PROBABILITY PRACTICE QUESTIONS : 41-50

Question 41.) A number is chosen at random from the set {1, 2, 3, ….., 2000}. Then, find the probability that the chosen number is a multiple of 11.

(A) `\frac{91}{2000}`

(B) `\frac{193}{2000}`

(D) `\frac{181}{1000}`

Answer. (C)

Solution.

In this question only one number is to be chosen from the whole set of 2000 numbers.

So, the total number of events = 2000

Now, the number of favorable events will be the number of multiples of 11.

In the given set of numbers the total number of multiples of 11 will be = `\frac{2000}{11}`

It will give a value in between 181 and 182. Since, it will be less than 182.

So, the number of favorable events will be = 181

Hence the required probability = `\frac{181}{2000}`

Question 42.) Consider the following data of 90 students. It was found that they play three games in following manner.

19 students play cricket, 22 students play football, 22 students play tennis, 33 students play cricket or football or both, 35 students play football or tennis or both, 34 students play cricket or tennis or both, 3 students play all three games (cricket, football and tennis).

If a student is chosen randomly from these 90 students, then the probability that the student plays at most one game is

(A) 0.5

(B) 0.4

(D) 0.65

Answer. (C)

Solution.

Let C denote the students who play cricket, F denote the students who play football and T denote the students who play tennis.

Given that

n(C) = 19, n(F) = 22, n(T) = 22

n(C∩F∩T) = 3

n(C∪F) = 33 n(FT) = 35 n(T C) = 34

n(C∪F) = n(C) + n(F) – n(C∩F)

33 = 19 + 22 – n(C∩F)

n(C∩F) = 8

Similarly n(F∩T) = 9 and n(T∩C) = 7

The Venn diagram representation will be like this,

n(C∪F∪T) = n(C) + n(F) + n(T) – n(C∩F) –n(F∩T) –n(T∩C) + n(C∩F∩T)

= 19 + 22 + 22 –8– 9 – 7 + 3 = 42

Number of students playing at most one game is = 90 – 42 + 7 + 8 + 9 = 72

So number of favorable events = 72

Total number of events = 90

Probability = `\frac{72}{90}` = 0.8

Question 43.) A bag contains 10 balls, of which 5 are red, 3 are blue, and 2 are green. 3 balls are drawn at random without replacement from the bag. Then the probability that the 3 balls have different colors is

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

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

(D) `\frac{1}{3}`

Answer. (A)

Solution.

Since there are total 10 balls and we need to draw 3 balls.

So, total number of events = ¹⁰C₃ = 120

Now the condition given is that all 3 balls drawn should be of different colours. Since there are 5 red, 3 blue and 2 green balls. So there are 3 different colours and we need to choose balls of 3 different colours. It means 1 red ball, 1 blue ball and 1 green ball.

So, number of favourable events = ⁵C_{1 ×}³C₁×²C₁= 5 × 3 × 2 = 30

Hence the required probability = `\frac{30}{120}` = `\frac{1}{4}`

Question 44.) Three numbers are chosen at random without replacement from {1, 2, 3, ...... 9}. The probability that their minimum is 4, given that their maximum is 8, is

(A) `\frac{1}{3}`

(B) `\frac{1}{9}`

(D) `\frac{1}{7}`

Answer. (D)

Solution.

As per the given condition in the question when maximum of the three numbers chosen is 8 then we have 7 numbers less than 8 and we need to choose only two numbers among these 7.

So, total number of events = ⁷C₂ = 21

Now the second condition will give us the number of favorable events. Since when the minimum of the three numbers chosen is 4 then only 3 numbers will left between 4 and 8, i.e.; {5, 6, 7}. And we need to choose only one of them.

Hence the number of favorable events = ³C₁ = 3

Hence the required probability = `\frac{3}{21}` = `\frac{1}{7}`

Question 45.) There are 5 multiple-choice questions in an examination with three alternative answers. Among these three alternatives, only one is correct. A student appearing in this exam is not able to solve any question, so he decides to guess the answers. Then find the probability that the student will get 3 correct answers.

(A) `\frac{10}{243}`

(B) `\frac{40}{243}`

(D) `\frac{25}{243}`

Answer. (B)

Solution.

Since there are 3 choices in each question and only 1 of them is correct.

So the probability of choosing a correct option is = `\frac{1}{3}`

Similarly the probability of choosing an incorrect option = `\frac{2}{3}`

Now, the condition is that the student answers correctly only 3 of them and wrong answer for rest of the two.

Hence the required probability = ⁵C₃ × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{1}{3}` × `\frac{2}{3}` × `\frac{2}{3}` = `\frac{40}{243}`

Question 46.) Given that A, B and C are three independent events and the probability of occurring of exactly one of the events among A or B is same as that for B or C and for C or A and is equals to `\frac{1}{3}`. If the probability of all the three events occur simultaneously is `\frac{1}{9}` then find the probability that at least one of the events occur.

(A) `\frac{11}{18}`

(B) `\frac{1}{3}`

(D) `\frac{4}{9}`

Answer. (A)

Solution.

This question is completely based on conditional probability. So here we will use probability distribution identities and rules.

Given that the probability of occurring of exactly one of the events among A or B is same as that for B or C and for C or A. it means

P(A) + P(B) – 2P(A∩B) = `\frac{1}{3}`

P(B) + P(C) – 2P(B∩C) = `\frac{1}{3}`

P(C) + P(A) – 2P(C∩A) = `\frac{1}{3}`

Adding all the three equations above we get,

∑P(A) – ∑P(A∩B) = `\frac{1}{2}`

It is also given that

P(A∩B∩C) = `\frac{1}{9}`

We need to find the probability that at least one of the events occur.

P(A∪B∪C) = ∑P(A) – ∑P(A∩B) + P(A∩B∩C) = `\frac{1}{2}` + `\frac{1}{9}` = `\frac{11}{18}`

Question 47.) Three persons A, B and C participate in a shooting competition and independently try to hit a target with a gun. All of them get 20 bullets each to hit the target. If A hits the target 4 times, B hits the target 9 times and C hits the target 10 times respectively using all their bullets. Then find the probability that the target is hit by only one among A or B and not by C.

(A) `\frac{53}{200}`

(B) `\frac{47}{200}`

(D) `\frac{83}{200}`

Answer. (B)

Solution.

Total number of events for each participant is = Number of bullets each participant get = 20

Since A hits the target only 4 times. So favorable events for A = 4

So, probability that A hits the target = `\frac{4}{20}` = `\frac{1}{5}`

And, probability that A does not hit the target = 1 – `\frac{1}{5}` = `\frac{4}{5}`

Similarly B hits the target 9 times. So favorable events for B = 9

So probability that B hits the target = `\frac{9}{20}`

And probability that B does not hit the target = 1 – `\frac{9}{20}` = `\frac{11}{20}`

Similarly C hits the target 10 times. So favorable events for C = 10

So probability that C hits the target = `\frac{10}{20}` = `\frac{1}{2}`

And probability that C does not hit the target = 1 – `\frac{1}{2}` = `\frac{1}{2}`

Now following the condition given in the question that the target gets hit by only one among A or B and not by C we will have two cases. One is that the target gets hit by A only but not by B and C and the second case is that the target gets hit by B only but not by A and C.

Hence the required probability = `\frac{1}{5}` × `\frac{11}{20}` × `\frac{1}{2}` + `\frac{4}{5}` × `\frac{9}{20}` × `\frac{1}{2}` = `\frac{47}{200}`

Question 48.) John performs an experiment in which the success rate is twice to the failure rate. If the experiment is performed 5 times by John then find the probability that John gets the success in exactly 4 trials of this experiment.

(A) `\frac{40}{243}`

(B) `\frac{100}{243}`

(D) `\frac{50}{243}`

Answer. (C)

Solution.

Let P be the probability of failure.

Then it is given that the rate of success is twice the rate of failure. Hence its probability will also be twice. i.e.; 2P.

Now P + 2P = 1

P = `\frac{1}{3}`

Now the probability of failure = P = `\frac{1}{3}`

And the probability of success = 2P = `\frac{2}{3}`

The probability that John gets success in exactly 4 out of 5 attempts is,

= ⁵C₄`\frac{2}{3}` × `\frac{2}{3}` × `\frac{2}{3}` × `\frac{2}{3}` × `\frac{1}{3}`= `\frac{80}{243}`

Question 49.) If A and B are two independent events. The probability that both A and B happen is `\frac{1}{6}` and the probability that neither A nor B happens is `\frac{1}{4}`. Then find the value of P(A)/P(B).

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

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

(D) `\frac{8}{3}`

Answer. (D)

Solution.

Let P(A) = x and P(B) = y

Now, P(A∩B) = P(A).P(B) = x.y = `\frac{1}{6}`

Also P(Aꞌ Bꞌ) = [1 – P(A)].[1 – P(B)] = (1 – x).(1 – y) = `\frac{1}{4}`

Solving both the above equations we get.

x = `\frac{2}{3}` and y = `\frac{1}{4}`

Hence, P(A)/P(B) = `\frac{x}{y}` = `\frac{8}{3}`

Question 50.) A bag contains 3 red and 2 black balls. A ball is drawn at random from the bag, its colour is observed and along with this ball one additional ball of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then find the probability that this drawn ball is black.

(A) `\frac{2}{5}`

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

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

Answer. (A)

Solution.

Here we will have two cases. The first case is when the ball drawn in first attempt is red. Then,

The probability of drawing a red ball is = `\frac{3}{5}`

When red ball is drawn then one additional red ball is returned into the bag. Then the bag will now have 4 red balls and 2 black balls.

So in this case the probability of drawing a black ball = `\frac{1}{3}`

Now we will consider the second case is when the ball drawn in first attempt is black. Then,

The probability of drawing a black ball in second attempt is = `\frac{2}{5}`

When black ball is drawn then one additional black ball is returned into the bag. Then the bag will now have 3 red balls and 3 black balls.