Probability Questions — JEE Mathematics

Q1

The mean and variance of a random variable

X

having binomial distribution are

4

and

2

respectively, then

P(X=1)

is :

A

{1 \over 4}

B

{1 \over 32}

C

{1 \over 16}

D

{1 \over 8}

Correct Answer

Option B

Solution

Mean =

np

= 4 Variance =

npq

= 2

\Rightarrow 4 \times q

= 2 $\Rightarrow$

q

=

{1 \over 2}

p = 1 - q = 1 -

{1 \over 2}

=

{1 \over 2}

n = 8 Now P(X = 1) =

{}^8{C_1}{\left( {{1 \over 2}} \right)^1}{\left( {{1 \over 2}} \right)^{8 - 1}}

=

8 \times {1 \over {{2^8}}}

=

{1 \over {32}}

Q2

Let A and B be two non-null events such that A

\subset

B . Then, which of the following statements is always correct?

A P(A|B) = 1

B P(A|B) = P(B) – P(A)

C P(A|B)

\le

P(A)

D P(A|B)

\ge

P(A)

Correct Answer

Option D

Solution

P\left( {{A \over B}} \right) = {{P\left( {A \cap B} \right)} \over {P\left( B \right)}}

As A $\subset$ B, then P(A $\cap$ B) = P(A) $\therefore$

P\left( {{A \over B}} \right) = {{P\left( A \right)} \over {P\left( B \right)}}

As P(B) $\le$ 1 $\therefore$

{{P\left( A \right)} \over {P\left( B \right)}}

$\ge$ P(A)

Q3

A dice is tossed

5

times. Getting an odd number is considered a success. Then the variance of distribution of success is :

A

8/3

B

3/8

C

4/5

D

5/4

Correct Answer

Option D

Solution

Total no of trials = 5 $\therefore$ n = 5 Odd no possibe are = {1, 3, 5} =3 Sample space = {1, 2, 3, 4, 5, 6} = 6 Probablity of getting odd number =

{3 \over 6}

=

{1 \over 2}

$\therefore$ p =

{1 \over 2}

Formula for variance = npq where q = 1 - p = 1 -

{1 \over 2}

=

{1 \over 2}

$\therefore$ Variance =

5 \times {1 \over 2} \times {1 \over 2}

=

{5 \over 4}

Q4

A problem in mathematics is given to three students

A,B,C

and their respective probability of solving the problem is

{1 \over 2},{1 \over 3}

and

{1 \over 4}.

Probability that the problem is solved is :

A

{3 \over 4}

B

{1 \over 2}

C

{2 \over 3}

D

{1 \over 3}

Correct Answer

Option A

Solution

Given

P\left( A \right) = {1 \over 2}

,

P\left( B \right) = {1 \over 3}

,

P\left( C \right) = {1 \over 4}

So,

P\left( {\overline A } \right) = {1 \over 2}

(Probablity that the problem can't be solve by A)

P\left( {\overline B } \right) = {2 \over 3}

(Probablity that the problem can't be solve by B) and

P\left( {\overline C } \right) = {3 \over 4}

(Probablity that the problem can't be solve by C) Now the probablity that the problem is solved by any one student of A, B and C = 1 - the probablity that the problem is solved by none of the students of A, B and C

P\left( {A \cup B \cup C} \right)

= 1 -

P\left( {\overline A } \right)P\left( {\overline B } \right)P\left( {\overline C } \right)

= 1 - {1 \over 2} \times {2 \over 3} \times {3 \over 4}

$=$

{3 \over 4}

Q5

The probability that

A

speaks truth is

{4 \over 5},

while the probability for

B

is

{3 \over 4}.

The probability that they contradict each other when asked to speak on a fact is :

A

{4 \over 5}

B

{1 \over 5}

C

{7 \over 20}

D

{3 \over 20}

Correct Answer

Option C

Solution

The probability of speaking truth by A, P(A) =

{4 \over 5}

. The probability of not speaking truth by A, P(

\overline A

) = 1 $-$

{4 \over 5} = {1 \over 5}

. The probability of speaking truth by B, P(B) =

{3 \over 4}

. The probability of not speaking truth of B, P(

\overline B

)

= {1 \over 4}

. The probability that they contradict each other

= P(A) \times P(\overline B ) + P(\overline A ) \times P(B)

= {4 \over 5} \times {1 \over 4} + {1 \over 5} \times {3 \over 4}

= {1 \over 5} + {3 \over {20}} = {7 \over {20}}

Q6

At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of

5

phone calls during

10

minute time intervals. The probability that there is at the most one phone call during a

10

-minute time period is :

A

{6 \over {{5^e}}}

B

{5 \over 6}

C

{6 \over 55}

D

{6 \over {{e^5}}}

Correct Answer

Option D

Solution

Required probability

= P(X = 0) + P(X = 1)

= {{{e^{ - 5}}} \over {0!}}{5^0} + {{{e^{ - 5}}} \over {1!}}{5^1}

= {e^{ - 5}} + 5{e^{ - 5}}

= {6 \over {{e^5}}}

Q7

Four persons can hit a target correctly with probabilities

{1 \over 2}

,

{1 \over 3}

,

{1 \over 4}

and

{1 \over 8}

respectively. if all hit at the target independently, then the probability that the target would be hit, is :

A

{{25} \over {32}}

B

{{25} \over {192}}

C

{{1} \over {192}}

D

{{7} \over {32}}

Correct Answer

Option A

Solution

Let four persons are A, B, C and D. Probablity of hitting a target by them, P(A) =

{1 \over 2}

P(B) =

{1 \over 3}

P(C) =

{1 \over 4}

P(D) =

{1 \over 8}

Probablity of hitting target atleast once = 1 - Probablity of not hitting by anybody P(Hit) = 1 -

P\left( {\overline A \cap \overline B \cap \overline C \cap \overline D } \right)

= 1 -

P\left( {\overline A } \right).P\left( {\overline B } \right).P\left( {\overline C } \right).P\left( {\overline D } \right)

= 1 -

{1 \over 2}.{2 \over 3}.{3 \over 4}.{7 \over 8}

=

{{25} \over {32}}

Q8

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly

9

twice is :

A

8/729

B

8/243

C

1/729

D

8/9.

Correct Answer

Option B

Solution

Probability of getting score 9 in a single throw

= {4 \over {36}} = {1 \over 9}.

Probability of getting score 9 exactly twice

= {}^3{C_2} \times {\left( {{1 \over 9}} \right)^2} \times {8 \over 9} = {8 \over {243}}.

Q9

The mean and the variance of a binomial distribution are

4

and

2

respectively. Then the probability of

2

successes is :

A

{28 \over 256}

B

{219 \over 256}

C

{128 \over 256}

D

{37 \over 256}

Correct Answer

Option A

Solution

Given that mean = 4 = np = 4 and variance = 2 npq = 2 $\Rightarrow$ 4q = 2

\Rightarrow q = {1 \over 2}

$\therefore$

p = 1 - q = 1 - {1 \over 2} = {1 \over 2}

Also, n = 8 Probability of 2 successes

= P(X = 2) = {}^8{C_2}{p^2}{q^6}

= {{8!} \over {2! \times 6!}} \times {\left( {{1 \over 2}} \right)^2} \times {\left( {{1 \over 2}} \right)^6} = 28 \times {1 \over {{2^8}}}

= {{28} \over {256}}

Q10

A die is thrown. Let

A

be the event that the number obtained is greater than

3.

Let

B

be the event that the number obtained is less than

5.

Then

P\left( {A \cup B} \right)

is :

A

{3 \over 5}

B

0

C

1

D

{2 \over 5}

Correct Answer

Option C

Solution

No of outcome for a die = { 1, 2, 3, 4, 5, 6 } According to the question, A = { 4, 5, 6 } $\therefore$ P(A) =

{3 \over 6}

B = { 1, 2, 3, 4 } $\therefore$ P(A) =

{4 \over 6}

A $\cap$ B = { 4 } So P(A $\cap$ B) =

{1 \over 6}

We know,

P\left( {A \cup B} \right)

=

P\left( A \right)

+

P\left( B \right)

-

P\left( {A \cap B} \right)

$\therefore$

P\left( {A \cup B} \right)

=

{3 \over 6}

+

{4 \over 6}

-

{1 \over 6}

=

{{7 - 1} \over 6}

=

{6 \over 6}

= 1 $\therefore$ Option (C) is correct.