PROBABILITY PRACTICE QUESTIONS : 81-90

Question 81.) In a company there are 40 workers, among which 14 people work in accounts department while 26 people work in the sales department. Among the 14 workers of the accounts department 9 are male and 5 are female. Similarly among the 26 workers of sales department 15 are male and 11 are female. If a worker is selected at random then find the probability that the selected worker is a male worker working in accounts department.

(A) `\frac{9}{40}`

(B) `\frac{7}{20}`

(D) `\frac{13}{20}`

Answer. (C)

Solution.

Total number of workers working in the company = 40

Total number of male workers working in the company = 9 + 15 = 24

Now the probability of selecting a male worker at random, P1 = `\frac{24}{40}`

Now there is one more condition that the randomly selected worker should be working in the accounts department.

So, Total number of male workers working in accounts department of the company = 9

And the probability that the randomly selected male worker should be working in accounts department, P2 = `\frac{9}{40}`

Hence the required probability, P = `\frac{P2}{P1}` = `\frac{3}{8}`

Question 82.) A helmet making company has two factories F1 and F2. 25% of the helmets made by the company are produced in factory F1 and rest 75% of the helmets made by the company are produced in factory F2. 6% of all the helmets made by the company turn out to be defective. It is given that the probability that a helmet produced in factory F1 turns out to be defective is 4 times of the probability that a helmet produced in factory F2 turns out to be defective. If a helmet is chosen at random and not found defective then find the probability that it is produced in the factory F2.

(A) `\frac{507}{700}`

(B) `\frac{169}{334}`

(D) `\frac{94}{169}`

Answer. (C)

Solution.

Let the probability that a helmet produced in factory F1 turns out to be defective is P1 and the probability that a helmet produced in factory F2 turns out to be defective is P2. Then as per given in the question,

P1 = 4 × P2

Now it is given in the question that 25% of the helmets made by the company are produced in factory F1 and rest 75% of the helmets made by the company are produced in factory F2.

Hence the percentage of defective helmets in factory F1 = `\frac{25}{100}` × P1

And the percentage of defective helmets in factory F2 = `\frac{75}{100}` × P2

Now it is given that the total percentage of defective helmets made by the company = 6% = `\frac{6}{100}`

Hence, `\frac{25}{100}` × P1 + `\frac{75}{100}` × P2 = `\frac{6}{100}`

Now putting P1 = 4 × P2 in the above equation and further solving we get P2 = `\frac{6}{175}` and P1 = `\frac{24}{175}`

Probability that a helmet produced in factory F2 found non-defective = 1 – `\frac{6}{175}` = `\frac{169}{175}`

Now we need to find that a helmet chosen at random is found to be non-defective is from factory F2. Then for such a case the number of total events will be,

The percentage of non-defective helmets produced by the company = 100 – 6% = 94% = `\frac{94}{100}`

And the number of favorable events will be the percentage of non-defective helmets produced in factory F2 and that will be,

`\frac{75}{100}` × `\frac{169}{175}` = `\frac{507}{700}`

Hence the required probability = `\frac{507/700}{94/100}` = `\frac{507}{658}`

Question 83.) If A and B are two dependent events such that P(A) = `\frac{2}{3}`, P(B̅) = `\frac{1}{4}`, and P(B/A) = `\frac{4}{5}` then which of the following is incorrect.

(A) P(A∪B) = `\frac{7}{60}`

(B) P(B) = `\frac{3}{4}`

(D) P(A∩B) = `\frac{8}{15}`

Answer. (A)

Solution.

It is given in the question that,

P(A) = `\frac{2}{3}`, P(B̅) = `\frac{1}{4}` and P(B/A) = `\frac{4}{5}`

Now by the relation of conditional probability we know that,

P(B/A) = P(A∩B)/P(A)

P(A∩B) = P(B/A) × P(A) = `\frac{4}{5}` × `\frac{2}{3}` = `\frac{8}{15}`

Now, we will calculate P(B) and it is given by,

P(B) = 1 – P(B̅) = 1 – `\frac{1}{4}` = `\frac{3}{4}`

Now, we will calculate P(A/B) and it is given by,

P(A/B) = P(A∩B)/P(B) = `\frac{32}{45}`

Now, we will calculate P(A∪B) and it is given by,

P(A∪B) = P(A) + P(B) – P(A∩B) = `\frac{2}{3}` + `\frac{3}{4}` – `\frac{8}{15}` = `\frac{53}{60}`

From all the above calculation it is clear that option A is the correct answer.

Question 84.) Two players X and Y are playing chess. The probabilities of X for winning, drawing, and losing a game are `\frac{1}{2}`, `\frac{2}{3}`, and `\frac{1}{4}` respectively. Each of them gets 3 points on winning, and 0 points on losing a game, and when the game draws then both the players get 1 point each. If 3 games are played between them then find the probability that X gets more points than Y.

(A) `\frac{1}{6}`

(B) `\frac{37}{72}`

(D) `\frac{83}{144}`

Answer. (D)

Solution.

The condition given in the question that the player X will get more points than the player Y is possible only if he wins more number of games than other. So here four different cases will occur. Let’s discuss about them,

Case 1, among the three games played X wins only 1 game and rest of the two games get draw. Probability for this case is given by,

P1 = `\frac{1}{2}` × `\frac{2}{3}` × `\frac{2}{3}` = `\frac{2}{9}`

Case 2, among the three games played X wins only 2 games and the last one game gets draw. Probability for this case is given by,

P2 = `\frac{1}{2}` × `\frac{1}{2}` × `\frac{2}{3}` = `\frac{1}{6}`

Case 3, among the three games played X wins 2 games and loses the last one. Probability for this case is given by,

P3 = `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{4}` = `\frac{1}{16}`

Case 4, among the three games played X wins all 3 games. Probability for this case is given by,

P4 = `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` = `\frac{1}{8}`

Since only one case will occur. Hence the required probability is given by,

P = P1 + P2 + P3 + P4 = `\frac{2}{9}` + `\frac{1}{6}` + `\frac{1}{16}` + `\frac{1}{8}` = `\frac{83}{144}`

Question 85.) A person buys stocks of 3 companies A, B, and C and decides to sell them after one month. The probabilities that their prices will rise after one month are `\frac{1}{2}`, `\frac{1}{4}`, and `\frac{1}{4}` respectively. The person will make a profit only if the prices of at least two out of three stocks will rise after one month. Find the probability that the person makes profit and the price of the stock of company B also rises.

(A) `\frac{21}{32}`

(B) `\frac{5}{8}`

(D) `\frac{5}{16}`

Answer. (B)

Solution.

Let P(A), P(B), and P(C) are the probabilities that the prices of stocks of companies A, B, and C will rise. Then P(A̅), P(B̅), and P(C̅) are the probabilities that the prices of stocks of companies A, B, and C will fall.

Now it is given that,

P(A) = `\frac{1}{2}`

P(B) = `\frac{1}{4}`

P(C) = `\frac{1}{4}`

Similarly the values of P(A̅), P(B̅), and P(C̅) will be given by,

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

P(B̅) = 1 – P(B) = 1 – `\frac{1}{4}` = `\frac{3}{4}`

P(C̅) = 1 – P(C) = 1 – `\frac{1}{4}` = `\frac{3}{4}`

Now the person will make profit if the price of at least two stocks will rise. It means that either the price of any two stocks will rise and that of third will fall or the price of all three stocks will rise.

Let P1 is the probability that the person makes profit. Then P1 is given by,

P1 = P(A) × P(B) × P(C̅) + P(A) × P(B̅) × P(C) + P(A̅) × P(B) × P(C) + P(A) × P(B) × P(C)

P1 = `\frac{1}{2}` × `\frac{1}{4}` × `\frac{3}{4}` + `\frac{1}{2}` × `\frac{3}{4}` × `\frac{1}{4}` + `\frac{1}{2}` × `\frac{1}{4}` × `\frac{1}{4}` + `\frac{1}{2}` × `\frac{1}{4}` × `\frac{1}{4}` = `\frac{8}{32}`

Now the next condition is that the price of stock of company B must rise.

Let P2 is the probability that the person makes profit. Then P2 is given by,

P2 = P(A) × P(B) × P(C̅) + P(A̅) × P(B) × P(C) + P(A) × P(B) × P(C)

P2 = `\frac{1}{2}` × `\frac{1}{4}` × `\frac{3}{4}` + `\frac{1}{2}` × `\frac{1}{4}` × `\frac{1}{4}` + `\frac{1}{2}` × `\frac{1}{4}` × `\frac{1}{4}` = `\frac{5}{32}`

Hence the required probability = `\frac{P2}{P1}` = `\frac{5}{8}`

Question 86.) 4 numbers are randomly chosen without replacement from a set of first 12 natural numbers. Find the probability that the chosen 4 numbers do not form a sequence of arithmetic progression.

(A) `\frac{53}{55}`

(B) `\frac{1}{11}`

(D) `\frac{37}{55}`

Answer. (A)

Solution.

The set of first 12 natural numbers is as following,

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Now we need to choose 4 numbers randomly without replacement. So the total number of events will be the number of ways of selecting 4 numbers from set of 12 numbers and that is given by,

¹²C₄ = 495

Now if we consider the condition given in the question that the selected 4 numbers should never form a sequence of arithmetic progression then we need to eliminate all those possible sets that can make sequences of arithmetic progression. The arithmetic progression means that the selected numbers follow a common difference between any two consecutive numbers in the whole sequence. So, here we will make such sequences with 3 different common differences; 1, 2, and 3. The common difference of 4 cannot be taken because it will not allow us to make a set of 4 numbers. Hence here we will have 3 different cases for the favourable events,

Case 1, when the common difference is 1 then we will have following sets of 4 selected numbers,

(1,2,3,4), (2,3,4,5), (3,4,5,6), (4,5,6,7), (5,6,7,8), (6,7,8,9), (7,8,9,10), (8,9,10,11), (9,10,11,12)

We have got total 9 sets here.

Case 2, when the common difference is 2 then we will have following sets of 4 selected numbers,

(1,3,5,7), (2,4,6,8), (3,5,7,9), (4,6,8,10), (5,7,9,11), (6,8,10,12)

We have got total 6 sets here.

Case 3, when the common difference is 3 then we will have following sets of 4 selected numbers,

(1,4,7,10), (2,5,8,11), (3,6,9,12)

We have got total 3 sets here.

So, the total number of sets of selected 4 numbers that will make a sequence of arithmetic progression = 9 + 6 + 3 = 18

Now the favorable events will be those sets of selected 4 numbers that will not make a sequence of arithmetic progression and they will be = 495 – 18 = 477

Hence required probability = `\frac{477}{495}` = `\frac{53}{55}`

Question 87.) Two people A and B appear for the interview of a job in a company. But both of them appear on two different days. A appears on first day along with 3 more people while B appears on second day along with 4 more people. Find the probability that only one of them gets selected for the job.

(A) `\frac{3}{5}`

(B) `\frac{7}{20}`

(D) `\frac{3}{4}`

Answer. (B)

Solution.

Since the candidate A has appeared for the interview along with 3 more people. It means total 4 people appeared for the interview.

So the probability for the selection of A, P(A) = `\frac{1}{4}`

And the probability for rejection of A, P(A̅) = 1 – `\frac{1}{4}` = `\frac{3}{4}`

Similarly the candidate B has appeared for the interview along with 4 more people. It means total 5 people appeared for the interview.

So the probability for the selection of B, P(B) = `\frac{1}{5}`

And the probability for rejection of B, P(B̅) = 1 – `\frac{1}{5}` = `\frac{4}{5}`

Now as per the condition given in the question we need to find the probability that only one among A and B will be selected. So the required probability is given by,

P(A) × P(B̅) + P(A̅) × P(B) = `\frac{1}{4}` × `\frac{4}{5}` + `\frac{3}{4}` × `\frac{1}{5}` = `\frac{7}{20}`

Question 88.) A shop has 12 cookie packets among which 7 packets are of chocolate cookies and 5 packets are of sugar cookies. If a person asks for 4 packets of cookies and the shop owner chooses the packets at random then find the probability that the buyer gets more than 1 packet of sugar cookies?

(A) `\frac{50}{99}`

(B) `\frac{43}{99}`

(D) `\frac{23}{33}`

Answer. (C)

Solution.

Since, we need to select 4 cookie packets out of 12 cookie packets.

Hence total number of events = ¹²C₄

Now the condition given in the question is that the person gets more than 1 packet of sugar cookies. It means the person has got 2 or more than 2 packets of sugar cookies and this selection is possible in 3 ways. So we have following 3 cases,

Case 1, selecting 2 packets of sugar cookies and 2 packets of chocolate cookies. It is given by,

⁵C₂×⁷C₂

Case 2, selecting 3 packets of sugar cookies and 1 packet of chocolate cookies. It is given by,

⁵C₃×⁷C₁

Case 3, selecting all 4 packets of sugar cookies and no packet of chocolate cookies. It is given by,

⁵C₄

Now total number of favorable events = ⁵C₂×⁷C₂ + ⁵C₃×⁷C₁ + ⁵C₄

Now putting all values and further calculating we get that the required probability = `\frac{19}{33}`

Question 89.) There are two set of natural numbers. First set is of first 8 prime numbers and second set is of first 11 odd numbers. If a number is chosen from both the sets at random then find the probability that both the chosen numbers are same.

(A) `\frac{15}{88}`

(B) `\frac{15}{22}`

(D) `\frac{7}{88}`

Answer. (D)

Solution.

The set of first 8 prime numbers is as following,

{2, 3, 5, 7, 11, 13, 17, 19}

The probability of selecting a number from this set = `\frac{1}{8}`

And the set of first 11 odd numbers is as following,

{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21}

The probability of selecting a number from this set = `\frac{1}{11}`

The common numbers in both the sets are as following,

{3, 5, 7, 11, 13, 17, 19}

Since there are total 7 common numbers in both the sets. So there are 7 ways to select a common number at random. Hence the required probability = 7 × `\frac{1}{8}` × `\frac{1}{11}` = `\frac{7}{88}`

Question 90.) 4 people are standing at the four corners of a square shaped ground. If all of them start moving with the same speed together at the same time to complete a round of the ground and return back to their initial position then find the probability that none of them collide with each other.

(A) `\frac{1}{8}`

(B) `\frac{1}{2}`

(D) `\frac{1}{4}`

Answer. (A)

Solution.

As per the question all the 4 people have to complete a round and return back to their initial position and all are standing on the four corners of the ground. So when a person stands in a corner of a square shaped ground and he needs to complete a round of the ground then he has two options to move; one towards his left and other towards his right.

So the probability for each person to choose a path to complete a round of the ground = `\frac{1}{2}`

Now according to the condition given in the question none of the 4 people collide with each other while moving. It is possible only if all choose the same option to move. i.e., either all decide to move towards their left or all decide to move towards their right.

Hence the required probability = `\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}{8}`