PROBABILITY PRACTICE QUESTIONS : 91-100

Question 91.) A retail store gives 3 reward coupons in a month to its regular customers. All customers use all the 3 reward coupons but two customers A and B do not use all of them. Both the customers A and B use only 1 out of 3 reward coupons. If the store gives an exclusive reward coupon to all its customers and it is known that among the customers A and B at least one of will use the reward coupon then find the probability that both of them use the reward coupon.

(A) `\frac{1}{5}`

(B) `\frac{1}{9}`

(C) `\frac{5}{9}`

(D) `\frac{3}{5}`

Answer. (A)

Solution.

Since, it is given that both the customers A and B use only 1 reward coupon out of 3.

So the probability of using a reward coupon = `\frac{1}{3}`

And the probability of not using a reward coupon = 1 – `\frac{1}{3}` = `\frac{2}{3}`

Now it is not a compulsory event that both A and B use the same coupon at a time. They may use the same coupon or different coupon. But the condition given in the question is that both of them must use the same coupon.

So now we will find the probability that any one of them uses the reward coupon and it is given by = `\frac{1}{3}` × `\frac{2}{3}` + `\frac{2}{3}` × `\frac{1}{3}` = `\frac{4}{9}`

Now the probability that both of them use the coupon = `\frac{1}{3}` × `\frac{1}{3}` = `\frac{1}{9}`

So the probability that at least one of them uses the reward coupon = `\frac{4}{9}` + `\frac{1}{9}` = `\frac{5}{9}`

Since we need to find the probability that both of them use the reward coupon when it is given that at least one of them use the reward coupon.

Hence the required probability = (Probability that both of them use the coupon) / (Probability that at least one of them uses the reward coupon) = `\frac{1}{5}`

Question 92.) A person invests in stocks of 3 different companies. If there probability of making profit are `\frac{2}{5}`, `\frac{1}{3}`, and `\frac{5}{7}` respectively. Then find the probability that at least two of them give the same result.

(A) `\frac{22}{105}`

(B) `\frac{83}{105}`

(C) 1

(D) `\frac{41}{105}`

Answer. (C)

Solution.

Let the three companies are A, B, and C. Then their probabilities of making profit are P(A), P(B), and P(C) and their probabilities of making loss are P(A̅), P(B̅), and P(C̅). Now it is given that,

P(A) = `\frac{2}{5}`

P(B) = `\frac{1}{3}`

P(C) = `\frac{5}{7}`

Now we will calculate P(A̅), P(B̅), and P(C̅)

P(A̅) = 1 – `\frac{2}{5}` = `\frac{3}{5}`

P(B̅) = 1 – `\frac{1}{3}` = `\frac{2}{3}`

P(C̅) = 1 – `\frac{5}{7}` = `\frac{2}{7}`

Now as per the condition given in the question we need to find the probability that at least two of the three companies give the same result. It means that either both of them make a profit or both of them bear a loss. So here we will have two different cases because the condition of at least is present here,

Case 1, when only two companies give the same result and the third company gives result opposite to them. It means if two companies are making profit then the third company will bear a loss and vice-versa. The probability for this case is,

P(A) × P(B) × P(C̅) + P(A̅) × P(B̅) × P(C) + P(A̅) × P(B) × P(C) + P(A) × P(B̅) × P(C̅) + P(A) × P(B̅) × P(C) + P(A̅) × P(B) × P(C̅)

= `\frac{2}{5}` × `\frac{1}{3}` × `\frac{2}{7}` + `\frac{3}{5}` × `\frac{2}{3}` × `\frac{5}{7}` + `\frac{3}{5}` × `\frac{1}{3}` × `\frac{5}{7}` + `\frac{2}{5}` × `\frac{2}{3}` × `\frac{2}{7}` + `\frac{2}{5}` × `\frac{2}{3}` × `\frac{5}{7}` + `\frac{3}{5}` × `\frac{1}{3}` × `\frac{2}{7}` = `\frac{83}{105}`

Case 2, when all the three companies give the same result. It means either all companies will make profit or all companies will bear loss. The probability for this case is,

P(A) × P(B) × P(C) + P(A̅) × P(B̅) × P(C̅) = `\frac{2}{5}` × `\frac{1}{3}` × `\frac{5}{7}` + `\frac{3}{5}` × `\frac{2}{3}` × `\frac{2}{7}` = `\frac{22}{105}`

Hence the required probability = `\frac{83}{105}` + `\frac{22}{105}` = `\frac{105}{105}` = 1

Question 93.) A company has two retail stores in a city: Store 1 and Store 2. It is given that 55% of its total customers shop at store 1, 45% of its total customers shop at store 2, and 20% of its total customers shop at both stores. The company launches a scratch card scheme for its customers to increase sales. If every customer gets a scratch card and there is only one prize-based scratch card, then find the probability that the winner is from store 2 but shops at both stores.

(A) `\frac{4}{9}`

(B) `\frac{4}{11}`

(C) `\frac{9}{11}`

(D) `\frac{5}{9}`

Answer. (A)

Solution.

Since it is given that the prize winner is from store 2 which means that the winner has received the scratch card from store 2. So it is clear that the winner do the shopping from the store 2.

So, total number of events will be the total number of customers doing shopping from store 2 and that is = 45%

But there is a condition given in the question that the winner does shopping from both the stores. So, the favorable events will be the number of customers doing shopping from both the stores and that is = 20%

Hence the required probability = `\frac{20%}{45%}` = `\frac{4}{9}`

Question 94.) If a number is chosen at random from a set of all 3 digit numbers then find the probability that the sum of the digits of the chosen 3 digit number is even.

(A) `\frac{1}{3}`

(B) `\frac{1}{2}`

(C) `\frac{1}{6}`

(D) `\frac{1}{9}`

Answer. (B)

Solution.

Total number of events = Total number of 3 digit numbers = 900

As we know that all 3 digit numbers are formed from the digits of the following set,

{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Now we have two types of numbers, first those which have a zero as a digit and second those which do not have a zero as a digit. So here we will have following three cases to satisfy the condition given in the question.

Case 1, when there are 2 zero digits in the number, well in this case the 3^rd digit should be even to make the sum of the digits even and there are only 4 three digit numbers having an even digit and 2 zero digits; 200, 400, 600, and 800.

So number of events for this case = 4

Case 2, when there is only 1 zero digit in the number, well in this case the remaining two digits should either be even or be odd to make the sum of the digits even because the sum of two odd numbers and two even numbers is always even. So, the combination of 4 even digits and 5 odd digits for this case is as following,

Combination for 2 even digits with one zero digit = 4 × 4 × 1 × 2 = 32

And the combination for 2 odd digits with one zero digit = 5 × 5 × 1 × 2 = 50

So number of events for this case = 32 + 50 = 82

Case 3, when there is no any zero digit in the number, well in this case among the three digits either two are odd and one is even or all the three digits are even to make the sum of the digits even because the sum of two odd numbers and even numbers is always even. So, the combination of 4 even digits and 5 odd digits for this case is as following,

Combination for 2 odd digits with one even digit = 5 × 5 × 4 × 3 = 300

And the combination for 3 even digits = 4 × 4 × 4 = 64

So number of events for this case = 300 + 64 = 364

Now total number of favorable events = 32 + 50 + 364 = 450

Hence the required probability = `\frac{450}{900}` = `\frac{1}{2}`

Question 95.) 5 boys and 3 girls participate in different games organized by a school. If they get a rank from 1 to 8 after completion of each game then find the probability that all the 3 girls always get ranks together in a consecutive manner.

(A) `\frac{1}{8}`

(B) `\frac{9}{28}`

(C) `\frac{3}{14}`

(D) `\frac{3}{28}`

Answer. (D)

Solution.

Since, it is given in the question that there are 5 boys and 3 girls participating in games.

So, total number of participants = 5+3 = 8

Now 8 participants can get ranks from 1 to 8 in 8  ways.

So total number of events = 8！

Now the condition given in the question that all the 3 girls always get ranks together in a consecutive manner means that all the three girls will always get ranks together like 1^st, 2^nd, 3^rd or 5^th, 6^th, 7^th, etc. So, we can consider it as 1 group of fixed 3 members and in such a case we have total 6 participants to rank.

So favorable events = Number of ways to arrange the participants = 6！× 3！

3！have been multiplied because the 3 girls will be arranged in a total 3！number of ways.

Hence the required probability = `\frac{3}{28}`

Question 96.) A retail store releases two types of coupons to increase its sales. One coupon is of 5% discount and other is of 10% discount. The number of coupons of 5% discount is double the number of coupons of 10% discount. Every customer gets only one coupon per month after shopping. The customers can redeem the coupons only when they have exactly 6 coupons. If a regular customer X collects all the 6 coupons and asks the store to redeem them then find the probability that he gets a discount more than 40%.

(A) `\frac{10}{243}`

(B) `\frac{25}{243}`

(C) `\frac{5}{243}`

(D) `\frac{50}{243}`

Answer. (C)

Solution.

Let x is the number of coupons of 10% discount then as per the question 2x is the number of coupons of 5% discount.

So total number of coupons = x + 2x = 3x

So the probability of getting a coupon of 10% discount by a customer = `\frac{x}{3x}` = `\frac{1}{3}`

And the probability of getting a coupon of 5% discount by a customer = `\frac{2x}{3x}` = `\frac{2}{3}`

Now the minimum discount a customer can redeem = 6 × 5% = 30%

And the maximum discount a customer can redeem = 6 × 10% = 60%

But we need to find the probability that the customer has got a discount more than 40%; i.e., either 45% or 50 % or 55% or 60%.

So here we have 4 different cases,

Case 1, the customer gets a discount of 45%, it is possible only when 3 coupons are of 5% discount and rest 3 are of 10% discount.

Probability for this case = `\frac{2}{3}` × `\frac{2}{3}` × `\frac{2}{3}` × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` = `\frac{8}{729}`

Case 2, the customer gets a discount of 50%, it is possible only when 2 coupons are of 5% discount and rest 4 are of 10% discount.

Probability for this case = `\frac{2}{3}` × `\frac{2}{3}` × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` = `\frac{4}{729}`

Case 3, the customer gets a discount of 55%, it is possible only when 1 coupon is of 5% discount and rest 5 are of 10% discount.

Probability for this case = `\frac{2}{3}` × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` = `\frac{2}{729}`

Case 4, the customer gets a discount of 60%, it is possible only when all 6 coupons are of 10% discount.

Probability for this case = `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` = `\frac{1}{729}`

Hence required probability = `\frac{8}{729}` + `\frac{4}{729}` + `\frac{2}{729}` + `\frac{1}{729}` = `\frac{15}{729}` = `\frac{5}{243}`

Question 97.) Suppose a man applies for the job of a manager in two different companies, X and Y. If the probability that he gets the job offer from only one of the companies X and Y is `\frac{8}{21}` and the number of applicants for the job of company X is 7 then find the number of applicants for the job of company Y.

(A) 6

(B) 3

(C) 9

(D) 7

Answer. (B)

Solution.

Let ‘a’ be the number of applicants for the job of the company Y, and P(X) and P(Y) be the probabilities that the man gets the job offer from company X and company Y.

Now it is given that there are total 7 applicants for the job of company X and since they need only one manager.

So probability of getting a job offer from company X by the applicants = P(X) = `\frac{1}{7}`

And the probability of not getting a job offer from company X by the applicants will be,

P(X̅) = 1 – P(X) = 1 – `\frac{1}{7}` = `\frac{6}{7}`

Similarly for the company Y,

The probability of getting a job offer from company Y by the applicants = P(Y) = `\frac{1}{a}`

And the probability of not getting a job offer from company X by the applicants will be,

P(Y̅) = 1 – P(Y) = 1 – `\frac{1}{a}`

Now it is given that the man gets job offer from any one of the company. It means that either he will get the job offer from company X or from company Y but not from both.

Hence the required probability = P(X) × P(Y̅) + P(X̅) × P(Y)

Now it is given that the required probability = `\frac{8}{21}`

Hence, P(X) × P(Y̅) + P(X̅) × P(Y) = `\frac{8}{21}`

Now putting values and further solving we get, a = 3

Question 98.) 8 people including 4 boys and 4 girls are playing a chess tournament. The tournament includes 3 rounds and 7 matches. 1^st round is elimination round then 2^nd round is semi final round and finally the 3^rd round which is the final round. If 4 players get eliminated in 1^st round, 2 more players get eliminated in the 2^nd round and finally the winner of the 3^rd round is the winner of tournament then find the probability that the two finalists are of same gender. Given that the probability of winning or losing a game for players of both genders is same. (No match results in a draw.)

(A) `\frac{7}{16}`

(B) `\frac{7}{32}`

(C) `\frac{7}{64}`

(D) `\frac{7}{48}`

Answer. (B)

Solution.

Since two players will play one game and the probability of their winning is same.

So the probability of winning a game by each player = `\frac{1}{2}`

And the probability of losing a game by each player = 1 – `\frac{1}{2}` = `\frac{1}{2}`

Now the condition given here is that the two finalists should be of same gender will be satisfied only when both the finalists are either male or female and it is possible only when at least two males or two females win the matches in the first round and also win the semi final round. So here we will consider the case of two female finalists and for that we will find the probability that at least two females win the 1^st elimination round and also win the semi final round and it is given by,

`\frac{1}{2}` × `\frac{1}{2}` + `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` + `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` = `\frac{7}{16}`

`\frac{1}{2}` × `\frac{1}{2}` = `\frac{1}{4}`

`\frac{7}{16}` × `\frac{1}{4}` = `\frac{7}{64}`

We will get the same probability even if we consider the case of two male finalists. But as per the condition the two finalists will be either males or females.

Hence the required probability = `\frac{7}{64}` × 2 = `\frac{7}{32}`

Question 99.) If a shooter gets 5 chances to hit a target and the probability of not hitting the target at least one time is greater than or equal to `\frac{31}{32}` then find the maximum value of the probability of hitting the target in every chance by the shooter.

(A) `\frac{1}{2}`

(B) `\frac{15}{32}`

(C) `\frac{7}{32}`

(D) `\frac{1}{8}`

Answer. (A)

Solution.

Let P is the probability of no failure.

Hitting the target in every chance means event of no failure. This is the case of Bernoulli’s distribution. Here we will use the following relation,

1 – `P^{5}` ≤≥ Probability of at least one failure

Where n is the number of times of happening of the event.

Let, P is the probability of no failure in all 5 chances and the probability of at least one failure = `\frac{31}{32}`

Now putting these values in above relation we get,

1 – `P^{5}` ≥ `\frac{31}{32}`

1 – `\frac{31}{32}` ≥ `P^{5}`

`P^{5}` ≤ `\frac{1}{32}`

P ≤ `\frac{1}{2}`

Hence the maximum value of the probability of hitting the target by the shooter = `\frac{1}{2}`

Question 100.) 10 people appear for an exam and seven of them pass the exam in 1^st attempt. Now the rest 3 people get 2 more attempts to pass the exam with a condition that they will be declared pass only when all 3 of them pass the exam. If they pass the exam in the 2^nd attempt of the 2 attempts given to them then find the probability that one person among them passes the exam in both attempts.

(A) `\frac{1}{2}`

(B) `\frac{1}{4}`

(C) 1

(D) `\frac{3}{4}`

Answer. (D)

Solution.

It is given that the 3 people will pass the exam only when all of them pass the exam. It means that if only one among them gets failed then all 3 will be considered as fail. So when all three will pass the exam together then they will be declared as pass.

Now there are two attempts and the people pass the exam only in one attempt.

So the probability of passing the exam = `\frac{1}{2}`

And the probability of not passing the exam = 1 – `\frac{1}{2}` = `\frac{1}{2}`

But one person passes the exam in both attempts.

So probability of passing the exam for this person = 1

Hence the required probability = `\frac{1}{2}` × `\frac{1}{2}` × 1 × 3 = `\frac{3}{4}`