Question 1

Question 1: There are 5 girls and 2 boys, then the probability that no two boys are sitting together for a photograph is (A) (B) (C) (D)

Accepted Answer

(D) Hint: There are 5 girls and 2 boys. They can be arranged among themselves in = 7! ways. ∴ Girls can be arranged among themselves in = 5! ways. No two boys should sit together. Let girls be denoted by the letter G. – G – G – G – G – G – There are 6 places, marked by ‘-’ where boys can sit. ∴ Boys

Question 2

Question 2: In a jar, there are 5 black marbles and 3 green marbles. Two marbles are picked randomly one after the other without replacement. What is the possibility that both the marbles are

Accepted Answer

(A)

Question 3

Question 3: Two dice are thrown simultaneously. Then the probability of getting two numbers whose product is even is (A) (B) (C) (D)

Accepted Answer

(A) Hint: Two dice are thrown. ∴ n(S) = 36 Getting two numbers whose product is even, i.e., one of the two numbers must be even. Let event A: Getting even number on first dice, event B: Getting even number on second dice. n(A) = 18, n(B) = 18, n(A ∩ B) = 9 Required probability = P(A ∩ B) = = =

Question 4

Question 4: In a set of 30 shirts, 17 are white and the rest are black. 4 white and 5 black shirts are tagged as ‘PARTY WEAR’. If a shirt is chosen at random from this set, the possibility of

Accepted Answer

(D) Hint: 17 white + 13 black = 30 shirts 4 white and 5 black are ‘PARTY WEAR’ A: Choosing a black shirt ∴ P(A) = B: Choosing a ‘PARTY WEAR’ shirt. ∴ P(B) = There are 5 black ‘PARTY WEAR’ shirts. ∴ P(A ∩ B) = ∴ Required probability P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = + – =

Question 5

Question 5: There are 2 shelves. One shelf has 5 Physics and 3 Biology books and the other has 4 Physics and 2 Biology books. The probability of drawing a Physics book is (A) (B) (C) (D)

Accepted Answer

(B) Hint: Let event S1: First shelve is selected, event S2: Second shelve is selected, event P: Drawing a physics book. ∴ P(S1) = and P(S2) = First shelve has 5 physics and 3 biology books, i.e., total 8 books. ∴ P(P/S1) = Similarly, P(P/S2) = ∴ P(P) = P(S1). P(P/S1) + P(S2). P(P/S2) = =

Question 6

Question 6: 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. The probability that both of them get selected is (A) (B) (C

Accepted Answer

(C)

Question 7

Question 7: The probability that a student knows the correct answer to a multiple-choice question is . If the student does not know the answer, then the student guesses the

Accepted Answer

The probability of the guessed answer being correct is . Given that the student has answered the question correctly, the probability that the student knows the correct answer is (A) (B) (C) (D) Answer: (D) Hint: Let event A: Student knows the correct answer, event A’: Student guesses the answer, eve

Question 8

Question 8: The bag I contain 3 red and 4 black balls while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. The probability

Accepted Answer

(D)

Question 9

Question 9: A fair die is tossed twice. What are the odds in favour of getting 4, 5, or 6 on the first toss and 1, 2, 3, or 4 on the second toss? (A) 1 : 3 (B) 3 : 1 (C) 1 : 2 (D) 2 : 1

Accepted Answer

(C) 1 : 2 Hint: A fair dice is tossed twice. ∴ n(S) = 36 A: Getting 4, 5, or 6 on the first toss and Getting 1, 2, 3, or 4 on the second toss. ∴ A = {(4, 1), (4, 2), (4, 3), (4, 4), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)} ∴ P(A) = ∴ Required answer = P(A) : P(A’) = 1 : 2

Question 10

Question 10: The odds against an event are 5 : 3 and the odds in favour of another independent event are 7 : 5. The probability that at least one of the two events will occur is (A) (B) (C) (D)

Accepted Answer

(B)

Question 11

Question 1: The letters of the word ‘EQUATION’ are arranged in a row. Find the probability that (i) all the vowels are together (ii) arrangement starts with a vowel and ends with a consonant.

Accepted Answer

The letters of the word EQUATION can be arranged in 8! ways. ∴ n(S) = 8! There are 5 vowels and 3 consonants. (i) A: all vowels are together we need to arrange (E, U, A, I, O), Q, T, N Let us consider all vowels as one unit. So, there are 4 units, which can be arranged in 4! ways. Also, 5 vowels can

Question 12

Question 2: There are 6 positive and 8 negative numbers. Four numbers are chosen at random, without replacement, and multiplied. Find the probability that the product is a positive number.

Accepted Answer

Let event A: Four positive numbers are chosen, event B: Four negative numbers are chosen, event C: Two positive and two negative numbers are chosen. Since four numbers are chosen without replacement, n(A) = 6 × 5 × 4 × 3 = 360 n(B) = 8 × 7 × 6 × 5 = 1680 In event C, four numbers are to be chosen wit

Question 13

Question 3: Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly, and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is t

Accepted Answer

S = {1, 2,…., 10} ∴ n(S) = 10 A: Number is more than 3. A = {4, 5, 6, 7, 8, 9, 10} ∴ n(A) = 7 ∴ P(A) = B: Number is even. B = {2, 4, 6, 8, 10} ∴ A ∩ B = {4, 6, 8, 10} ∴ n(A ∩ B) = 4 ∴ P(A ∩ B) = Required probability = P(B/A) = = =

Question 14

Question 4: If A, B and C are independent events, P(A ∩ B) = , P(B ∩ C) = , P(C ∩ A) = , then find P(A), P(B) and P(C).

Accepted Answer

Since A and B are independent events, P(A ∩ B) = P(A). P(B) ∴ P(A). P(B) = ……(i) B and C are independent events. ∴ P(B ∩ C) = P(B). P(C) ∴ P(B). P(C) = ……(ii) A and C are independent events. ∴ P(A ∩ C) = P(A). P(C) ∴ P(A). P(C) = ……(iii) Dividing (i) by (ii), we get P(A) = P(C) ……(iv) Substituting e

Question 15

Question 5: If the letters of the word ‘REGULATIONS’ be arranged at random, what is the probability that there will be exactly 4 letters between R and E?

Accepted Answer

There are 11 letters in the word ‘REGULATIONS’ which can be arranged among themselves in 11! ways. ∴ n(S) = 11! Let event A: There will be exactly 4 letters between R and E. R, E can occur at (1, 6), (2, 7), ….,(6, 11) positions. So, there are 6 possibilities. Also, R and E can interchange their pos

Question 16

Question 6: In how many ways can the letters of the word ARRANGEMENTS be arranged? (i) Find the chance that an arrangement chosen at random begins with the letters EE. (ii) Find the probabilit

Accepted Answer

The word ‘ARRANGEMENTS’ has 12 letters in which 2A, 2E, 2N, 2R, G, M, T, S are there. n(S) = Total number of arrangements = (i) A: Arrangement chosen at random begins with the letters EE. If the first and second places are filled with EE, there are 10 letters left in which 2A, 2N, 2R, G, M, T, S are

Question 17

Question 7: A letter is taken at random from the letters of the word ‘ASSISTANT’ and another letter is taken at random from the letters of the word ‘STATISTICS’. Find the probability that the

Accepted Answer

Word ASSISTANT has 2A, I, N, 3S, 2T, and word STATISTICS has A, C, 2I, 3S, 3T. C and N are uncommon letters. In the words ASSISTANT, there are 9 letters out of which 2 letters are ‘A’, and in the word STATISTICS, there are 10 letters, out of which 1 letter is A. ∴ Probability of choosing A from both

Question 18

Question 8: A die is loaded in such a way that the probability of the face with j dots turning up is proportional to j for j = 1, 2,….., 6. What is the probability, in one roil of the die, tha

Accepted Answer

According to the given condition, the probability of the face with 1, 2, 3, 4, 5, 6 dots turning up is proportional to 1, 2, 3, 4, 5, 6. Let k be the common ration of proportionality. ∴ The probabilities of the faces with 1, 2, 3, 4, 5, 6 dots turning up are 1k, 2k, 3k, 4k, 5k, 6k respectively. Sinc

Question 19

Question 9: An urn contains 5 red balls and 2 green balls. A ball is drawn. If it’s green, a red ball is added to the urn, and if it’s red, a green ball is added to the urn. (The original ball

Accepted Answer

A: Event of drawing a red ball and placing a green ball in the urn B: Event of drawing a green ball and placing a red ball C: Event of drawing a red ball in the second draw P(A) = P(B) = P(C/A) = P(C/B) = Required probability P(C) = P(A) P(C/A) + P(B) P(C/B) = =

Question 20

Question 10: The odds against A solving a certain problem are 4 to 3 and the odds in favour of B solving the same problem are 7 to 5, find the probability that the problem will be solved.

Accepted Answer

The odds against A solving the problems are 4 : 3. Let P(A’) = P(A does not solve the problem) = So, the probability that A solves the problem = P(A) = 1 – P(A’) = 1 – = Similarly, let P(B) = P(B solves the problem) Since odds in favour of B solving the problem are 7 : 5. ∴ P(B) = Required probabili

Question 21

Question 11: If P(A) = P(A/B) = , P(B/A) = , then find (i) P(A’/B) (ii) P(B’/A’)

Accepted Answer

Since P(A) = P(A/B) = P(A) = and ∴ P(A) = ……(i) P(B) = 5 P(A ∩ B) ……..(ii) Since P(B/A) = ∴ P(A) = 3 P(A ∩ B) ………(iii)

Question 22

Question 12: Let A and B be independent events with P(A) = and P(A ∪ B) = 2P(B) – P(A). Find (i) P(B) (ii) P(A/B) (iii) P(B’/A)

Accepted Answer

A and B are independent events.. ∴ P(A ∩ B) = P(A) × P(B) (i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ∴ P(A ∪ B) = P(A) + P(B) – P(A) × P(B) ∴ 2P(B) – P(A) = P(A) + P(B) – P(A) × P(B) ……[∵ P(A ∪ B) = 2P(B) – P(A)] ∴ 2P(B) – = + P(B) – × P(B) ∴ 2P(B) – P(B) + P(B) = +

Question 23

Question 13: Find the probability that a year selected will have 53 Wednesdays.

Accepted Answer

A leap year comes after 3 years. ∴ The probability of a year being a leap year = ∴ Probability of a year being a non-leap year = 1 – = In a non-leap year, there are 52 weeks and one extra day, whereas a leap year has 52 weeks and 2 extra days. ∴ 53rd Wednesday’s chance in a non-leap year = Two extra

Question 24

Question 14: The chances of P, Q and R, getting selected as principal of a college are , , respectively. Their chances of introducing IT in the college are , , respectively. Find the probabilit

Accepted Answer

Let event P: P become principal, event Q: Q become principal, event R: R become principal, event E: Subject IT is introduced. Given, P(P) = P(Q) = P(R) = P(E/P) = P(E/Q) = P(E/R) = (a) Required probability P(E) = P(P) P(E/P) + P(Q) P(E/Q) + P(R) P(E/R) = = = =

Question 25

Question 15: Suppose that five good fuses and two defective ones have been mixed up. To find the defective fuses, we test them one-by-one, at random and without replacement What is the probabil

Accepted Answer

Number of fuses = 5 + 2 = 7 Testing two fuses one-by-one at random, without replacement from 7 can be done in ways. ∴ n(S) = = 7 × 6 = 42 Let event A: Getting defective fuses in the first two tests without replacement. There are two defective fuses. ∴ n(A) = = 2 × 1 = 2 ∴ P(A) =

Question 26

Question 16: For three events A, B and C, we know that A and C are independent, B and C are independent, A and B are disjoint, P(A ∪ C) = , P(B ∪ C) = , P(A ∪ B ∪ C) = . Find P(A), P(B) and P(C

Accepted Answer

Let P(A) = x, P(B) = y, P(C) = z Since A, B are disjoint, A ∩ B = Φ and A ∩ B ∩ C = Φ ∴ P(A ∩ B) = 0, P(A ∩ B ∩ C) = 0 ……(i) Since A and C are independent, P(A ∩ C) = P(A) P(C) = xz Since B and C are independent, P(B ∩ C) = P(B) P(C) = yz P(A ∪ C) = P(A) + P(C) – P(A ∩ C) ∴ = x + z – xz ……..(ii) P(B

Question 27

Question 17: The ratio of boys to girls in a college is 3 : 2 and 3 girls out of 500 and 2 boys out of 50 of that college are good singers. A good singer is chosen. What is the probability that

Accepted Answer

Let event S: The student is a good singer, event B: The student is a boy, event G: The student is a girl. Since the ratio of boys to girls is 3 : 2 and 3 girls out of 500 and 2 boys out of 50 are good singers.

Question 28

Question 18: A and B throw a die alternatively till one of them gets a 3 and wins the game. Find the respective probabilities of winning. (Assuming A begins the game).

Accepted Answer

Since P(getting 3) = , P(not getting 3) = 1 – = In 1st throw if A gets 3, A wins ∴ P(A win) = In 2nd throw by B (i.e., A does not get 3), ∴ P(B wins) = In 3rd throw by A, P(A wins) = (3rd throw by A shows that B has lost in 2nd throw) and so on.

Question 29

Question 19: Consider independent trials consisting of rolling a pair of fair dice, over and over. What is the probability that a sum of 5 appears before a sum of 7?

Accepted Answer

When two dice are thrown, the sample space is S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (

Question 30

Question 20: A machine produces parts that are either good (90%), slightly defective (2%), or obviously defective (8%). Produced parts get passed through an automatic inspection machine, which

Accepted Answer

Let event G: The event that machine produces a good part, event S: The event that machine produces a slightly defective part, event D: The event that machine produces an obviously defective part.

Chapter 9 Probability Miscellaneous Exercise 9 Solutions