Question 1

Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.

Accepted Answer

The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl. So the required probability: = 8 17 * 9 16 + 9 17 * 8 16 + 8 17 * 7 16 = 9 34 + 9 34 + 7 34 = 25 34

Question 2

A bag contains 5 red smileys, 6 yellow smileys and 3 green smileys. If two smileys are picked at random, what is the probability that both are red or both are green in colour?

Accepted Answer

Given total number of smileys = 5 + 6 + 3 = 14 Now, required probability = 5 C 2 14 C 2 + 3 C 2 14 C 2 = 10 91 + 3 91 = 13 91 = 1 7

Question 3

If the difference between the mode and median is 2, then the difference between the median and mean is:

Accepted Answer

1

Question 4

A coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed. If 8 possible outcomes are equally likely. Find the probability that the die shows a number greater than 4, if it is known that the first t…

Accepted Answer

Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 } Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then, A = { T5, T6 } B = { T1, T2, T3, T4, T5, T6 } Therefore, Required probability = P A B = P A ∩ B P B = 2 8 6 8 = 1 3

Question 5

A room contains 3 brown, 5 black and 4 white chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is white ?

Accepted Answer

Total number of chairs = (3 + 5 + 4) = 12. Let S be the sample space. Then, n(s)= Number of ways of picking 2 chairs out of 12 = 12×11/2×1 = 66 Let n(E) = number of events of selecting 2 chairs for selecting no white chairs. => 8C2 = 8×7/2×1 = 28 Therefore required probability = 28/66 = 14/33.

Question 6

From a pack of 52 cards, 3 cards are drawn. What is the probability that one is ace, one is queen and one is jack?

Accepted Answer

Required probability: 4 C 1 * 4 C 1 * 4 C 1 52 C 3 = 4 * 4 * 4 22100 = 16 5525

Question 7

fifteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together ?

Accepted Answer

In a circle of n different persons, the total number of arrangements possible = (n - 1)! Total number of arrangements = n(S) = (15 – 1)! = 14 ! Taking three persons as a unit, total persons = 13 (in 4 units) Therefore no. of ways for these 13 persons to around the circular table = (13 - 1)! = 12! In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit = n(E) = 12! X 3! Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14! = 12!x3x2 / 14x13x12! = 6/14x13 = 3/91

Question 8

The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?

Accepted Answer

The 8 letters can be written in 8! ways. n(S) = 8! Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down. Now the letters a, (bcde), f, g and h can be arranged in 5! ways. The letters b,c,d and e can be arranged themselves in 4! ways. n(E) = 5! x 4! Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14 Hence the answer is 1/14.

Question 9

In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?

Accepted Answer

We know that, Total probability = 1 Given probability of black stones = 1/4 => Probability of blue and white stones = 1 - 1/4 = 3/4 But, given blue + white stones = 9 + 6 = 15 Hence, 3/4 ----- 15 1 ----- ? => 15 x 4/3 = 20. Hence, total number of stones in the box = 20.

Question 10

In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin?

Accepted Answer

Total coins 30 In that, 1 rupee coins 20 50 paise coins 10 Probability of total 1 rupee coins = 20C11 Probability that 11 coins are picked = 30C11 Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.

Question 11

The probability of success of three students X, Y and Z in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.

Accepted Answer

P(X) = 1 5 , P(Y) = 1 4 , P(Z) = 1 3 Required probability: = [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C) = 1 4 * 1 3 * 4 5 + 3 4 * 1 3 * 1 5 + 2 3 * 1 4 * 1 5 + 1 4 * 1 3 * 1 5 = 4 60 + 3 60 + 2 60 + 1 60 = 10 60 = 1 6

Question 12

I forgot the last digit of a 7-digit telephone number. If 1 randomly dial the final 3 digits after correctly dialing the first four, then what is the chance of dialing the correct number?

Accepted Answer

It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways. Hence required probability = 1 10 3 = 1/1000

Question 13

Out of 3 girls and 6 boys a group of three members is to be formed in such a way that at least one member is a girl. In how many different ways can it be done?

Accepted Answer

Total number of possible ways = 9C3 = 84 ways Required atleast one girl in the group of three = total possible ways - ways in which none is girl None of the members in the group is girl = 6C3 = 20 Therefore, number of ways that at least one member is a girl in the group of three = 84 - 20 = 64 ways.

Question 14

In a single throw with 2 dices, what is probability of neither getting an even number on one and nor a multiple of 3 on other?

Accepted Answer

We first calculate the probability of getting an even number on one and a multiple of 3 on other,Here, n(s) = 6x6 = 36 and E = (2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4)(3,6) (6,2)(6,4) n(E) = 11P(E) = 11/36Required probability = 1-11/36 = 25/36

Question 15

If x is chosen at random from the set {1,2,3,4} and y is to be chosen at random from the set {5,6,7}, what is the probability that xy will be even?

Accepted Answer

S ={(1,5),(1,6),(1,7),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7)} Total element n(S)=12 xy will be even when even x or y or both will be even. Events of x, y being even is E. E ={(1,6),(2,5),(2,6),(2,7),(3,6),(4,5),(4,6),(4,7)} n(E) = 8 P(E)=n(E)n(S)=812 = 4/3

Question 16

The probability that a number selected at random from the first 100 natural numbers is a composite number is ?

Accepted Answer

The number of exhaustive events = 100 C₁ = 100. We have 25 primes from 1 to 100. Number of favourable cases are 75. Required probability = 75/50 = 3/2.

Question 17

The probability that a bowler bowled a ball from a point will hit by the batsman is ¼. Three such balls are bowled simultaneously towards the batsman from that very point. What is the probability that the batsman will hit the ball ?

Accepted Answer

37/64

Question 18

If an unbiased dice is rolled once, the odds in favour of getting a point which is multiple of 3 is:

Accepted Answer

Total number =6 Getting a 'multiple of 3' = 2 . So, probability = 2/6 =1/3

Question 19

Three dice are thrown together.Find the probability of getting a total of atleast 6.

Accepted Answer

Total number of events=6 x 6 x 6=216 Let A be the event of getting a total of atleast 6 and B denoted event of getting a total of less than 6 i.e.,3,4,5. So,B={(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)} Favourable number of cases=10 Therefore,P(B)=10/216 => P(A) = 1 - P(B) = 1 - (10/216) = 103/108

Question 20

In a box, there are four marbles of white color and five marbles of black color. Two marbles are chosen randomly. What is the probability that both are of the same color?

Accepted Answer

Number of white marbles = 4 Number of Black marbles = 5 Total number of marbles = 9 Number of ways, two marbles picked randomly = 9C2 Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2 = 1/6 + 5/18 = 4/9.

Probability Questions