PROBABILITY PRACTICE QUESTIONS : 51-60

Question 51.) There are two bags A & B. Bag 'A' contains 3 white, 3 red and 4 black balls. Another bag 'B' contains 4 white, 2 red and 2 black balls. If two balls are drawn at random, without replacement, from a randomly selected bag and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from bag 'B' is

(A) `\frac{10}{17}`

(B) `\frac{5}{12}`

(C) `\frac{7}{16}`

(D) `\frac{11}{13}`

Answer. (A)

Solution.

Let’s consider the case that bag A have been selected,

Since it has total 10 balls and we need to draw 2 balls from it one after another without replacement. Then the total number of events = ¹⁰C₂ = 45

We need to draw two balls, one white ball and then one red ball.

So number of favorable events = ³C_1 × ³C₁ = 9

Probability of drawing a white and a red ball from bag A is = `\frac{9}{45}` = `\frac{1}{5}`

Let’s consider the case that bag B have been selected,

Since it has total 8 balls and we need to draw 2 balls from it one after another without replacement. Then the total number of events = ⁸C₂ = 28

We need to draw two balls, one white ball and then one red ball.

So number of favorable events = ⁴C₁ × ²C₁ = 8

Probability of drawing a white and a red ball from bag B is = `\frac{8}{28}` = `\frac{2}{7}`

Hence, the required probability = `\frac{(2/7)}{(2/7) + (1/5)}` = `\frac{10}{17}`

Question 52.) Two different students’ groups A and B have equal number of students. There are 3 pencils to be distributed amongst the students of these groups so that no student gets more than one pencil. If the probability that all the pencils go to the students of the group B is `\frac{1}{12}`, then the number of students in each group is

(A) 3

(B) 8

(C) 5

(D) 4

Answer. (C)

Solution.

Let the number of students in each group is x. Then total number of students will be 2x.

Now it is given that 3 pencils are to be distributed among 2x students.

So, number of total events = ^2xC₃ = `\frac{(2x).(2x-1).(2x-2)}{6}`

Now the condition is that all 3 pencils go to students of group B.

So number of favourable events = ^xC₃ = `\frac{(x).(x-1).(x-2)}{6}`

Hence the probability that all the pencils go to the students of group B is

= `\frac{(x).(x-1).(x-2)}{(2x).(2x-1).(2x-2)}` = `\frac{1}{12}`

On solving it we get x = 5

Question 53.) An urn contains 6 white and 3 black balls. A ball is drawn at random from the urn. If the drawn ball is black, then two white balls are added to the urn and if the drawn ball is white, then two black balls are added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. Then find the probability that the second ball is white.

(A) `\frac{3}{5}`

(B) `\frac{4}{5}`

(C) `\frac{7}{9}`

(D) `\frac{4}{9}`

Answer. (A)

Solution.

Since there are 6 white and 3 black balls
Probability of drawing one white ball = `\frac{6}{9}` = `\frac{2}{3}`

Similarly the probability of drawing one black ball = `\frac{3}{9}` = `\frac{1}{3}`

Now as per the given condition in the question, when first ball drawn is white then two black balls are added to the urn. After it the urn will have 5 white and 5 black balls.

Then the probability that the second ball drawn is white = `\frac{5}{10}` = `\frac{1}{2}`

And if the first ball drawn is black then two white balls are added to the urn. After it the urn will have 8 white and 2 black balls.

Then the probability that the second ball drawn is white = `\frac{8}{10}` = `\frac{4}{5}`

Hence required probability = `\frac{2}{3}` × `\frac{1}{2}` + `\frac{1}{3}` × `\frac{4}{5}` = `\frac{3}{5}`

Question 54.) In a game of throwing a dice, a man will win fifty bucks if he gets either 1 or 6 as the output on a throw of the dice otherwise he will lose twenty five bucks. A person can take maximum 3 chances only and these extra chances will be given only when the player loses in a chance. That means if the player wins in first chance then he will not get the second chance but if he loses the first chance only then he will get the second chance. If a man Peter plays this game and want to win then his expected gain will be

(A) 100 bucks

(B) 0 bucks

(C) 50 bucks

(D) 25 bucks

Answer. (B)

Solution.

Let P(W) is the probability of winning and P(L) is the probability to lose the game.

When the dice will be thrown only one time then it can produce 6 outputs.

So number of total events will be = 6

But as per the condition the person will win only when he will get either 1 or 6 as output.

So number of favorable events will be = 2

Hence probability of winning, P(W) = `\frac{2}{6}` = `\frac{1}{3}`

And the probability to lose the game, P(L) = 1 – P(W) = 1 – `\frac{1}{3}` = `\frac{2}{3}`

Now when Peter plays the game then four different cases will appear.

Case 1, if Peter wins in first chance then his gain is = 50 bucks

Case 2, if Peter loses in first chance and wins in second chance then his gain will be

= (– 25) + 50 = 25 bucks

Case 3, if Peter loses in first two chances and wins in third chance then his gain will be

= (– 25) + (– 25) + 50 = 0 bucks

Case 4, if Peter loses in all the three attempts then his gain will be

= (– 25) + (– 25) + (– 25) = (–75) bucks

Now the expected gain of Peter will be

= P(W) × 50 + P(L) × P(W) × 25 + [P(L)]² × P(W) × 0 + [P(L)]³ × (–75)

= `\frac{1}{3}` × 50 + `\frac{2}{3}` × `\frac{1}{3}` × 25 + [`\frac{2}{3}`]² × `\frac{1}{3}` × 0 + [`\frac{2}{3}`]³ × (–75) = 0

Question 55.) In a group of 90 people, 50 people like black shoes, 45 people like white shoes and 20 people like both black and white shoes. If one of these people is selected at random, then the probability that the person selected doesn’t likes any of the black or white shoes.

(A) `\frac{2}{9}`

(B) `\frac{5}{9}`

(C) `\frac{5}{6}`

(D) `\frac{1}{6}`

Answer. (D)

Solution.

The complete representation of the question has been shown in the following venn diagram

From the above diagram it is clear that the number of people who don’t like any of the black or white shoes is 15.

Along with it we will also use following mathematical formula,

Total no. of people = No. of people who like black shoes + No. of people who like white shoes – No. of people who like both black and white shoes + No. of people who like neither black nor white shoes

Now let the number of people who like neither black nor white shoes is ‘n’

90 = 50 + 45 – 20 + n

n = 15

Hence the required probability = `\frac{15}{90}` = `\frac{1}{6}`

Question 56.) An unbiased coin is tossed. If the outcome is a head, then a pair of unbiased dice is rolled and the sum of the numbers obtained is observed. If the toss of the coin results in tail then a number is chosen from a set of first 9 natural numbers. If the coin is tossed one time then find the probability that the final resultant number (either it is the sum of the numbers obtained in rolling the dice or chosen from the set of first 9 natural numbers) is a prime number.

(A) `\frac{23}{36}`

(B) `\frac{7}{9}`

(C) `\frac{31}{72}`

(D) `\frac{5}{12}`

Answer. (C)

Solution.

When the coin is tossed once then the probability of getting a head = `\frac{1}{2}`

And the probability of getting a tail = 1 – `\frac{1}{2}` = `\frac{1}{2}`

Now as per the first condition given in the question when head will appear as the result of toss then the pair of dice will be rolled and the sum of their outputs will be observed.

Here total number of events = 36

Now the sum of outputs of two dice rolled together cannot exceed 12 and the prime numbers under 12 are 2,3,5,7, and 11. So to get a prime number in sum of results we will have following combinations,

(1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5)

These are total 15 combinations. So number of favorable events = 15

So the probability for this condition is = `\frac{15}{36}`

Now as per the second condition given in the question when tails will appear as the result of toss then a number is chosen from the set of first 9 natural numbers.

Here total number of events = 9

Now the prime numbers under 9 are 2, 3, 5, and 7. So to get a prime number in result we will have 4 numbers. So number of favorable events = 4

So the probability for this condition is = `\frac{4}{9}`

Hence the required probability = `\frac{1}{2}` × `\frac{15}{36}` + `\frac{1}{2}` × `\frac{4}{9}` = `\frac{31}{72}`

Question 57.) There are two shooters, A and B. A hits the target once in every 3 shots while B hits the target once in every 4 shots. If they get total 24 bullets to hit a target together, then how many bullets shot by them will hit the target.

(A) 14

(B) 7

(C) 12

(D) 10

Answer. (B)

Solution.

Shooter A hits the target once in every 3 shots. So probability that the target gets hit by A = `\frac{1}{3}

Shooter B hits the target once in every 4 shots. So probability that the target gets hit by B = `\frac{1}{4}

Now 24 bullets will be distributed among them equally. So, each one will get 12 bullets to hit the target individually.

So, target hit by A = 12 × `\frac{1}{3} = 4

Similarly, target hit by B = 12 × `\frac{1}{4} = 3

Hence number of bullets shot by them that will hit the target = 4 + 3 = 7

Question 58.) Two numbers are selected at random from a set of first 11 multiples of 3. Given that the sum of selected numbers is even, then the conditional probability that both the numbers are odd is

(A) `\frac{2}{5}`

(B) `\frac{3}{5}`

(C) `\frac{4}{5}`

(D) `\frac{1}{2}`

Answer. (B)

Solution.

The set of first 11 multiples of 3 is {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33}. This set has 5 even numbers and 6 odd numbers.

Now the sum of two selected numbers will be even only if both of the numbers are even or both are odd.

So the total number of events will be the number of selecting two even numbers or two odd numbers. Which is = ⁵C₂ + ⁶C₂ = 10 + 15 = 25

Now as per the condition given in the question the selected two numbers should be odd.

So number of favorable events = ⁶C₂ = 15

Hence the required probability = `\frac{15}{25}` = `\frac{3}{5}`

Question 59.) If a fair dice is rolled by a person until two threes are obtained in succession. Then find the probability that the person gets the desired output in fourth throw.

(A) `\frac{5}{216}`

(B) `\frac{5}{36}`

(C) `\frac{25}{216}`

(D) `\frac{91}{216}`

Answer. (A)

Solution.

The probability of getting exactly 4 when dice is rolled once = `\frac{1}{6}`

Hence the probability of not getting 4 when the dice is rolled = 1 – `\frac{1}{6}` = `\frac{5}{6}`

Here the condition is that when two threes are obtained in succession in rolling the dice then the person will stop rolling the dice. But it is given that the dice is rolled for four times. So here two cases will appear.

Case 1, we get 3 as an output in first throw but not in second throw and then again we get 3 in next two throws successively.

So the probability for this case = `\frac{1}{6}` × `\frac{5}{6}` × `\frac{1}{6}` × `\frac{1}{6}`

Case 2, we do not get 3 in first two throws but we get 3 in next two throws successively.

So the probability for this case = `\frac{5}{6}` × `\frac{5}{6}` × `\frac{1}{6}` × `\frac{1}{6}`

We can never consider the case of getting 3 in second throw. Because in that case when we again get 3 in third throw then the condition of getting two successive 3 will be fulfilled and we will not be allowed to throw the dice for fourth time.

Hence required probability = `\frac{1}{6}` × `\frac{5}{6}` × `\frac{1}{6}` × `\frac{1}{6}` + `\frac{5}{6}` × `\frac{5}{6}` × `\frac{1}{6}` × `\frac{1}{6}` = `\frac{5}{216}`

Question 60.) The official records of a hospital show that only one out of three patients suffering from a particular disease survives. What is the probability that 4 out of the 5 randomly selected patients will survive?

(A) `\frac{20}{243}`

(B) `\frac{16}{243}`

(C) `\frac{25}{243}`

(D) `\frac{10}{243}`

Answer. (D)

Solution.

Since only 1 among 3 patients survive.

So, the probability of survival, P(S) = `\frac{1}{3}`

And, the probability of dying, P(D) = 1 – `\frac{1}{3}` = `\frac{2}{3}`

Now the given condition is that, exactly 4 out of the 5 randomly selected patients will survive.

The probability for survival of 4 patients = [P(S)]⁴

The probability for dying of 1 patient = [P(D)]¹

Hence required probability = ⁵C₄ × [P(S)]⁴ × [P(D)]¹ = 5 × (`\frac{1}{3}`)⁴ × `\frac{2}{3}` = `\frac{10}{243}`