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JEE Mathematics 2024 Question with Solution

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

  • A

    37153\frac{37}{153}

  • B

    57153\frac{57}{153}

  • C

    47153\frac{47}{153}

  • D

    40153\frac{40}{153}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: There are 33 rotten apples and 1515 good apples, so total apples =18= 18. Let XX be the number of rotten apples in a draw of 22 apples.

Find: The variance of XX.

Total number of ways to draw 22 apples from 1818 apples is

(182)=153\binom{18}{2} = 153

The probability distribution of XX is:

P(X=0)=(152)(182)=105153P(X=0)=\frac{\binom{15}{2}}{\binom{18}{2}}=\frac{105}{153} P(X=1)=(31)(151)(182)=45153P(X=1)=\frac{\binom{3}{1}\binom{15}{1}}{\binom{18}{2}}=\frac{45}{153} P(X=2)=(32)(182)=3153P(X=2)=\frac{\binom{3}{2}}{\binom{18}{2}}=\frac{3}{153}

Now compute the expected value:

E(X)=0⋅105153+1⋅45153+2⋅3153=51153=13E(X)=0\cdot \frac{105}{153}+1\cdot \frac{45}{153}+2\cdot \frac{3}{153}=\frac{51}{153}=\frac{1}{3}

Next,

E(X2)=02⋅105153+12⋅45153+22⋅3153=57153E(X^2)=0^2\cdot \frac{105}{153}+1^2\cdot \frac{45}{153}+2^2\cdot \frac{3}{153}=\frac{57}{153}

Using

Var⁡(X)=E(X2)−[E(X)]2\operatorname{Var}(X)=E(X^2)-[E(X)]^2

we get

Var⁡(X)=57153−(13)2=57153−19=40153\operatorname{Var}(X)=\frac{57}{153}-\left(\frac{1}{3}\right)^2=\frac{57}{153}-\frac{1}{9}=\frac{40}{153}

Therefore, the variance of XX is 40153\frac{40}{153}, so the correct option is D.

Using the distribution directly

Given: XX denotes the number of rotten apples when 22 apples are drawn from 33 rotten and 1515 good apples.

Find: Var⁡(X)\operatorname{Var}(X).

Possible values of XX are 0,1,20, 1, 2.

For X=0X=0, both apples must be good:

P(X=0)=(152)(182)=105153P(X=0)=\frac{\binom{15}{2}}{\binom{18}{2}}=\frac{105}{153}

For X=1X=1, one rotten and one good apple are chosen:

P(X=1)=(31)⋅(151)(182)=45153P(X=1)=\frac{\binom{3}{1}\cdot \binom{15}{1}}{\binom{18}{2}}=\frac{45}{153}

For X=2X=2, both apples are rotten:

P(X=2)=(32)(182)=3153P(X=2)=\frac{\binom{3}{2}}{\binom{18}{2}}=\frac{3}{153}

Now,

E(X)=0⋅105153+1⋅45153+2⋅3153=51153\begin{aligned} E(X) &= 0\cdot \frac{105}{153}+1\cdot \frac{45}{153}+2\cdot \frac{3}{153}\\ &= \frac{51}{153} \end{aligned}

Also,

E(X2)=02⋅105153+12⋅45153+22⋅3153=57153\begin{aligned} E(X^2) &= 0^2\cdot \frac{105}{153}+1^2\cdot \frac{45}{153}+2^2\cdot \frac{3}{153}\\ &= \frac{57}{153} \end{aligned}

Hence,

Var⁡(X)=E(X2)−[E(X)]2=57153−(51153)2=57153−260123409=612023409−260123409=351923409=40153\begin{aligned} \operatorname{Var}(X) &= E(X^2)-[E(X)]^2 \\ &= \frac{57}{153}-\left(\frac{51}{153}\right)^2 \\ &= \frac{57}{153}-\frac{2601}{23409} \\ &= \frac{6120}{23409}-\frac{2601}{23409} \\ &= \frac{3519}{23409}=\frac{40}{153} \end{aligned}

Thus, the correct option is D.

Common mistakes

  • Using permutations instead of combinations. Here only selection matters, not order, so use (182)\binom{18}{2}, not arrangements of two apples.

  • Computing E(X)E(X) correctly but forgetting that variance needs E(X2)−[E(X)]2E(X^2) - [E(X)]^2. Do not replace E(X2)E(X^2) with [E(X)]2[E(X)]^2.

  • Missing the factor 22 in the term for X=2X=2 while finding E(X)E(X), or the factor 222^2 while finding E(X2)E(X^2). Each probability must be multiplied by the corresponding value of the random variable.

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