Class 12th PCMB

Q101. A bag X contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black balls. One bag is

selected at random and a ball is drawn from it. Then, the probability chosen to be white is,

1.

2

15

2.

7

15

3.

8

15

4.

14

15

Ans: 4.

8

15

Solution:

A white ball can be drawn in two mutually exclusive ways:

( )

( )

( )

( )

( )

1. Selecting bag X and then drawing a white ball from it.

2. Selecting bag Y ane then drawing a white ball from it.

Let E , E and A be the three evenes as defined below:

E = Selecting abg X

E = Selecting bag Y

A = Drawing a white ball

We know that one bag is selected randomly.

Q102. A rifleman is firing at a distant target and has only 10% chance of hiting it. the least number of round he must fire

in order to have more than 50% chance of hitting it at least once is:

1. 11

2. 9

3. 7

4. 5

Ans: 3. 7

Solution:

Given p =

1

10 ⇒ q =

9

10

Let n be the number of rounds.

P(X ≥ 1) = 1 − P(X = 0)

⇒ P(X ≥ 1) ≥ 0.5

⇒ 1 − P(X = 0) ≥ 0.5

⇒ P(X = 0) ≤ 0.5

⇒ 0.9n ≤ 0.5

Using log table,

n ≤ 6.572 ≈ 7

He must fire in order to have more than

50% chance of hitting the target at least once.

Q103.

If P(A ∪ B) = 0.8 and P(A ∩ B) = 0.3 then P(

¯A

) = P(

¯B

) =

1. 0.3

2. 0.5

3. 0.7

4. 0.9

Ans: 4. 0.9

Solution:

If P(A ∪ B) = 0.8 P(A ∩ B) = 0.3,

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

⇒ P(A) + P(B) = P(A ∪ B) − P(A ∩ B)

⇒ P(A) + P(B) = 1.1

⇒ P(

¯A

) + P(

¯B

) = 1 − P(A) + 1 − P(B)

⇒ P(

¯A

) + P(

¯B

) = [P(A) + P(B)]

⇒ P(

¯A

) + P(

¯B

) = 2 − 1.1

⇒ P(

¯A

) + P(

¯B

) = 0.9

Q104. Choose the correct answer from the given four options.

A and B are two students. Their chances of solving a problem correctly are

1

3 and

1

4 , respectively. If the

probability of their making a common error is,

1

20 and they obtain the same answer, then the probability of their

answer to be correct is:

1.

1

12

2.

1

40

3.

13

120

1 2

1

2

4.

10

13

Ans: 4.

10

13

Solution:

Let E = Event that both A and B solve the problem

∴ P(E1) =

1

3 ×

1

4 =

1

12

Let E = Event that both A and B got incorrect solution of the problem

∴ P(E2) =

2

3 ×

3

4 =

1

2

Here, P

E1

E = 1, P

E

E2

=

1

20

∴ P

E1

E =

P ( E1 ∩ E )

P ( E ) =

P ( E1 ) ⋅ P

E1

E

P ( E1 ) ⋅ P

E1

E + P ( E2 ) ⋅ P

E1

E

=

1

12 × 1

1

12 × 1 +

1

2 ×

1

20

=

10

30

Q105. Two persons A and B take turns in throwing a pair of dice.The first person to throw 9 from both dice will be

awarded the prize. If A throws first, then the probability that B wins the game is.

1.

9

17

2.

8

17

3.

8

9

4.

1

9

Ans: 2.

8

17

Solution:

9 can be obtained from throw of two dice in only 4 cases as given below:

{(3, 6), (4, 5), (5, 4), (6, 3)]

⇒ P(getting 9) =

4

36 =

1

9

P(not getting 9) =

32

36 =

8

9

Now,

P(B is winning) = P(getting 9 in 2 throw) + P(getting p in 4 throw) + P(getting 9 in 6 throw) + …..

=

8

9 ×

1

9 +

8

9 ×

8

9 ×

8

9 ×

1

9 + . . . . .

=

8

81 1 +

64

81 +

64

81

2 + . . . . . .

=

8

81 ×

1

1 −

64

81

=

8

81 ×

81

17

=

8

17

Q106. Choose the correct answer from the given four options.

A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an

even number on the die and a spade card is:

1.

1

2

2.

1

4

3.

1

8

4.

3

4

Ans: 3.

1

8

1

2

( ) ( )

( ) ( )

( ) ( )

nd th th

[ ( ) ]

Solution:

Let E = Event for getting an even number on the die

And E = Event that a spade card is selected.

∴ P(E1) =

3

6 =

1

2 and P(E2) =

13

52 =

1

4

Then, P(E1 ∩ E2) = P(E1) ⋅ P(E2)

=

1

2 ⋅

1

4 =

1

8

Q107. Let X be a discrete random variable. Then the variance of X is:

1. E(X )

2. E(X ) + (E(X))

3. E(X ) – (E(X))

4. √E(X2) − (E(X))2

Ans: 4. E(X ) – (E(X))

Solution:

Since, the variance of a discrete random variable X is given by:

Var(X) = E(X ) – (E(X))

Hence, the correct alternative is option (c).

Q108.

If P(B) =

3

5 , P(A | B) =

1

2 and P(A ∪ B) =

4

5 , then P(

¯

A ∩ B) + P(

¯A

∩ B) =

1.

1

5

2.

4

5

3.

1

2

4. 1

Ans: 4. 1

Solution:

P(B) =

3

5 , P

A

B =

1

2 , P A ∪ B =

4

5

Consider,

P

A

B =

1

2

⇒

P ( A ∩ B )

P(B) =

1

2

⇒

P ( A ∩ B )

3

5

=

1

2

⇒ P(A ∩ B) =

3

10

P(

¯

A ∪ B) + P(

¯A

∩ B)

= 1 − P(A ∩ B) + P(B) − P(A ∩ B)

= 1 −

3

10 +

3

5 −

3

10

= 1

Q109. In a college 30% students fail in Physics, 25% fail in Mathenatics and 10% fail in both. One student is chosen at

random. The probability that she fails in Physics if she failed in Mathematics is.

1.

1

10

2.

1

3

3.

2

5

4.

9

20

Ans: 3.

2

5

Solution:

Let A be the event that students failed in Physics. B be the event that students failed in Mathematics.

1

2

2

2 2

2 2

2 2

2 2

( ) ( )

( )

Given that, P(A) = 30% =

30

100

P(B) = 25% =

25

100

P(A ∩ B) = 10% =

10

100

Required probability is given by P

A

B

⇒ P

A

B =

P ( A ∩ B )

P(B) =

10

100

25

100

=

2

5

Q110. Choose the correct answer from the given four options.

If P(A) =

2

5 , P(B) =

3

5 and P(A ∩ B) =

1

5 , then P

A ′

B ′ ⋅ P

B ′

A ′ is equas:

1.

5

6

2.

5

7

3.

25

42

4. 1

Ans: 1.

5

6

Solution:

Here, P(A) =

2

5 , P(B) =

3

5 and P(A ∩ B) =

1

5

P

A ′

B ′ =

P ( A ′ ∩ B ′ )

P ( B ′ )

+

1 − P ( A ∩ B )

1 − P ( B )

=

1 − [ P ( A ) + P ( B ) − A ( A ∩ B ) ]

1 − P ( B )

=

1 −

2

5 +

3

10 −

1

5

1 −

3

10

=

1 −

4 + 3 − 2

10

7

10

−

1 −

1

2

7

10

=

5

7

And P

B ′

A ′ =

P ( B ′ ∩ A ′ )

P ( A ′ )

=

1 − P ( A ∪ B )

1 − P ( A )

=

1 −

1

2

1 −

2

5

∵ P(A ∪ B) =

1

2

=

1

2

3

5

=

5

6

∴ P

A ′

B ′ ⋅ P

B ′

A ′ =

5

7 ⋅

5

6 =

25

42

Q111. Choose the correct answer from the given four options.

If two events are independent, then:

1. They must be mutually exclusive.

2. The sum of their probabilities must be equal to 1.

3. (a) and (b) both are correct.

4. None of the above is correct.

Ans: 4. None of the above is correct.

Solution:

If two events A and B are independent, then we know that

P(A ∩ B) = P(A) ⋅ P(B), P(A) ≠ 0, P(B) ≠ 0

Since, A and B have a common outcome.

Further, mutually exclusive events never have a common outcome.

( )

( )

( ) ( )

( )

( )

( )

( )

[ ]

( ) ( )

In other words, two independents events having non-zero probabilities of occurrence cannot be mutually exclusive and conversely,

i.e., two mutually exclusive events having non-zero probabilities of outcome cannot be independent.

Q112. Two events A and B will be independent, if

1. A and B are mutually exclusive

2. P(A ′ B ′ ) = [1 − P(A)][1 − P(B)]

3. P(A) = P(B)

4. P(A) + P(B) = 1

Ans: Two events A and B are said to be independent, if P(AB) = P(A) × P(B)

Distracter Rationale.

1. Let P(A) = m, P(B) = n, 0 < m, n < 1

A and B are mutually exclusive.

∴ A ∩ B = ϕ

⇒ P(AB) = 0

However, P(A) ⋅ P(B) = mn ≠ 0

∴ P(A). P(B) ≠ P(AB)

2. Consider the result given in alternative.

P(A ′ B ′ ) = [1 − P(A)][1 − P(B)]

⇒ P(A ′ ∩ B ′ ) = 1 − P(A) − P(B) + P(A). P(B)

⇒ 1 − P(A ∪ B) = 1 − P(A) − P(B) + P(A). P(B)

⇒ P(A ∪ B) = P(A) + P(B) − P(A) ⋅ P(B)

⇒ P(A) + P(B) − P(AB) = P(A) + P(B) − P(A). P(B)

⇒ P(AB) = P(A). P(B)

This implies that A and B are independent, if P(A ′ B ′ ) = [1 − P(A)][1 − P(B)]

3. Let A: Event of getting an odd number on throw of a die = {1, 3, 5}

⇒ P(A) =

3

6 =

1

2

B: Event of getting an even number on throw of a die = {2, 4, 6}

P(B) =

3

6 =

1

2

Here, A ∩ B = ϕ

∴ P(AB) = 0

P(A). P(B) =

1

4 ≠ 0

⇒ P(A). P(B) ≠ P(AB)

4. From the above example, it can be seen that,

P(A) + P(B) =

1

2 +

1

2 = 1

However, it cannot be inferred that A and B are independent.

Thus, the correct answer is B.

Q113. If X follows a binomial distribution with parameter n = 100 and p =

1

3 , then P(X = r) is maximum when r =

1. 32

2. 34

3. 33

4. 31

Ans: 3. 33

Solution:

n = 100, p =

1

3 ⇒ q =

2

3

np =

100

3 = 33 +

1

3

⇒ Probability is maximum at 33.

Q114.

If A and B are such that P(A ∪ B) =

5

9 and P(

¯A

∪

¯B

) =

2

3 , then P(

¯A

) + P(

¯B

) =

1.

9

10

2.

10

9

3.

8

9

4.

9

8

Ans: 2.

10

9

Solution:

P(A ∪ B) =

5

9 , P(

¯A

∪

¯B

) =

2

3

Consider,

P(

¯A

∪

¯B

) = P(

¯

A ∪ B)

⇒ P(

¯A

∪

¯B

) =

2

3

⇒ 1 − P(A ∩ B) =

2

3

⇒ P(A ∩ B) =

1

3

⇒ P(A) + P(B) − P(A ∪ B) =

1

3

⇒ P(A) + P(B) −

5

9 =

1

3

⇒ P(A) + P(B) =

8

9

P(

¯A

) + P(

¯B

) = 1 − P(A) + 1 − P(B)

⇒ P(

¯A

) + P(

¯B

) = 2 − [P(A) + P(B)]

⇒ P(

¯A

) + P(

¯B

) = 2 −

8

9

⇒ P(

¯A

) + P(

¯B

) =

10

9

Q115. Mark the correct alternative in the following question:

Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If

P(X = r )

P(X = n − r ) is independent of n and r, then p equals:

1.

1

2

2.

1

3

3.

1

5

4.

1

7

Ans: 1.

1

2

Solution:

Consider,

P(X = r) = kP(X = n−r)

Using nCr = nCn − r, q = 1 − p

prqn − r = kpn − rqr

pr(1 − p)n − r = kpn − r(1 − p)r

p2r − n = k(1 − p)2r − n

(p

q )2r − n = k

when p = q then k = 1

⇒ p = q =

1

2

Q116. Choose the correct answer from the given four options.

Two events E and F are independent. If P(E) = 0.3, P(E ∪ F) = 0.5, then P

E

F − P

F

E equal:

1.

2

7

2.

3

35

( ) ( )

3.

1

70

4.

1

7

Ans: 3.

1

70

Solution:

We have, P(E) = 0.3, P(E ∪ F) = 0.5

Also E and F are independent.

Now P(E ∪ F) = P(E) + P(F) − P(E ∩ F)

⇒ 0.5 = 0.3 + P(F) − 0.3P(F)

⇒ P(F) =

0.5 − 0.3

0.7 =

2

7

∴ P

E

F − P

F

E

= P(E) − P(F) (as E and F are indepandent)

=

3

10 −

2

7 =

1

70

Q117. The probability distribution of a discrete random variable X is given below:

X: 1 2 3 4

P(X): 1

10

1

5

3

10

2

5

The value of E(X ) is:

1. 3

2. 5

3. 7

4. 10

Ans: 4. 10

Solution:

X 1 2 3 4

P(X) 1

10

1

5

3

10

2

5

X2P(X)

1

10

4

5

27

10

32

5 E(X2) = 10

Q118. Choose the correct answer from the given four options.

The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is:

1.

7

64

2.

7

128

3.

45

1024

4.

7

41

Ans: 2.

7

128

Solution:

We know that, P(x = r) = nCr(P)r. (q)6n − r

Here, n = 10, p =

1

2 , q =

1

2

and r ≥ 8 i.e, r = 8, 9, 10

⇒ P(X = r) = P(r = 8) + P(r = 9) + P(r = 10)

= 10C8

1

2

8 1

2

10 − 8 + 10C9

1

2

9 1

2

10 − 9 + 10C10

1

2

10 1

2

10 − 10

=

10 !

8 ! 2 ! +

10 !

9 ! 1 ! + 1

1

2

10

= [45 + 10 + 1]

1

2

10

= 56

1

2

10 =

7

128

Q119.

( ) ( )

2

( ) ( ) ( ) ( ) ( ) ( )

( )( )

( )

( )

If A and B are two events such that P(A) ≠ 0 and P(B) ≠ 1, then P(

¯A

|

¯B

) =

1. 1 − P(A | B)

2. 1 − P(

¯A

| B)

3.

1 − P ( A ∪ B )

P (

¯B

)

4. =

P (

¯A

)

P (

¯B

)

Ans: 3.

1 − P ( A ∪ B )

P (

¯B

)

Solution:

We have,

P(A) ≠ 0 and P(B) ≠ 1

Now,

P(

¯A

|

¯B

) =

P (

¯A

∩

¯B

)

P (

¯B

)

=

P (

¯

A ∩ B )

P (

¯B

)

=

1 − P ( A ∪ B )

P (

¯B

)

Hence, the correct alternative is option (C).

Q120. A and B are two students. Their chances of solving a problem correctly are

1

3 and

1

4 respectively. If the probability

of their making common error is

1

20 and they obtain the same answer, then the probability of their answer to be

correct is.

1.

10

13

2.

13

120

3.

1

40

4.

1

12

Ans: 1.

10

13

Solution:

Let E be the event that Both A and B solve the problem.

A and B are independent,

⇒ P(E1) = P(A) × P(B)

⇒ P(E1) =

1

3 ×

1

4 =

1

12

Let E both A and B got wrong solution.

P(E2) = 1 −

1

3 × 1 −

1

4 =

1

2

Let E be the event getting same answer.

P

E

E1

= 1, P

E

E2

=

1

20

⇒ P

E1

E =

1

12 × 1

1

12 × 1 +

1

2 ×

1

20

=

10

13

Q121. Choose the correct answer from the given four options.

A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested,

the probability that both are dead is:

1.

33

56

2.

9

64

1

2

( ) ( )

( ) ( )

( )

3.

1

14

4.

3

28

Ans: 4.

3

28

Solution:

Required probability = PD ⋅ PD

=

3

8 ⋅

2

7 =

3

28

Q122. If two events are independent, then.

1. They must be mutually exclusive.

2. The sum of their probabilities must be equal to 1.

3. (a) and (b) both are correct.

4. None of the above is correctIf two. events are independent, then.

Ans: 4. None of the above is correctIf two. events are independent, then.

Solution:

Let A and B are two independent events, Then,

P(A ∩ B) = P(A) × P(B)

As, P(A ∩ B) ≠ 0 or P(A) + P(B) ≠ 1

So, both are neither mutually exclisive nor their sum of probability is 1.

Hence, the correct alternative is option (d).

Q123. Three persons, A, B and C fine a target in turn starting with A. Their probability of hitting the target are 0.4, 0.2

and 0.2, respectively. The probability of two hits is

1. 0.024

2. 0.452

3. 0.336

4. 0.188

Ans: 4. 0.188

Solution:

Let:

A be the event of hitting the target by the person A,

B be the event of hitting the target by the person B and

C be the event of hitting the target by the person C

We have,

P(A) = 0.4, P(B) = 0.3 and P(C) = 0.2

Also,

P(

¯A

) = 1 − P(A) = 1 − 0.4 = − 0.6,

P(

¯B

) = 1 − 0.3 = 0.7 and

P(

¯C

) = 1 − 0.2 = 0.8

Now,

P(Two hits) = P(AB

¯C

) + P(A

¯B

C) + P(

¯A

BC)

= P(A) × P(B) × P(

¯C

) + P(A) × (

¯B

) × P(C) + P(

¯A

) × P(B) × P(C)

= 0.4 × 0.3 × 0.8 + 0.4 × 0.7 × 0.2 + 0.6 × 0.3 × 0.2

= 0.096 + 0.056 + 0.036

= 0.188

Hence, the correct alternative is option (d).

Q124. Choose the correct answer in each of the following:

If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then

1. A ⊂ B

2. B ⊂ A

3. B = ϕ

4. A = ϕ

Ans: A ⊂ B

P(B|A) = 1

⇒

P ( B ∩ A )

P ( A ) = 1 P(B ∩ A) = P(A)

∴ (A) is correct answer.

Q125. 1∫0

√x(1 − x) dx equals:

1.

π

2

2.

π

4

3.

π

6

4.

π

8

Ans: 4.

π

8

Solution:

I =

1∫0

√x(1 − x) dx

I =

1∫0

√x−x2dx

I =

1∫0

1

4 + x − x2 +

1

4 dx

I =

1∫0

1

4 − x2 − x +

1

4 dx

I =

1∫0

1

2

2 − x −

1

2

2dx

I =

x −

1

2

2 √x(1 − x) +

1

2 ×

1

4 sin − 1(2x − 1) 10

I = 0 +

1

8 (sin − 1(1) − sin − 1( − 1))

I =

1

8

π

2 −

π

2

I =

π

8

Q126. A coin is tossed 4 times. The probability that at least one head turns up is:

1.

1

16

2.

2

16

3.

14

16

4.

15

16

Ans: 4.

15

16

Solution:

n = 4, p = q =

1

2

P(X ≥ 1) = 1 − P(X = 0)

P(X ≥ 1) = 1 − (1

2 )4

P(X ≥ 1) =

15

16

Q127. A fair coin is tossed 100 times. The probability of getting tails an odd nimber of times is:

1.

1

2

√√

( )

√( ) ( )

[ ]

( ( ))

2.

1

8

3.

3

8

4. None of these

Ans: 1.

1

2

Solution:

Here, n = 100

Let X denote the number of times a tail is obtained.

Here, p = q =

1

2

P(X = odd) = P(X = 1, 3, 5, …99)

= ( 100C1 + 100C3 + ⋯ + 100C99 )(1

2 )100

= Sum of odd coefficients in binomial expansion in (1 + x)100 (1

2 )100

=

2 ( 100 − 1 )

2100

=

1

2

Q128. Five persone entered the lift cabin on the ground floor of an 8 floor house. Suppose that each of them

independently and with equal probability can leave the cabin at any flor beginning with the first, then the

probability of all 5 persons leaving at different floors is,

1.

7P5

75

2.

75

7P5

3.

6

6P5

4.

5P5

5

Ans:

1.

7P5

75

Solution:

Five persons can leave different floors

By P ways.

Possible ways of leavinf the lift = 7

Required probability =

7P5

75

Q129. A die is thrown and a card is selected ar random from a deck pf 52 playing cards. The probability of getting an

even number of the die and a spade card is

1.

1

2

2.

1

4

3.

1

8

4.

3

4

Ans: 3.

1

8

Solution:

A Sample space when a die is thrown,

S = {1, 2, 3, 4, 5, 6} ⇒ n(S ) = 6

Let A be the event that getting even number.

A = {2, 4, 6} ⇒ n(A) = 3

⇒ P(A) =

3

6 =

1

2

A card is selected from a deck of 52 cards.

7

5

5

1 1

n(S2) = 52C2 = 52

Let B be the event that getting spade card.

n(B) = 13C2 = 13 ⇒ P(B) =

13

52 =

1

4

Required probability = P(A) × P(B)

=

1

2 ×

1

4 =

1

8

Q130. Choose the correct answer from the given four options.

Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If

P ( x = r )

P ( x = n – r ) is independent of n and r, then p equals:

1.

1

2

2.

1

3

3.

1

5

4.

1

7

Ans: 1.

1

2

Solution:

P(x = r) = nCr(p)r(q)n − r

=

n !

( n − r ) ! r ! (p)r(1 − p)n − r[ ∴ q = 1 − p]

Now,

P ( x = r )

P ( x = n − r ) =

nCrpr ( 1 − p )n − r

nCn − rpn − r ( 1 − p )r

=

Pr ( 1 − P )n − r

pn − r ( 1 − p )r [asnCr = nCn − r ]

=

1 − p

p

n − 2r

Above expression is independent of n and r, if

1 − p

p = 1 ⇒ p =

1

2

Q131. Choose the correct answer in each of the following:

The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on

one face is

1. 1

2. 2

3. 5

4.

8

3

Ans: xi pi pixi

1

2

5

3

62

61

6

3

64

65

6

Σ pixi =

12

6 = 2

Therefore, option (B) is correct.

Q132. An urn contains 9 balls two of which are red, three blue and four black. Three balls are drawn at random. The

probability that they are of the same colour is,

1.

5

84

2.

3

9

3.

3

7

4.

7

17

( )

Ans: 1.

5

84

Solution:

Given:

Red balls = 2

Blue balls = 3

Black balls = 4

P(All three balls are of same colour) = P(all three are blue) + P(all three are black)

=

3

9 ×

2

8 ×

1

7 +

4

9 ×

3

8 ×

2

7

=

1

84 +

4

84

=

5

84

Q133. The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least

0.8, is:

1. 7

2. 6

3. 5

4. 3

Ans: 4. 3

Solution:

A fair coin is tossed ⇒ p = q =

1

2

P(X ≥ 1) ≥ 0.8

⇒ 1 − P(0) ≥ 0.8

⇒ P(0) = 0.2

⇒ (1

2 )n = 0.2

⇒ 2 − n = 0.2

⇒ 2n ≥ 5

⇒ n ≥ 3

Q134. A bag contains 5 brown and 4 white socks. A man pulls out two socks. The probability that these are of the sane

colour is.

1.

5

108

2.

18

108

3.

30

108

4.

48

108

Ans: 3.

48

108

Solution:

Total number of balls = 5brown + 4white = 9

Required probability =

5

9 ×

4

8 +

4

9 ×

3

8 =

4

9

⇒

4 × 12

9 × 12 =

48

108

Q135. If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value

greater than 1 is:

1.

2

3

2.

4

5

3.

7

8

4.

15

16

Ans: 4.

15

16

Solution:

E(X) = 2, V(X) = 1

np = 2, npq = 1

⇒ q =

1

2 = p

⇒ n = 4

P(X ≥ 1) = 1 − P(X = 0)

P(X ≥ 1) = 1 − (1

2 )4

P(X ≥ 1) = 1 −

1

16 =

15

16

Q136. Choose the correct answer from the given four options.

If the events A and B are independent, then P(A ∩ B) is equal to:

1. P(A) + P(B)

2. P(A) − P(B)

3. P(A) ⋅ P(B)

4.

P ( A )

P ( B )

Ans: 3. P(A) − P(B)

Solution:

If A and B are independent, then P(A ∩ B) = P(A) ⋅ P(B)

Q137. If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six

successes in this distribution is:

1. 16C6 (1

4 )10 (3

4 )6

2. 16C6 (1

4 )6 (3

4 )10

3. 12C6 ( 1

20 )(3

4 )6

4. 12C6 ( 1

20 )6 (3

4 )6

Ans: 2. 16C6 (1

4 )6 (3

4 )10

Solution:

np = 4, npq = 3

⇒ q =

3

4 , p =

1

4 , n = 16

P(X = 6) = 16C6 (1

4 )6 (3

4 )10

Q138.

If A and B are two events such that P(A) =

3

8 , P(B) =

5

4 . and P(A | B) × P(

¯A

∩ B) is equals to.

1.

2

5

2.

3

8

3.

3

20

4.

6

25

Ans: 4.

6

25

Solution:

P(A) =

3

8 , P(B) =

5

8 , P A ∪ B =

3

4

P(A ∩ B) = P(A) + P(B) − P(A ∩ B)

⇒

3

4 =

3

8 +

5

8 − P(A ∩ B)

⇒ P(A ∩ B) =

1

4

( )

P

A

B P

¯A

B =

P ( A ∩ B )

P(B) ×

P (

¯A

∩ B )

P(B)

P

A

B P

¯A

B =

P ( A ∩ B )

P(B) ×

P(B) − P ( A ∩ B )

P(B)

P

A

B P

¯A

B =

1

4

5

8

×

( 5

8 −

1

4 )

5

8

P

A

B P

¯A

B =

6

25

Q139. A five-digit number is written down at raddom. The probability that the number is divisible by 5, and no two

consecutive digits are identical, is:

1.

1

5

2.

1

5 ( 9

10 )3

3. (3

5 )4

4. None of these

Ans: 4. None of these

Solution:

If last digit is either O or 5 then the number is divisible by 5.

Case : 1

Last digit is 0.

First three places can be selected by 9 × 9 × 9 = 729 ways.

If c = 0 then three places can be selected by 9 × 8 × 1 = 72

If C ≠ 0 then 729 – 72 = 657

Fourth place has 8 choices = 657 × 8 = 5256

Total = 72 + 5256 = 5904

Case : 2

If C = 5

First place other than 5

then first three places can be filled in 8 × 8 × 1 = 64

If first place is 5 then first three places can be filled in 1× 9 × 1 = 9 ways.

If third place is other than 5 then 729 – 64 – 9 = 656 ways.

For fourth place has 8 choices.

As per required condition = (64 + 9) × 9 + 656 × 8 = 5905

required probability =

5904 + 5905

9 × 10 × 10 × 10 × 10 =

11809

90000

NOTE: Answer not matching with back answer.

Q140. Choose the correct answer from the given four options.

X -4 -3 -2 -1 0

P(X) 0.1 0.2 0.3 0.2 0.2

For the following probability distribution E(X) is equal to:

1. 0

2. -1

3. -2

4. -1.8

Ans: 4. -1.8

Solution:

E(X) = Σ P(X)

= − 4 × (0.1) + ( − 3 × 0.2) + ( − 2 × 0.3) + ( − 1 × 0.2) + (0 × 0.2)

= − 0.4 − 0.6 − 0.6 − 0.2 = − 1.8

Q141. Choose the correct answer from the given four options.

If A and B are two independent events with P(A) =

3

5 and P(A) =

4

9 , then P(A’ ∩ B’) equals:

( ) ( )

( ) ( )

( ) ( )

( ) ( )

1.

4

15

2.

8

45

3.

1

3

4.

2

9

Ans: 4.

2

9

Solution:

Since A and B are independent events, A’ And B’ are aslo independent.

∴ P(A ′ ∩ B ′ ) = P(A) ⋅ P(B)

= [1 − P(A)][1 − P(B)]

= 1 −

3

5 1 −

4

9

=

2

5 ⋅

5

9 =

2

9

Q142. Choose the correct answer from the given four options.

X 1 2 3 4

P(X) 1

10

1

5

3

10

2

5

For the following probability distribution E(X ) is equal to:

1. 3.

2. 5.

3. 7.

4. 10.

Ans: 4. 10.

solution:

E(X2) = Σ X2P(X)

= 1 ⋅

1

10 + 4 ⋅

1

5 + 9 ⋅

3

10 + 16 ⋅

2

5

=

1

10 +

4

5 +

27

10 +

32

5

=

1 + 8 + 27 + 64

10 = 10

Q143. In a box containing 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs, none is

defective?

1. ( 9

10 )5

2.

9

10

3. 10 − 5

4. (1

2 )2

Ans: 1. ( 9

10 )5

Solution:

Let X denote the number of defective bulbs.

Hence, the binomial distribution is given by

n = 5, p =

10

100 =

1

10

& q =

90

100 =

9

10

Hence, the distribution is given by

P(X = r) = 5Cr ( 1

10 )r ( 9

10 )5 − r

∴ P(X = 0) = ( 9

10 )5

Q144. Choose the correct answer in each of the following:

( )( )

2

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then

the value of E(X) is

1.

37

221

2.

5

13

3.

1

13

4.

2

13

Ans: n(S) = 52, n(A) = 4

P(X = 0) =

48C2

52C2

=

48 × 47

52 × 51 =

188

221

P(X = 1) =

48C2 ×4C1

52C2

=

2 × 48 × 4

52 × 51 =

32

221

P(X = 2) =

4C2

52C2

=

4 × 3

52 × 51 =

1

221

xi pi pixi

0

1

2

188

221

32

221

1

221

0

32

221

2

221

Σ pixi =

34

221 =

2

13

Now E(X) =

2

13

Therefore, option (D) is correct.

Q145. If X is a binomial variate with parameters n and p, where 0 < p < 1 such that

P(X = r)

P(X = n – r ) is independent of n and r,

then p equals:

1.

1

2

2.

1

3

3.

1

4

4. None of these

Ans: 1.

1

2

Solution:

Consider,

P(X = r) = kP(X = n − r)

Using nCr = nCn − r, q = 1 − p

prqn − r = kpn − rqr

pr(1 − p)n − r = kpn − r(1 − p)r

P2r − n = k(1 − p)2r − n

(p

q )2r − n = k

when p = q then k = 1

⇒ p = q =

1

2

Q146. Choose the correct answer from the given four options.

In a college, 30% students fail in physics, 25% fail in mathematics and 10% fail in both. One student is chosen at

random. The probability that she fails in physics if she has failed in mathematics is:

1.

1

10

2.

2

5

3.

9

20

4.

1

3

Ans: 2.

2

5

Solution:

Here, P ( Ph ) =

30

100 =

3

10

P ( M ) =

25

100 =

1

4

And P ( M ∩ Ph ) =

10

100 =

1

10

∴ P

Ph

M =

P ( Ph ∩ M )

P ( M )

=

1

10

1

4

=

2

5

( )