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Probability Part 2

Section: Mathematics

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Question : 23 of 100

Marks: +1, -0

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

[31-Jan-2024 Shift 1]

$\;\frac{37}{153}$
$\;\frac{57}{153}$
$\;\frac{47}{153}$
$\;\frac{40}{153}$

Solution: 👈: Video Solution

3 bad apples, 15 good apples.

Let

X

be no of bad apples

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

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

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

\;E(X)=0 \times \;\frac{105}{153}+1 \times \;\frac{45}{153}

+2 \times \;\frac{3}{153}=\;\frac{51}{153}

\;=\;\frac{1}{3}

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

\;=0 \times \;\frac{105}{153}+1 \times \;\frac{45}{153}+4 \times \;\frac{3}{153}

-\left(\;\frac{1}{3}\right)^2

=\;\frac{57}{153}-\;\frac{1}{9}=\;\frac{40}{153}

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