday) (Friday, Sat).
So there are total 7 cases out of which (Sat, Sun) (Sun, Mon) are two favorable cases.
So, P(53 Sundays) = 2/7

Now,
P(52 Sundays) + P(53 Sundays) = 1
So, P(52 Sundays) = 1 - P(53 Sundays) = 1 - (2/7) = 5/7

Question: How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

Solution:
'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time = ¹⁰P₄ = 5040

Question: Tickets numbered 1 to 20 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 5?

Solution:
Here,
S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

∴ P(E) = n(E)/n(S) = 9/20.

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

Solution:
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = 8!/(2! × 2!) = 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4!/2! = 12

∴ Required number of words = (10080 × 12) = 120960

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

Question: How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Solution:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

∴ Required number of numbers = (1 × 5 × 4) = 20.

Question: In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Solution:
Total number of outcomes possible, n(S) = 10 + 25 = 35
Total number of prizes, n(E) = 10

∴ P(E) = n(E)/n(S) =10/35 =2/7

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

Question: In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is-

Solution:
Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then,
n(S) = Number ways of selecting 3 students out of 25
= ²⁵C₃
= 2300

n(E) = (¹⁰C₁ × ¹⁵C₂) = 10 × 105
= 1050

∴ P(E) = n(E)/n(S) = 1050/2300 = 21/46

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]

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

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