Probability — Page 10 | JEE Mathematics

Q91

If a random variable X follows the Binomial distribution B(5, p) such that P(X = 0) = P(X = 1), then

{{P(X = 2)} \over {P(X = 3)}}

is equal to :

A 1

B 10

C 25

D 5

Correct Answer

Option D

Solution

Given,

n = 5

and

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

We know,

p(x = r) = {}^n{C_r}\,.\,{p^r}\,.\,{q^{n - r}}

where

p + q = 1

$\therefore$

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

\Rightarrow {}^5{C_0}\,.\,{p^0}\,.\,{q^5} = {}^5{C_1}\,.\,{p^1}\,.\,{q^4}

\Rightarrow 1\,.\,1\,.\,{(1 - p)^5} = 5\,.\,p\,.\,{(1 - p)^4}

\Rightarrow 1 - p = 5p

\Rightarrow 6p = 1

\Rightarrow p = {1 \over 6}

$\therefore$

q = 1 - p = 1 - {1 \over 6} = {5 \over 6}

Now,

{{P(X = 2)} \over {P(X = 3)}} = {{{}^5{C_2}\,.\,{p^2}\,.\,{q^3}} \over {{}^5{C_3}\,.\,{p^3}\,.\,{q^2}}}

= {{10\,.\,q} \over {10\,.\,p}}

= {5 \over 6} \times {6 \over 1}

= 5

Q92

Out of

60 \%

female and

40 \%

male candidates appearing in an exam,

60 \%

candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :

A

\dfrac{2}{3}

B

\dfrac{11}{16}

C

\dfrac{23}{32}

D

\dfrac{13}{16}

Correct Answer

Option A

Solution

P (Female)

= {{60} \over {100}} = {3 \over 5}

P (Male)

= {2 \over 5}

P (Female/Qualified)

= {{40} \over {60}} = {2 \over 3}

P (Male/qualified)

= {{20} \over {60}} = {1 \over 3}

Q93

If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

A

\dfrac{33}{2^{32}}

B

\dfrac{33}{2^{29}}

C

\dfrac{33}{2^{28}}

D

\dfrac{33}{2^{27}}

Correct Answer

Option C

Solution

If $n$ is number of trails, $p$ is probability of success and $q$ is probability of unsuccess then,

\begin{aligned} & \text{ Mean }=n p \text{ and variance }=n p q \text{. } \\\\ & \text{ Here }\\\\ & n p+n p q=24 \quad \dots(i)\\\\ & n p . n p q=128 \quad \dots(ii)\\\\ &\text{ and } q=1-p \quad \dots(iii) \end{aligned}

from eq. (i), (ii) and (iii) : $p=q=\dfrac{1}{2}$ and $n=32$ . $\therefore$ Required probability $=p(X=1)+p(X=2)$

\begin{aligned} & ={ }^{32} C_{1} \cdot\left(\frac{1}{2}\right)^{32}+{ }^{32} C_{2} \cdot\left(\frac{1}{2}\right)^{32} \\\\ & =\left(32+\frac{32 \times 31}{2}\right) \cdot \frac{1}{2^{32}} \\\\ & =\frac{33}{2^{28}} \end{aligned}

Q94

If the numbers appeared on the two throws of a fair six faced die are

\alpha

and

\beta

, then the probability that

x^{2}+\alpha x+\beta>0

, for all

x \in \mathbf{R}

, is :

A

\dfrac{17}{36}

B

\dfrac{4}{9}

C

\dfrac{1}{2}

D

\dfrac{19}{36}

Correct Answer

Option A

Solution

For $x^{2}+\alpha x+\beta>0 \forall x \in R$ to hold, we should have $\alpha^{2}-4 \beta<0$ If $\alpha=1, \beta$ can be 1, 2, 3, 4, 5, 6 i.e., 6 choices If $\alpha=2, \beta$ can be 2, 3, 4, 5, 6 i.e., 5 choices If $\alpha=3, \beta$ can be $3,4,5,6$ i.e., 4 choices If $\alpha=4, \beta$ can be 5 or 6 i.e., 2 choices If $\alpha=6$ , No possible value for $\beta$ i.e., 0 choices Hence total favourable outcomes

\begin{aligned} &=6+5+4+2+0+0 \\\\ &=17 \end{aligned}

Total possible choices for $\alpha$ and $\beta=6 \times 6=36$ Required probability $=\dfrac{17}{36}$

Q95

If

A

and

B

are two events such that

P(A)=\dfrac{1}{3}, P(B)=\dfrac{1}{5}

and

P(A \cup B)=\dfrac{1}{2}

, then

P\left(A \mid B^{\prime}\right)+P\left(B \mid A^{\prime}\right)

is equal to :

A

\dfrac{3}{4}

B

\dfrac{5}{8}

C

\dfrac{5}{4}

D

\dfrac{7}{8}

Correct Answer

Option B

Solution

P(A) = {1 \over 3},\,P(B) = {1 \over 5}

and

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

$\therefore$

P\left( {A \cap B} \right) = {1 \over 3} + {1 \over 5} - {1 \over 2} = {1 \over {30}}

Now,

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

= {{{9 \over {30}}} \over {{4 \over 5}}} + {{{5 \over {30}}} \over {{2 \over 3}}} = {5 \over 8}

Q96

The mean and variance of a binomial distribution are

\alpha

and

\dfrac{\alpha}{3}

respectively. If

\mathrm{P}(X=1)=\dfrac{4}{243}

, then

\mathrm{P}(X=4

or 5

)

is equal to :

A

\dfrac{5}{9}

B

\dfrac{64}{81}

C

\dfrac{16}{27}

D

\dfrac{145}{243}

Correct Answer

Option C

Solution

Given, mean

= np = \alpha

. and variance

= npq = {\alpha \over 3}

\Rightarrow q = {1 \over 3}

and

p = {2 \over 3}

P(X = 1) = n.{p^1}.{q^{n - 1}} = {4 \over {243}}

\Rightarrow n.{2 \over 3}.{\left( {{1 \over 3}} \right)^{n - 1}} = {4 \over {243}}

\Rightarrow n = 6

P(X = 4\,\mathrm{or}\,5) = {}^6{C_4}\,.\,{\left( {{2 \over 3}} \right)^4}\,.\,{\left( {{1 \over 3}} \right)^2} + {}^6{C_5}\,.\,{\left( {{2 \over 5}} \right)^5}\,.\,{1 \over 3}

= {{16} \over {27}}

Q97

Let

\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3}

be three mutually exclusive events such that

\mathrm{P}\left(\mathrm{E}_{1}\right)=\dfrac{2+3 \mathrm{p}}{6}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\dfrac{2-\mathrm{p}}{8}

and

\mathrm{P}\left(\mathrm{E}_{3}\right)=\dfrac{1-\mathrm{p}}{2}

. If the maximum and minimum values of

\mathrm{p}

are

\mathrm{p}_{1}

and

\mathrm{p}_{2}

, then

\left(\mathrm{p}_{1}+\mathrm{p}_{2}\right)

is equal to :

A

\dfrac{2}{3}

B

\dfrac{5}{3}

C

\dfrac{5}{4}

D 1

Correct Answer

Option B

Solution

0 \le {{2 + 3P} \over 6} \le 1 \Rightarrow P \in \left[ { - {2 \over 3},{4 \over 3}} \right]

0 \le {{2 - P} \over 8} \le 1 \Rightarrow P \in [ - 6,2]

0 \le {{1 - P} \over 2} \le 1 \Rightarrow P \in [ - 1,1]

0

0

P \in \left[ {{2 \over 3},{{26} \over 3}} \right]

Taking intersection of all

P \in \left[ {{2 \over 3},1} \right)

{P_1} + {P_2} = {5 \over 3}

Q98

Let

X

be a binomially distributed random variable with mean 4 and variance

\dfrac{4}{3}

. Then,

54 \,P(X \leq 2)

is equal to :

A

\dfrac{73}{27}

B

\dfrac{146}{27}

C

\dfrac{146}{81}

D

\dfrac{126}{81}

Correct Answer

Option B

Solution

Mean

= 4 = \mu = np

Variance

= {\sigma ^2} = np(1 - P) = {4 \over 3}

4(1 - P) = {4 \over 3}

P = {2 \over 3}

n \times {2 \over 3} = 4

n = 6

P(X = k) = {}^n{C_k}\,{P^k}{(1 - P)^{n - k}}

P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)

= {}^6{C_0}{P^0}{(1 - P)^6} + {}^6{C_1}{P^1}{(1 - P)^5} + {}^6{C_2}{P^2}{(1 - P)^4}

= {}^6{C_0}{\left( {{1 \over 3}} \right)^6} + {}^6{C_1}\left( {{2 \over 3}} \right){\left( {{1 \over 3}} \right)^5} + {}^6{C_2}{\left( {{2 \over 3}} \right)^2}{\left( {{1 \over 3}} \right)^4}

= {\left( {{1 \over 3}} \right)^6}[1 + 12 + 60] = {{73} \over {{3^6}}}

54P\,(X \le 2) = {{73} \over {{3^6}}} \times 54 = {{146} \over {27}}

Q99

Let

S

be the sample space of all five digit numbers. It

p

is the probability that a randomly selected number from

S

, is a multiple of 7 but not divisible by 5 , then

9 p

is equal to :

A 1.0146

B 1.2085

C 1.0285

D 1.1521

Correct Answer

Option C

Solution

Among the 5 digit numbers, First number divisible by 7 is 10003 and last is 99995.

$\Rightarrow$ Number of numbers divisible by 7.

= {{99995 - 10003} \over 7} + 1

= 12857

First number divisible by 35 is 10010 and last is 99995. $\Rightarrow$ Number of numbers divisible by 35

= {{99995 - 10010} \over {35}} + 1

= 2572

Hence number of number divisible by 7 but not by 5

= 12857 - 2572

= 10285

9P. = {{10285} \over {90000}} \times 9

= 1.0285

Q100

Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If

P(X>n-3)=\dfrac{k}{2^{n}}

, then k is equal to :

A 528

B 529

C 629

D 630

Correct Answer

Option B

Solution

Mean

= np = 16

Variance

= npq = 8

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

and

n = 32

P(x > n - 3) = p(x = n - 2) + p(x = n - 1) + p(x = n)

= \left( {{}^{32}{C_2} + {}^{32}{C_1} + {}^{32}{C_0}} \right)\,.\,{1 \over {{2^n}}}

= {{529} \over {{2^n}}}