Probability — Page 16 | JEE Mathematics

Q151

Two number

\mathrm{k}_1

and

\mathrm{k}_2

are randomly chosen from the set of natural numbers. Then, the probability that the value of

\mathrm{i}^{\mathrm{k}_1}+\mathrm{i}^{\mathrm{k}_2},(\mathrm{i}=\sqrt{-1})

is non-zero, equals

A

\dfrac{3}{4}

B

\dfrac{1}{2}

C

\dfrac{1}{4}

D

\dfrac{2}{3}

Correct Answer

Option A

Solution

$i^{k_1}+i^{k_2} \neq 0 \quad i^{k_1} \rightarrow 4$ option for $\mathrm{i},-1,-\mathrm{i}, 1$ Total cases $\Rightarrow 4 \times 4=16$ Unfovourble cases $\Rightarrow \mathrm{i}^{\mathrm{k}_1}+\mathrm{i}^{\mathrm{k}_2}=0$

\left\{\begin{array}{c} 1,-1 \\ -1,1 \\ i,-i \\ -i, i \end{array}\right\}

4 Cases $\Rightarrow$ Probability $=\dfrac{16-4}{16}=\dfrac{3}{4}$

Q152

Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If

x

denote the number of defective oranges, then the variance of

x

is

A

26 / 75

B

14/25

C

18 / 25

D

28 / 75

Correct Answer

Option D

Solution

\begin{aligned} &\text{ Probability distribution }\\ &\begin{array}{c|c} \mathrm{x}_{\mathrm{i}} & \mathrm{p}_{\mathrm{i}} \\ \\ \mathrm{x}=0 & \frac{{ }^7 \mathrm{C}_2}{{ }^{10} \mathrm{C}_2}=\frac{42}{90} \\ \\ \mathrm{x}=1 & \frac{{ }^7 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^{10} \mathrm{C}_2}=\frac{42}{90} \\ \\ \mathrm{x}=2 & \frac{{ }^3 \mathrm{C}_2}{{ }^{10} \mathrm{C}_2}=\frac{6}{90} \end{array} \end{aligned}

\begin{aligned} &\text{ Now, }\\ &\begin{aligned} & \mu=\sum \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=\frac{42}{90}+\frac{12}{90}=\frac{54}{90} \\ & \sigma^2=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_1^2-\mu^2=\frac{42}{90}+\frac{24}{90}-\left(\frac{54}{90}\right)^2 \\ & \Rightarrow \frac{66}{90}-\left(\frac{54}{90}\right)^2 \\ & \sigma^2 \Rightarrow \frac{28}{75} \therefore(1) \end{aligned} \end{aligned}

Q153

A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is

\dfrac{m}{n}

,

\gcd(m, n) = 1

, then

n^2 - m^2

is equal to :

A 64

B 80

C 60

D 72

Correct Answer

Option B

Solution

Define the Events: Let $A$ be the event that an unbiased coin is drawn.

Let $B$ be the event that the coin with heads on both sides is drawn.

Let $H$ be the event that a head turns up when the coin is tossed.

State the Given Probabilities: $P(A) = \dfrac{19}{20}$ (since there are 19 unbiased coins out of 20 total coins) $P(B) = \dfrac{1}{20}$ (since there is 1 coin with two heads out of 20 total coins) $P(H|A) = \dfrac{1}{2}$ (probability of getting a head given an unbiased coin is drawn) $P(H|B) = 1$ (probability of getting a head given the coin with two heads is drawn) Use Bayes' Theorem: We want to find $P(A|H)$ , which is the probability that the drawn coin was unbiased given that a head turned up.

Bayes' Theorem states:

P(A|H) = \frac{P(H|A) \cdot P(A)}{P(H)}

Calculate $P(H)$ : We can find $P(H)$ using the law of total probability:

P(H) = P(H|A) \cdot P(A) + P(H|B) \cdot P(B)

P(H) = \left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right) + (1) \cdot \left(\frac{1}{20}\right) = \frac{19}{40} + \frac{1}{20} = \frac{19}{40} + \frac{2}{40} = \frac{21}{40}

Apply Bayes' Theorem to find $P(A|H)$ :

P(A|H) = \frac{P(H|A) \cdot P(A)}{P(H)} = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{19}{20}\right)}{\frac{21}{40}} = \frac{\frac{19}{40}}{\frac{21}{40}} = \frac{19}{21}

Determine $m$ and $n$ : Since $P(A|H) = \dfrac{m}{n}$ and $\gcd(m, n) = 1$ , we have $m = 19$ and $n = 21$ .

Calculate $n^2 - m^2$ :

n^2 - m^2 = 21^2 - 19^2 = (21 + 19)(21 - 19) = (40)(2) = 80

Therefore, $n^2 - m^2 = 80$ . So, the correct answer is: Option B: 80

Q154

Let a random variable X take values 0, 1, 2, 3 with P(X=0)=P(X=1)=p, P(X=2)=P(X=3) and E(X2)=2E(X). Then the value of 8p−1 is :

A 2

B 0

C 3

D 1

Correct Answer

Option A

Solution

\begin{aligned} & 2 p+2 q=\frac{1}{2} \\ & p+q \\ & E\left(x^2\right)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 \cdot p+1 . p+4 \cdot q+9 q \\ & =p+13 q \\ & E(x)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 . p+1 . p+2 q+3 q=p+5 q \\ & p+13 q=2(p+5 q) \\ & p=3 q \\ & \text{ So, } q=\frac{1}{8} \& p=\frac{3}{8} \\ & \text{ So, } 8 p-1=2\quad\text{Option (2)} \end{aligned}

Q155

\text{ Given three indentical bags each containing } 10 \text{ balls, whose colours are as follows : }

\begin{array}{lccc} & \text{ Red } & \text{ Blue } & \text{ Green } \\ \text{ Bag I } & 3 & 2 & 5 \\ \text{ Bag II } & 4 & 3 & 3 \\ \text{ Bag III } & 5 & 1 & 4 \end{array}

A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is p and if the ball is Green, the probability that it is from bag III is

q

, then the value of

\left(\dfrac{1}{p}+\dfrac{1}{q}\right)

is:

A 6

B 9

C 7

D 8

Correct Answer

Option C

Solution

Probability that a Red ball comes from Bag I ( $p$ ): $p(B_1 / R) = \dfrac{p(B_1) \cdot p(R / B_1)}{p(R)}$ Substituting the values, $p(B_1 / R) = \dfrac{\dfrac{1}{3} \times \dfrac{3}{10}}{\dfrac{1}{3} \times \dfrac{3}{10} + \dfrac{1}{3} \times \dfrac{4}{10} + \dfrac{1}{3} \times \dfrac{5}{10}}$ Simplify to find $p$ : $= \dfrac{\dfrac{1}{10}}{\dfrac{1}{3}(1.2)} = \dfrac{1}{4}$ So, $p = \dfrac{1}{4}$ .

Probability that a Green ball comes from Bag III ( $q$ ): $p(B_3 / G) = \dfrac{p(B_3) \cdot p(G / B_3)}{p(G)}$ Substitute the values, $p(B_3 / G) = \dfrac{\dfrac{1}{3} \times \dfrac{4}{10}}{\dfrac{1}{3} \times \dfrac{5}{10} + \dfrac{1}{3} \times \dfrac{3}{10} + \dfrac{1}{3} \times \dfrac{4}{10}}$ Simplify to find $q$ : $= \dfrac{\dfrac{2}{15}}{\dfrac{1}{3} \times 1.2} = \dfrac{1}{3}$ So, $q = \dfrac{1}{3}$ .

Calculation of $\left(\dfrac{1}{p} + \dfrac{1}{q}\right)$ : $\dfrac{1}{p} = 4, \quad \dfrac{1}{q} = 3$ Therefore, $\left(\dfrac{1}{p} + \dfrac{1}{q}\right) = 4 + 3 = \boxed{7}$

Q156

The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is

A

\dfrac{129}{182}

B

\dfrac{17}{26}

C

\dfrac{19}{26}

D

\dfrac{103}{182}

Correct Answer

Option A

Solution

\begin{array}{lll} 3 E, & 1 D, & 8 P \\ 3 E, & 2 D, & 7 P \\ 4 E, & 1 D, & 7 P \\ 4 E, & 2 D, & 6 P \end{array}

\begin{aligned} & P=\frac{{ }^4 C_3 \cdot{ }^2 C_1 \cdot{ }^{10} C_8+{ }^4 C_3 \cdot{ }^2 C_2 \cdot{ }^{10} C_7+{ }^4 C_4 \cdot{ }^2 C_1 \cdot{ }^{10} C_7+{ }^4 C_4 \cdot{ }^2 C_2 \cdot{ }^{10} C_6}{{ }^{10} C_{12}} \\ & =\frac{129}{182} \end{aligned}

Q157

If the probability that the random variable

X

takes the value

x

is given by

P(X=x)=k(x+1) 3^{-x}, x=0,1,2,3 \ldots

, where

k

is a constant, then

P(X \geq 3)

is equal to

A

\dfrac{1}{9}

B

\dfrac{8}{27}

C

\dfrac{7}{27}

D

\dfrac{4}{9}

Correct Answer

Option A

Solution

To find $P(X \geq 3)$ , we first determine the constant $k$ using the total probability for $X$ .

The probability $P(X = x)$ is given by: $P(X = x) = k(x+1) \cdot 3^{-x}, \quad x = 0, 1, 2, 3, \ldots$ The total probability must equal 1: $s = \sum_{x = 0}^{\infty} k(x+1) \cdot 3^{-x}$ Calculating that series: $s = \dfrac{k}{3^0} + \dfrac{2k}{3} + \dfrac{3k}{3^2} + \ldots$ Therefore, dividing the series by 3: $\dfrac{s}{3} = \dfrac{k}{3} + \dfrac{2k}{3^2} + \ldots$ Subtracting these: $s - \dfrac{s}{3} = k + \dfrac{k}{3} + \dfrac{k}{3^2} + \ldots$ The resulting series is a geometric series: $\dfrac{2s}{3} = k \left( 1 + \dfrac{1}{3} + \dfrac{1}{3^2} + \ldots \right)$ The sum of the infinite geometric series is: $\dfrac{2s}{3} = k \cdot \dfrac{1}{1-\dfrac{1}{3}} = \dfrac{3k}{2}$ Equating: $s = \dfrac{9k}{4} = 1$ Thus, solving for $k$ : $k = \dfrac{4}{9}$ Next, compute $P(X \geq 3)$ : $P(X \geq 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))$ Calculating these: $P(X = 0) = k = \dfrac{4}{9}$ $P(X = 1) = \dfrac{2k}{3} = \dfrac{8}{27}$ $P(X = 2) = \dfrac{3k}{9} = \dfrac{4}{27}$ Adding these probabilities: $P(X = 0) + P(X = 1) + P(X = 2) = \dfrac{4}{9} + \dfrac{8}{27} + \dfrac{4}{27} = \dfrac{8}{9}$ Finally, calculate $P(X \geq 3)$ : $P(X \geq 3) = 1 - \dfrac{8}{9} = \dfrac{1}{9}$

Q158

A random variable X has the following probability distribution : .tg .tg X: 1 2 3 4 5 P(X): K2 2K K 2K 5K2 Then P(X > 2) is equal to :

A

{1 \over {6}}

B

{7 \over {12}}

C

{1 \over {36}}

D

{23 \over {36}}

Correct Answer

Option D

Solution

\sum\limits_{i = 1}^5 {P(X)}

= 1 $\Rightarrow$ K2 + 2K + K + 2K + 5K2 = 1 $\Rightarrow$ 6K2 + 5K – 1 = 0 $\Rightarrow$ (6K - 1)(k + 1) = 0 $\Rightarrow$ K =

{1 \over 6}

and K = -1(rejected) $\therefore$ P(X

>

2) = K + 2K + 5K2 =

{1 \over 6} + {2 \over 6} + {5 \over {36}}

=

{{23} \over {36}}

Q159

It is given that the events

A

and

B

are such that

P\left( A \right) = {1 \over 4},P\left( {A|B} \right) = {1 \over 2}

and

P\left( {B|A} \right) = {2 \over 3}.

Then

P(B)

is :

A

{1 \over 6}

B

{1 \over 3}

C

{2 \over 3}

D

{1 \over 2}

Correct Answer

Option B

Solution

Given that,

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

$\Rightarrow$

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

=

{1 \over 2}

.............. equation (1)

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

$\Rightarrow$

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

=

{2 \over 3}

.............. equation (2) Dividing equation (1) by equation (2) we get,

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

=

{3 \over 4}

$\Rightarrow$

{P\left( B \right)}

=

{4 \over 3}

$\times$

{P\left( A \right)}

=

{4 \over 3}

$\times$

{1 \over 4}

=

{1 \over 3}

$\therefore$ Option (B) is correct.

Q160

A random variable X has the following probability distribution : .tg .tg X 0 1 2 3 4 P(X) k 2k 4k 6k 8k The value of P(1

A

{4 \over 7}

B

{2 \over 3}

C

{3 \over 7}

D

{4 \over 5}

Correct Answer

Option A

Solution

$\because$ x is a random variable $\therefore$

k + 2k + 4k + 6k + 8k = 1

$\therefore$

k = {1 \over {21}}

Now, $$P(1