Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable x to be the number of rotten apples in a draw of two apples, the variance of x is

Probability PYQ — JEE Main 2024 · 31 Jan · Shift 1 · Q38 with Solution

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Correct answer: B

Identify the distribution

We have:

Total apples $= 18$
Rotten apples $= 3$
Good apples $= 15$

Two apples are drawn without replacement.

Let $X$ be the number of rotten apples in the draw of two apples.

So $X$ can take values $0,1,2$ .

This is a hypergeometric distribution.

Find the probabilities

Total number of ways to draw $2$ apples from $18$ :

\binom{18}{2} = 153

(i) $P(X=0)$

Both apples are good:

P(X=0)=\frac{\binom{15}{2}}{\binom{18}{2}}=\frac{105}{153}

(ii) $P(X=1)$

One rotten and one good:

P(X=1)=\frac{\binom{3}{1}\binom{15}{1}}{\binom{18}{2}}=\frac{45}{153}

(iii) $P(X=2)$

Both apples are rotten:

P(X=2)=\frac{\binom{3}{2}}{\binom{18}{2}}=\frac{3}{153}

Check:

\frac{105+45+3}{153}=1

Compute $E(X)$

E(X)=0\cdot \frac{105}{153}+1\cdot \frac{45}{153}+2\cdot \frac{3}{153}

E(X)=\frac{45+6}{153}=\frac{51}{153}=\frac{1}{3}

Compute $E(X^2)$

E(X^2)=0^2\cdot \frac{105}{153}+1^2\cdot \frac{45}{153}+2^2\cdot \frac{3}{153}

E(X^2)=\frac{45+12}{153}=\frac{57}{153}=\frac{19}{51}

Compute the variance

\operatorname{Var}(X)=E(X^2)-[E(X)]^2

\operatorname{Var}(X)=\frac{57}{153}-\left(\frac{1}{3}\right)^2

Since

\frac{57}{153}=\frac{19}{51}

we get

\operatorname{Var}(X)=\frac{19}{51}-\frac{1}{9}

Taking LCM $153$ :

\frac{19}{51}=\frac{57}{153}, \qquad \frac{1}{9}=\frac{17}{153}

Thus,

\operatorname{Var}(X)=\frac{57-17}{153}=\frac{40}{153}

Match with options

\boxed{\operatorname{Var}(X)=\frac{40}{153}}

So the correct option is:

B. $\frac{40}{153}$

Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

They match.

More from Probability

View written solutionFree

Correct answer: B

Identify the distribution

We have:

Total apples $= 18$
Rotten apples $= 3$
Good apples $= 15$

Two apples are drawn without replacement.

Let $X$ be the number of rotten apples in the draw of two apples.

So $X$ can take values $0,1,2$ .

This is a hypergeometric distribution.

Find the probabilities

Total number of ways to draw $2$ apples from $18$ :

\binom{18}{2} = 153

(i) $P(X=0)$

Both apples are good:

P(X=0)=\frac{\binom{15}{2}}{\binom{18}{2}}=\frac{105}{153}

(ii) $P(X=1)$

One rotten and one good:

P(X=1)=\frac{\binom{3}{1}\binom{15}{1}}{\binom{18}{2}}=\frac{45}{153}

(iii) $P(X=2)$

Both apples are rotten:

P(X=2)=\frac{\binom{3}{2}}{\binom{18}{2}}=\frac{3}{153}

Check:

\frac{105+45+3}{153}=1

Compute $E(X)$

E(X)=0\cdot \frac{105}{153}+1\cdot \frac{45}{153}+2\cdot \frac{3}{153}

E(X)=\frac{45+6}{153}=\frac{51}{153}=\frac{1}{3}

Compute $E(X^2)$

E(X^2)=0^2\cdot \frac{105}{153}+1^2\cdot \frac{45}{153}+2^2\cdot \frac{3}{153}

E(X^2)=\frac{45+12}{153}=\frac{57}{153}=\frac{19}{51}

Compute the variance

\operatorname{Var}(X)=E(X^2)-[E(X)]^2

\operatorname{Var}(X)=\frac{57}{153}-\left(\frac{1}{3}\right)^2

Since

\frac{57}{153}=\frac{19}{51}

we get

\operatorname{Var}(X)=\frac{19}{51}-\frac{1}{9}

Taking LCM $153$ :

\frac{19}{51}=\frac{57}{153}, \qquad \frac{1}{9}=\frac{17}{153}

Thus,

\operatorname{Var}(X)=\frac{57-17}{153}=\frac{40}{153}

Match with options

\boxed{\operatorname{Var}(X)=\frac{40}{153}}

So the correct option is:

B. $\frac{40}{153}$

Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

They match.

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