PROBABILITY PRACTICE QUESTIONS : 61-70

Question 61.) An unbiased coin is tossed 5 times. Find the probability that head appears at least 3 times.

(A) `\frac{1}{16}`

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

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

Answer. (C)

Solution.

This question is very simple to solve and understand if we consider the traditional counting method. But here I will use a new trick. So please read the solution carefully.

When a coin is tossed 5 times, then the number of total outputs = `2^{5}` = 32

Now as per the condition given in the question the favorable events will be the events when head will appear at least 3 times. It means that when it will appear 4 or 5 times then those events will also be the favorable events.

Now when head appears 3 times then tails will appear 2 times. Similarly when head will appear 4 times then tails will appear 1 time and when head will appear 5 times then tails will not appear. The same case will happen for tails also. If tails will appear 5 times then head will not appear, if tails will appear for 4 times then head will appear 1 time and if tails will appear for 3 times then head will appear for 2 times.

Now if we carefully observe all the conditions told above then it will cover all the possible events that can appear when a coin is tossed 5 times. But we are considering only half of the events following the condition given in the question. So, number of favorable events will also be half of the total events.

Hence total number of favorable events = `\frac{32}{2}`

Hence required probability = `\frac{1}{2}`

Question 62.) In a workshop, there are five machines and the probability of any one of them to be out of service on a day is `\frac{1}{3}`. Then find the probability that two machines are out of service on a day.

(A) `\frac{8}{27}`

(B) `\frac{160}{243}`

(D) `\frac{1}{81}`

Answer. (C)

Solution.

Given that the probability for a machine to be out of service on a day is = P = `\frac{1}{3}`

So the probability of a machine to be fit for work on a day is = P̅ = 1 – P = 1 – `\frac{1}{3}` = `\frac{2}{3}`

Now the required probability = ⁵C₂ × [P]²× [P̅]³ = 10 × `(\frac{1}{3})^{2}` × `(\frac{2}{3})^{3}`= `\frac{80}{243}`

Question 63.) In the shelf of a library, there are 16 books, out of which 8 are of history and the remaining 8 are of economics. A person comes near to the shelf takes out the books at random, one after the other with replacement, till the second history book is taken out. Find the probability that the second history book is taken out and placed back as soon as possible but it should happen before the third economics book is taken out and placed back.

(A) `\frac{11}{16}`

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

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

Answer. (A)

Solution.

Let H represents the event of taking out a history book then placing it back and P(H) represents its probability. Similarly E represents the event of taking out an economics book then placing it back and P(E) represents its probability.

P(H) = `\frac{8}{16}` = `\frac{1}{2}`

P(E) = `\frac{8}{16}` = `\frac{1}{2}`

Now as per the conditions given in the question the process of taking out the books and placing it back remains continue until the second history book is taken out and placed back. But this must happen before the third economics book is taken out and placed back. Just observe this statement and we will figure it out that this event will also get completed if second history book is taken out and placed back before the first economics book is taken out or before the second economics book is taken out. Because here the main event is the event of taking out the second history book and placing it back but this event has a barrier of the event of taking out the third economics book.

So it will have six possible cases favoring the above condition and that are as following,
1^st case, first history book is taken out and placed back in first attempt and the second history book is taken out and placed back in second attempt.

Probability for this event = P(H) × P(H) = `\frac{1}{2}` × `\frac{1}{2}` = `\frac{1}{4}`

2^nd case, first history book is taken out and placed back in first attempt and then an economics book is taken out and placed back and after it the second history book is taken out and placed back in third attempt.

Probability for this event = P(H) × P(E) × P(H) = `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` = `\frac{1}{8}`

3^rd case, an economics book is taken out in first attempt and after that the first history book is taken out and placed back in second attempt and the second history book is taken out and placed back in third attempt.

Probability for this event = P(E) × P(H) × P(H) = `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` = `\frac{1}{8}`

4^th case, first history book is taken out and placed back in first attempt and then two economics books are taken out and placed in next two consecutive attempts and after it the second history book is taken out and placed back in fourth attempt.

Probability for this event = P(H) × P(E) × P(E) × P(H) = `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` = `\frac{1}{16}`

5^th case, two economics books are taken out and placed in first two consecutive attempts and then first history book is taken out and placed back in third attempt and the second history book is taken out and placed back in fourth attempt.

Probability for this event = P(E) × P(E) × P(H) × P(H) = `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` = `\frac{1}{16}`

6^th case, an economics book is taken out and placed back in first attempt, then a history book is taken out and placed back in second attempt, then again an economics book is taken out and placed back in third attempt and finally the second history book is taken out and placed back in fourth attempt

Probability for this event = P(E) × P(H) × P(E) × P(H) = `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` × `\frac{1}{2}` = `\frac{1}{16}`

Now the required probability will be the sum of probabilities of all possible cases.

Hence required probability = `\frac{1}{4}` + `\frac{1}{8}` + `\frac{1}{8}` + `\frac{1}{16}` + `\frac{1}{16}` + `\frac{1}{16}` = `\frac{11}{16}`

Question 64.) There are two boxes A and B. Box A contains 30 cards numbered from 11 to 40 and box B contains 25 cards numbered 51 to 75. A box is selected at random and a card is drawn from it. If the number on the card is found to be a non-prime number then find the probability that the card is drawn from Box B.

(A) `\frac{6}{25}`

(B) `\frac{4}{15}`

(D) `\frac{9}{19}`

Answer. (D)

Solution.

Since here we have two boxes and the probability of selecting a box at random = `\frac{1}{2}`

Let’s consider the case for box A first.

Since the box A contains 30 cards numbered from 11 to 40.

So number of total events here = 30

And number of favorable events = Number of non prime numbered cards from 11 to 40 = 8

So probability for this case, P(A) = `\frac{1}{2}` × `\frac{8}{30}` = `\frac{2}{15}`

Now we will consider the case for box B.

Since the box B contains 25 cards numbered from 51 to 75.

So number of total events here = 25

And number of favorable events = Number of non prime numbered cards from 51 to 75 = 6

So probability for this case, P(B) = `\frac{1}{2}` × `\frac{6}{25}` = `\frac{3}{25}`

But in the question a condition is given that the box is selected at random and then the card is drawn.

Hence required probability = `\frac{P(B)}{P(A)+P(B)}` = `\frac{9}{19}`

Question 65.) A and B are two shooters. They decide to play a game of hitting a target alternatively. They set the target and decide that who will hit the target first will win the game and the game will stop there. If their probabilities of hitting the target are `\frac{2}{5}` and `\frac{4}{9}` respectively then find the probability that A wins the game.

(A) `\frac{38}{45}`

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

(D) `\frac{7}{15}`

Answer. (B)

Solution.

Let P(A) & P(B) are the probabilities of A & B for hitting the target and it is given that,

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

P(B) = `\frac{4}{9}`

Now if P̅(A) & P̅(B) are the probabilities of A & B for not hitting the target. Then,

P̅(A) = 1 – `\frac{2}{5}` = `\frac{3}{5}`

P̅(B) = 1 – `\frac{4}{9}` = `\frac{5}{9}`

Now as per the condition given in the question A will win only if he hits the target before B and in which chance it will happen we don’t know about it. A may hit the target in 1^st chance or in 10^th chance or in 100^th chance. In other words in which chance A will hit the target it is completely uncertain but one thing is made to sure that B should not hit the target.

So the required probability = P(A)+ P̅(A) × P̅(B) × P(A) + P̅(A) × P̅(B) × P̅(A) × P̅(B) × P(A) + P̅(A) × P̅(B) × P̅(A) × P̅(B) × P̅(A) × P̅(B) × P(A) + ………………………….∞

If we observe this complete series then we will find that it is an infinite series of geometric progression with a common ratio of P̅(A) × P̅(B) and the sum of this infinite geometric progression series is = [P(A)] / [1 - P̅(A) × P̅(B)]

Now on putting the respected values and solving the equation we will get the required probability = `\frac{3}{5}`

Question 66.) A building has two lifts and each lift operates independently but their operation is controlled by a single control unit. The probability of proper operation without any error of the first lift is 0.9 and that of the second lift is 0.8. If a problem occurs in the control unit that now only one of the lift will operate properly then find the probability that the first lift is not operating properly but the second lift is operating properly.

(A) `\frac{2}{7}`

(B) `\frac{8}{17}`

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

Answer. (A)

Solution.

Let L1 & L2 are the first and second lifts.

Now it is given that the probability of proper operation of first lift, P(L1) = 0.9

And the probability of proper operation of second lift, P(L2) = 0.8

We will also calculate the probabilities of non proper operations of the lifts.

The probability of non proper operation of first lift, P̅(L1) = 1 – 0.9 = 0.1

And the probability of non proper operation of second lift, P̅(L2) = 1 – 0.8 = 0.2

Now in order to calculate the required probability that satisfies the condition given in the question.

Required probability = [P̅(L1) × P(L2)] / [P̅(L1) × P(L2) + P(L1) × P̅(L2) + P̅(L1) × P̅(L2)] = `\frac{0.1×0.8}{0.1×0.8+0.9×0.2+0.1×0.2}` = `\frac{2}{7}`

Question 67.) A and B are two mutually exclusive events such that their probabilities are given by P(A) = `\frac{1-n}{2}` and P(B) = `\frac{4n-1}{2}`. Then the set for all possible values of n is,

(A) [0, `\frac{2}{3}`]

(B) [`\frac{1}{4}`, `\frac{2}{3}`]

(D) [`\frac{2}{3}`, `\frac{3}{4}`]

Answer. (B)

Solution.

As we know that the probability lies between 0 and 1.

So, 0 ≤ P(A) ≤ 1

Putting value of P(A) we get,

0 ≤ `\frac{1-n}{2}` ≤ 1

Solving this inequality we will get,

–1 ≤ n ≤ 1

And, 0 ≤ P(B) ≤ 1

Putting value of P(B) we get,

0 ≤ `\frac{4n-1}{2}` ≤ 1

Solving this inequality we will get,

`\frac{1}{4}` ≤ n ≤ `\frac{3}{4}`

Now, P(A∪B) = P(A) + P(B) – P(A∩B)

Now, it is given that A and B are two mutually exclusive events,

So, P(A∩B) = 0

Hence, P(A∪B) = P(A) + P(B)

And, 0 ≤ P(A∪B) ≤ 1

Putting value of P(A∪B) we get,

0 ≤ `\frac{1-n}{2}` + `\frac{4n-1}{2}` ≤ 1

Solving this inequality, we will get,

0 ≤ n ≤ `\frac{2}{3}`

Now on comparing all the inequalities of n and eliminating the common values we will get that

`\frac{1}{4}` ≤ n ≤ `\frac{2}{3}`

Hence the set for all possible values of n = [`\frac{1}{4}`, `\frac{2}{3}`]

Question 68.) In a shooting game, there are 10 targets to shoot and two chances are given to players to shoot the targets, 5 targets in first chance and 5 targets in second chance. Out of these 10 targets, a player shoots first 5 targets correctly with a probability of `\frac{2}{3}` for each target and the remaining 5 targets correctly with a probability of `\frac{1}{3}` for each target. If the probability that the player shoots exactly 8 out of 10 targets is 10m × `(\frac{1}{3})^{10}`, then find the value of 'm'.

(A) 128

(B) 80

(D) 184

Answer. (C)

Solution.

Given that the probability of shooting a target in 1^st chance = `\frac{2}{3}`

So the probability of missing a target in 1^st chance = 1 – `\frac{2}{3}` = `\frac{2}{3}`

Given that the probability of shooting a target in 2^nd chance = `\frac{2}{3}`

So the probability of missing a target in 2^nd chance = 1 – `\frac{2}{3}` = `\frac{2}{3}`

If the player shoots exactly 8 out of 10 targets then he will miss only 2 targets and he is getting two chances to shoot the target. So he may miss these 2 targets in ³C₁, i.e., 3 ways. So here we have 3 different cases.

1^st case, the player misses both the 2 targets in first chance. Then he will shoot only 3 targets in 1^st chance and all 5 targets in 2^nd chance. Then the probability for this case will be,

P1 = ⁵C₃×`(\frac{2}{3})^{3}` × `(\frac{1}{3})^{2}` ×⁵C₅ × `(\frac{1}{3})^{5}` = 80 × `(\frac{1}{3})^{10}`

2^nd case, the player misses both the 2 targets in second chance. Then he will shoot all 5 targets in 1^st chance and only 3 targets in 2^nd chance. Then the probability for this case will be,

P2 = ⁵C₅×`(\frac{2}{3})^{5}` ×⁵C₃ × `(\frac{1}{3})^{3}` × `(\frac{2}{3})^{2}` = 1280 × `(\frac{1}{3})^{10}`

3^rd case, the player miss 1 of the 2 targets in first chance and other target in second chance. Then he will shoot 4 targets in 1^st chance and 4 targets in 2^nd chance. Then the probability for this case will be,

P3 = ⁵C₄×`(\frac{2}{3})^{4}` ×`(\frac{1}{3})^{1}` × ⁵C₄ × `(\frac{1}{3})^{4}` × `(\frac{2}{3})^{1}` = 800 × `(\frac{1}{3})^{10}`

Now the required probability = Sum of the probabilities of all the three cases

P = P1 + P2 + P3 = 2160 × `(\frac{1}{3})^{10}`

Now it is given in the question that, P = 10m × `(\frac{1}{3})^{10}` = 2160 × `(\frac{1}{3})^{10}`

Hence, m = 216

Question 69.) Six numbers are randomly selected from the set of first 15 natural numbers and arranged in the increasing order. Then find the probability that the second lowest is 7 and highest is greater than 12.

(A) `\frac{72}{1001}`

(B) `\frac{18}{1001}`

(D) `\frac{36}{1001}`

Answer. (D)

Solution.

Since it is given that we need to select 6 numbers from the set first 15 natural numbers.

So total number of events = ¹⁵C₆ = 5005

Now as per the 1^st condition given in the question that the second lowest number is 7 so the first number will be chosen from first six natural numbers.

So number of ways choosing a number from first six natural numbers = ⁶C₁ = 6

And as per the 2^nd condition given in the question that the highest number is greater than 12 so the 3^rd, 4^th, and 5^th numbers will be chosen from five natural numbers in between 7 and 13.

So number of ways choosing these three numbers = ⁵C₃ = 10

And now the last number is chosen from last 3 numbers.

So number of ways choosing the last number = ³C₁ = 3

Hence number of favorable events = 6 × 1 × 10 × 3 = 180

Required probability = `\frac{180}{5005}` = `\frac{36}{1001}`

Question 70.) Find the probability, that in a randomly selected 3-digit number all digits are odd and also different.

(A) `\frac{1}{15}`

(B) `\frac{1}{10}`

(D) `\frac{29}{90}`

Answer. (A)

Solution.

Total number of events = The total number of three digit numbers = 900

Now as per the condition given in the question all the three digits of the selected must be odd and different from each other. It means that they are made using the set of odd digits, i.e.; from the following set of numbers,

(1, 3, 5, 7, 9)

These are total 5 digits and one digit will be used only once to make the three digit number.

Hence favourable events = Number of 3 digit numbers formed by these 5 digits without repetition

= 5 × 4 × 3 = 60

Hence required probability = `\frac{60}{900}` = `\frac{1}{15}`