PROBABILITY PRACTICE QUESTIONS : 71-80

Question 71.) If A and B are two events such that P(A) = `\frac{1}{6}`, P(B) = `\frac{2}{7}` and P(A∪B) = `\frac{3}{8}`, then find the ratio of P(`\frac{A}{B′}`) to P(`\frac{B}{A′}`) .

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

(B) `\frac{1}{8}`

(C) `\frac{1}{4}`

(D) `\frac{1}{5}`

Answer. (A)

Solution.

Now we know that P(`\frac{A}{B′}`) is given by,

P(`\frac{A}{B′}`) = P(A∩B′) / P(B′) = [P(A) - P(A∩B)] / [1 - P(B)]        ……….eq.(1)

And P(`\frac{B}{A′}`) is given by,

P(`\frac{B}{A′}`) = P(B∩A′) / P(A′) = [P(B) - P(A∩B)] / [1 - P(A)]         ……….eq.(2)

So, now we need to calculate the value of P(A∩B) first to proceed further and it is given by,

P(A∩B) = P(A) + P(B) – P(A∪B)

Now putting values of P(A), P(B) and P(A∪B) from the question and further evaluating we get,

P(A∩B) = `\frac{1}{6}` + `\frac{2}{7}` – `\frac{3}{8}` = `\frac{13}{168}`

Now putting this value of P(A∩B) and also the values of P(A) and P(B) in eq.(1) and eq.(2) we get,

P(`\frac{A}{B′}`) = `\frac{1}{8}` , and P(`\frac{B}{A′}`) = `\frac{1}{4}`

Now the required ratio = `\frac{1}{2}`

Question 72.) A number is randomly selected from a set of all four digit numbers. Find the probability that the randomly selected number is divisible by 2 but not by 4.

(A) `\frac{1}{2}`

(B) `\frac{1}{3}`

(C) `\frac{1}{4}`

(D) `\frac{3}{4}`

Answer. (C)

Solution.

Total number of events = The total number of four-digit numbers = 9000

Now let us consider the condition given in the question that the selected number should be divisible by 2 but not by 4. Now as we know that all numbers divisible by 4 are also divisible by 2 but all numbers divisible by 2 are not compulsory divisible by 4. So we will eliminate all those numbers which are divisible by 4.

Now the set of all 4 digit numbers will be,

(1000, 1001, 1002, 1003, 1004, ………………………………., 9996, 9997, 9998, 9999)

Now in order to find the number of numbers in the set divisible by a particular number we will use the following formula,

() + 1

So number of all 4 digit numbers divisible by 2 = `\frac{9998-1000}{2}` +1 = 4500

And number of all 4 digit numbers divisible by 4 = `\frac{9996-1000}{4}` +1 = 2250

Now number of favourable events = 4500 – 2250 = 2250

Hence required probability = `\frac{2250}{9000}` = `\frac{1}{4}`

Question 73.) A number is selected at random from a set of all natural 3 digit numbers. Find the probability that the selected number is divisible by either 3 or 5 but not by 30.

(A) `\frac{13}{30}`

(B) `\frac{2}{5}`

(C) `\frac{19}{30}`

(D) `\frac{11}{30}`

Answer. (A)

Solution.

Total number of events = The total number of three digit numbers = 9000

Now the set of all 3 digit numbers will be,

(100, 101, 102, 103, 104, ………………………………., 996, 997, 998, 999)

We will use the following formula,

(Difference of the largest and the smallest numbers in the set divisible by the particular number / The particular number) + 1

Number of all 3 digit numbers divisible by 3 = `\frac{999-102}{3}` + 1 = 300

And number of all 3 digit numbers divisible by 5 = `\frac{995-100}{5}` + 1 = 180

Since the first condition given in the question is that the selected number is divisible by either 3 or 5. It means that the selected number should not be divisible by both 3 and 5.

Now the LCM of 3 and 5 is 15 so we need to eliminate the multiples of 15 from the set of favorable events.

So number of all 3-digit numbers divisible by 15 = `\frac{990-105}{15}` + 1 = 60

Now as per the second condition given in the selected number should not be divisible by 30 so we need to eliminate them from the set of favorable events.

So number of all 3-digit numbers divisible by 30 = `\frac{990-120}{30}` + 1 = 30

Now number of favorable events = 300 + 180 – 60 – 30 = 390

Hence required probability = `\frac{390}{900}` = `\frac{13}{30}`

Question 74.) There are two bags A and B. Bag A contains 5 yellow and 3 red balls and bag B contains 5 tickets numbered from 1 to 5. The tickets drawn from the bag B at random checked for the number printed on it, the same number of balls are drawn from bag A at random with replacement. Then find the probability that all the balls drawn are yellow.

(A) `\frac{1}{5}`

(B) `\frac{1}{4}`

(C) `\frac{5}{8}`

(D) `\frac{1}{8}`

Answer. (B)

Solution.

Since it is given that only that much number of balls are drawn from the bag A which number appears on the ticket drawn from bag B and since the bag B have 5 tickets so here we will have 5 different cases,

1^st case, when number 1 appears on the ticket drawn from bag B then only 1 ball will be drawn from bag A. Then,

Number of ways of drawing a ball from bag A = ⁸C₁ = 8

And number of ways of drawing a yellow ball from bag A = ⁵C₁ = 5

Now the probability of drawing 1 yellow ball is = `\frac{5}{8}`

2^nd case, when number 2 appears on the ticket drawn from bag B then 2 balls will be drawn from bag A. Then,

Number of ways of drawing two balls from bag A = ⁸C₂ = 28

And number of ways of drawing two yellow balls from bag A = ⁵C₂ = 10

Now the probability of drawing two yellow balls is = `\frac{10}{28}` = `\frac{5}{14}`

3^rd case, when number 3 appears on the ticket drawn from bag B then three balls will be drawn from bag A. Then,

Number of ways of drawing three balls from bag A = ⁸C₃ = 56

And number of ways of drawing three yellow balls from bag A = ⁵C₃ = 10

Now the probability of drawing three yellow balls is = `\frac{10}{56}` = `\frac{5}{28}`

4^th case, when number 4 appears on the ticket drawn from bag B then four balls will be drawn from bag A. Then,

Number of ways of drawing four balls from bag A = ⁸C₄ = 70

And number of ways of drawing four yellow balls from bag A = ⁵C₄ = 5

Now the probability of drawing four yellow balls is = `\frac{5}{70}` = `\frac{1}{14}`

5^th case, when number 5 appears on the ticket drawn from bag B then five balls will be drawn from bag A. Then,

Number of ways of drawing five balls from bag A = ⁸C₅ = 56

And number of ways of drawing five yellow balls from bag A = ⁵C₅ = 1

Now the probability of drawing five yellow balls is = `\frac{1}{56}`

We also need the probability of drawing a ticket from bag B. Since it has total five tickets and we need to draw only one ticket. So the probability = `\frac{1}{5}`

Hence the required probability = `\frac{1}{5}` × (`\frac{5}{8}` + `\frac{5}{14}` + `\frac{5}{28}` + `\frac{1}{14}` + `\frac{1}{56}`)= `\frac{1}{5}` × `\frac{70}{56}` = `\frac{1}{4}`

Question 75.) There are 6 apples in a basket. If two apples are picked from it at random and both are found to be rotten, then find the probability that at least 5 out of 6 apples in the basket are rotten.

(A) `\frac{1}{3}`

(B) `\frac{5}{6}`

(C) `\frac{2}{5}`

(D) `\frac{5}{7}`

Answer. (D)

Solution.

Here we don’t know how many rotten apples are there in the basket. So the total events will be all those events that include picking two rotten apples from the basket. So here we will have 5 different cases,

1^st case, if there are 2 rotten apples in the basket, then the number of ways of picking 2 rotten apples from 2 rotten apples = ²C₂ = 1

2^nd case, if there are 3 rotten apples in the basket, then the number of ways of picking 2 rotten apples from 3 rotten apples = ³C₂ = 3

3^rd case, if there are 4 rotten apples in the basket, then the number of ways of picking 2 rotten apples from 4 rotten apples = ⁴C₂ = 6

4^th case, if there are 5 rotten apples in the basket, then the number of ways of picking 2 rotten apples from 5 rotten apples = ⁵C₂ = 10

5^th case, if there are 6 rotten apples in the basket, then the number of ways of picking 2 rotten apples from 6 rotten apples = ⁶C₂ = 15

So total number of events = 1+ 3 + 6 + 10 + 15 = 35

Note:- Here we have not considered the case of only one rotten apple in the basket because we are picking two apples at random.

Now we will calculate the number of favorable events on the basis of the condition given in the question that at least 5 apples in the basket are rotten. This condition means that either 5 apples are rotten or 6 apples are rotten. So the favorable will be the events of picking 2 rotten apples from either 5 rotten apples or 6 rotten apples. Therefore,

Number of favorable events = ⁵C₂ + ⁶C₂ = 10 + 15 = 25

Hence the required probability = `\frac{25}{35}` = `\frac{5}{7}`

Question 76.) A random variable A has the following probability distribution:

A	0	1	2	3	4	5
P(A)	0	N	2N	2N	2N2	4N2

Determine the value of N.

(A) `\frac{1}{4}`

(B) `\frac{4}{5}`

(C) `\frac{1}{6}`

(D) `\frac{1}{9}`

Answer. (C)

Solution.

As we know that the sum of probabilities of a probability distribution for random variables is always one.

So, `\sum_0^n` P(A) = 1

0 + N + 2N + 2N + 2N² + 4N² = 1

6N² + 5N – 1 = 0

6N² + 6N – N – 1 = 0

6N(N+1) – (N+1) = 0

(6N–1)(N+1) = 0

N = -1 or `\frac{1}{6}`

Since (-1) is not a valid value so N = `\frac{1}{6}`

Question 77.)  A random variable A has the following probability distribution:

A	0	1	2	3	4	5
P(A)	0	N	2N	2N	2N2	4N2

Determine the value of the variance of A.

(A) `\frac{47}{18}`

(B) `\frac{437}{324}`

(C) `\frac{147}{18}`

(D) `\frac{533}{324}`

Answer. (B)

Solution.

First we need to find the value of N and here the case is of a probability distribution for random variables, so we will use the following formula.

`\sum_0^n` P(A) = 1

0 + N + 2N + 2N + 2N² + 4N² = 1

On solving it we get, N = `\frac{1}{6}`

Now the formula for variance of A is,

Variance of A = Sum of all values of A²  P(A) – (Mean of A)²

So in order to calculate the variance of A we need the mean value of A and also the values of A  P(A) and A² × P(A) first.

A	0	1	2	3	4	5
P(A)	0	`\frac{1}{6}`	`\frac{1}{3}`	`\frac{1}{3}`	`\frac{1}{18}`	`\frac{1}{9}`
A  P(A)	0	`\frac{1}{6}`	`\frac{2}{3}`	1	`\frac{2}{9}`	`\frac{5}{9}`
A2  P(A)	0	`\frac{1}{6}`	`\frac{4}{3}`	3	`\frac{8}{9}`	`\frac{25}{9}`

Now the mean value of A will be given by sum of all values of A × P(A),

Mean of A = 0 + `\frac{1}{6}` + `\frac{2}{3}` + 1 + `\frac{2}{9}` + `\frac{5}{9}` = `\frac{47}{18}`

Sum of all values of A² × P(A) = 0 + `\frac{1}{6}` + `\frac{4}{3}` + 3 + `\frac{8}{9}` + `\frac{25}{9}` = `\frac{147}{18}`

Now the variance of A will be given by putting all values and further calculating we get,

Variance of A = `\frac{437}{324}`

Question 78.) A random variable A has the following probability distribution:

A	0	1	2	3	4	5
P(A)	0	N	2N	2N	2N2	4N2

Determine the value of P(1<A<5).

(A) `\frac{16}{27}`

(B) `\frac{13}{18}`

(C) `\frac{11}{18}`

(D) `\frac{19}{36}`

Answer. (B)

Solution.

First we need to find the value of N and here the case is of a probability distribution for random variables, so we will use the following formula.

`\sum_0^n` P(A) = 1 = 1

0 + N + 2N + 2N + 2N² + 4N² = 1

On solving it we get, N = `\frac{1}{6}`

Now we need to calculate, P(1<A<5) and it will be given by,

P(1<A<5) = P(A=2) + P(A=3) + P(A=4) = 2N + 2N + 2N² = 4N + 2N²

Now putting N = `\frac{1}{6}` and calculating further we get,

P(1<A<5) = `\frac{13}{18}`

Question 79.) Consider all the information shown in following venn diagram and calculate the value of P(A∩B∩C̅).

(A) `\frac{1}{15}`

(B) `\frac{1}{10}`

(C) `\frac{1}{5}`

(D) `\frac{1}{18}`

Answer. (A)

Solution.

First we will calculate the total number of events with the help of data shown in the venn diagram in the question, we will get,

Total number of events = 46 + 10 + 9 + 7 + 6 + 5 + 4 + 3 = 90

For calculating the number of favourable events we will find the number satisfying the relation of A B C̅ and this area has been shown in the following venn diagram,

Now if we relate it with the image shown in the question then we get that the total number of favorable events = 6

Hence P(A∩B∩C̅) = `\frac{6}{90}` = `\frac{1}{15}`

Question 80.) A perfume company does a survey on a group of 280 people for two types of their perfumes, type A and type B. Among which `\frac{1}{5}` of the total people do not like any perfume, 72 like only perfume of type A, `\frac{3}{7}` of the total people like both the type of perfumes. If a person is chosen at random then find the probability that the selected person likes only the perfume of type B.

(A) `\frac{1}{7}`

(B) `\frac{5}{28}`

(C) `\frac{19}{70}`

(D) `\frac{4}{35}`

Answer. (D)

Solution.

Total number of people = 280

Number of people who do not like any type of perfume = 280    = 56

Number of people who like only type A perfume = 72

Number of people who like both the types of perfumes = 280    = 120

So number of people who like only type B perfume = 280 – (56 + 72 + 120) = 280 – 248 = 32

Now the venn diagram representation of the above data is as following,

Now for the required probability we have,

Total number of events = Total number of people = 280

Number of favorable events = Number of people who like only type B perfume = 32

Hence required probability = `\frac{32}{280}` = `\frac{4}{35}`