Question 1

If the probability for A fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is?

Accepted Answer

∵ P(A) = 0.2 ∴ P(A) = 1 - 0.0 = 0.8 And P (B) = 0.3 ∴ P(B) = 1 - 0.3 = 0.7 Required probability = P(A ∪ B) = P(A ∩ B) = 1 - P (A ∩ B ) = 1 - P(A) P(B) = 1 - (0.8) (0.7) = 1-0.56 = 0.44

Question 2

A fair coin is tossed 100 times . The probability of getting tails an odd number of times is?

Accepted Answer

The total number of cause is 2100. The number of favourable cases are 100C1 + 100C3 + ..... + 100C99 = 2100 - 1 = 299 ∴ Reqd probability = 299/ 2100 = 1/2

Question 3

A natural number is selected at random from the set x = { x : 1 ≤ x ≤ 100 }. The probability that the number satisfies the inequality x 2 - 13x ≤ 30 is?

Accepted Answer

x2 - 13 x - 30 ≤ 0 ⇒ (x + 2) (x - 15) ≤ 0 ⇒ -2 ≤ x ≤ 15 But x is a natural number. ∴ 1 ≤ x ≤ 15 ∴ Reqd. Probability P = 15/100 = 3/20

Question 4

The probability that 9 students pass in Mathematics, Physics and Chemistry are m, p and c respectively, of these subject the students has a 75% chance of passing in at least one subject 50% chance of passing in at least two subject and 40% chance of…

Accepted Answer

P(M) = m, P (p) = p, P(c) = c ∵ The probability of at least one success = P (M ∪ P ∪ C) = m + p + c -mp - mc - pc + mcp = 3/4 ...(1) The probability of at least two successes = mcp + mcp + mcp + mcp = mc(1 - p) + mp (1 - c ) + (1 - m )cp + mcp = mc + mp + cp - 2mcp = 1/2 The probability of exactly two success = mcp + mcp + mcp = mc(1 - p) + mp (1 - c ) cp(1 - m ) = mc + mp + cp - 3 mcp = 2/5 (2) & (3) gives, ⇒ mcp = 1/2 - 2/5 = 1/10 ∴ mc + mp + cp = 2/10 + 1/2 = 1/5 + 1/2 = 7/10 From (1), m + p + c - 7/10 + 1/10 = 3/4 ⇒ m + p + c = 3/4 + 7/10 - 1/10 = 27/20 Thus, pmc = 1/10 is a true relation.

Question 5

If the integer m and n are chosen at random between 1 and 100, then the probability that a number of the from 7 m + 7 n is divisible by 5 equals?

Accepted Answer

Number of possible cases = 100 x 100 for 7m + 7n to be divisible by 5, One of the term has to end with 9 and other with 1 (7m cannot be divided by 9 ) ⇒ m can be 2, 6 , 10, 14, ..... 98 (25 values) and n can be 4, 8, 12 ...... 100(25 values) Since m and n can interchange ∴ Required probability = (2 x 25 x 25) / (100 x 100) = 1/8

Question 6

If n integer taken at random are multiplied together then the probability that the last digit of the product is 2, 4, 6, 8 is?

Accepted Answer

If the last digit in the product is to 2, 4, 6, 8 the last digit in all the n number should not be 0 and 5 and the last digit of all number should not be selected exclusively from the set of number {1, 3, 7, 9} ∴ Favourable number of cases = 8n - 4n But generally the last digit can be one of 0, 1, 2, 3, .... 9. Hence, the total number of ways = 10n Hence, the required probability = 8n - 4n / 10n = 4n - 2n / 5n

Question 7

Two dice and two coins are tossed. The probability that both the coins show head and the sum of the numbers found on the two dice is a prime number is?

Accepted Answer

The probability that head is show in one coin is 1/2. The probability that the sum of the number on the dice is a prime = the probability that the following pair of number on the dice is a getting on the dice, namely (1, 1), (1, 2), (2, 1), (1, 4 ), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5), (5, 6) = 15/36. ∴ The required probability = 1/2 x 1/2 x 15/36 = 5/48.

Question 8

A speaks truth in 60% of the cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other, narrating the same incident?

Accepted Answer

Let A = Event that A speaks the truth and B = Event that B speaks the truth Then, P(A) = 60/100 = 3/5 and P(A) = 80/100 = 4/5 ∴ P(A) = (1 - 3/5) = 2/5 and P(B) = (1 - 4/5) = 1/5 P(A and B contradict each other) = P [(A speaks the truth and B tells a lie) or (A tells a lie and B speak the truth)] =P[(A and B) or (A and B)] =P(A and B) + P (A and B) = P(A) x P(B) + P(A) x P(B) = (3/5 x 1/5) + ( 2/5 x 4/5) = (3/25 + 8/25 ) = 11/25 = (11/25 x 100)% = 44%

Question 9

The probability that a leap year selected at random contains 53 Sundays, is

Accepted Answer

In a leap year, there are 366 days. It means 52 full weeks + 2 odd days. These two days can be (Mon, Tues), (Tues, Wed), (Wed, Thurs), (Thurs, Fri), (Fri, Sat), (Sat, Sun) or (Sun, Mon), So, Required probability = 2/7

Question 10

From 4 children, 2 women and 4 men, 4 person are selected. The probability that there are exactly 2 children among the selected persons, is

Accepted Answer

Total number of cases = 10c4 Favorable number of cases = 4c2. 6c2 {Since, we are to select 2 children out of 4 and remaining 2 persons are to be selected from remaining 6 persons ( 2W + 4M)} ∴ required Probability = 4c2 . 6c2 / 10c4 = [{(4 x 3) / (2 x 1)} x {(6 x 5) / (2 x 1)}] / [ (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)] = [(12/2) x (30/2)] / 210 = 90 / 210 = 9 /12

Question 11

The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target atleast three times, is

Accepted Answer

Required Probability Given, n = 5 and r = 3 Then, Success P = 3/4 Failure, q = 1 - 3/4 = 1/4 Man hit the target thrice = 5c3 (3/4)3 (1/4)2 + 5c4 (3/4)3 (1/4) + 5c3(3/4)5 = (270/1024) + (405/1024) + (243/1024) = 918/1024 = 459/512

Question 12

A dice is rolled three times and sum of three numbers appearing on the uppermost face is 15. The chance that the first roll was four is

Accepted Answer

Total number of favourable outcomes n(S) = 63 = 216 Combinations of outcomes for getting sum of 15 on uppermost face = (4, 5, 6), (5, 4, 6), ( 6, 5, 4), (5, 6, 4), (4, 6, 5), (6, 4, 5), (5, 5, 5),(6, 6, 3), (6, 3, 6), (3, 6, 6) Now, outcomes on which first roll was a four, n(E) = (4, 5, 6),(4, 6, 5) ∴ P(E) = n(E)/n(S) = 2/216 =1/108

Question 13

If a 4- digit number is formed at random using the digit 1, 3, 5, 7 and 9 without repetition, then the probability that it is divisible by 5, is

Accepted Answer

Number of 4- digit numbers which are formed with 1, 3, 5, 7, 9 = 5P4 = 5 x 4 x 3 x 2 = 120 = n (s) Number of 4 digit number which are formed with 1, 3, 5, 7, 9 and are divisible by 5 = 4P3 = 4 x 3 x 2 = 24 = n (E) ∴ p(E) = n(E)/n( s) = 24/120 = 1/5

Question 14

If three marbles are picked at random, what is the probability that at least one is blue?

Accepted Answer

Total number of possible outcomes = 12C3 = 220 Number of events which do not contain blue marbles (3 marbles out of 7 marbles) = 7C3 = 35 ∴ Required probability = 1 - 35/220 = 37/44

Question 15

The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ?

Accepted Answer

Vowels are A I A I O, C A S T I G A T I O N (O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O) So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows : n ( E ) = 5 P 5 2 ! * 2 ! ( A , I a r e 2 t i m e s ) * 6 P 6 2 ! ( T i s 2 t i m e s ) = 21600 n ( S ) = 11 ! 2 ! * 2 ! * 2 ! ( A , I a r e 2 t i m e s ) = 4989600 21600 4989600 = 0 . 043

Question 16

In a throw of a coin, find the probability of getting a head.

Accepted Answer

Here, s={H,T} and E={H} P(E) = n(E)/n(S) = 1/2

Question 17

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

Accepted Answer

P(odd) = P (even) = 1 2 1(because there are 50 odd and 50 even numbers) Sum or the three numbers can be odd only under the following 4 scenarios: Odd + Odd + Odd = 1 2 * 1 2 * 1 2 = 1 8 Odd + Even + Even = 1 2 * 1 2 * 1 2 = 1 8 Even + Odd + Even = 1 2 * 1 2 * 1 2 = 1 8 Even + Even + Odd = 1 2 * 1 2 * 1 2 = 1 8 Other combinations of odd and even will give even numbers. Adding up the 4 scenarios above: = 1 8 + 1 8 + 1 8 + 1 8 = 4 8 = 1 2

Question 18

A number is selected at random from the numbers 1 to 30. What is the probability that it is divisible by either 3 or 7 ?

Accepted Answer

1/30

Question 19

A fair coin is tossed 11 times. What is the probability that only the first two tosses will yield heads ?

Accepted Answer

Probability of occurrence of an event, P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S) ⇒ Probability of getting head in one coin = ½, ⇒ Probability of not getting head in one coin = 1- ½ = ½, Hence, All the 11 tosses are independent of each other. ∴ Required probability of getting only 2 times heads = 1 2 2 × 1 2 9 = 1 2 11

Question 20

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn find the probability the card drawn is red.

Accepted Answer

We know that: When one card is drawn from a pack of 52 cards The numbers of possible outcomes n(s) = 52 We know that there are 26 red cards in the pack of 52 cards ⇒ The numbers of favorable outcomes n(E) = 26 Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S) ∴ required probability = 26/52 = 1/2.

Probability Questions