Search any question & find its solution
Question:
Answered & Verified by Expert
Two bad eggs are mixed accidentally with 10 good ones. If three eggs are drawn at random from this lot in succession without replacement, then the variance of the probability distribution of the number of bad eggs drawn is
Options:
Solution:
2581 Upvotes
Verified Answer
The correct answer is:
$\frac{15}{44}$
If $X$ is the number of bad eggs.
$$
\begin{aligned}
& \therefore \quad X=0,1,2 \\
& P(X=0)=P(\text { all } 3 \text { good eggs })=\frac{{ }^{10} C_3}{{ }^{12} C_3}=\frac{12}{22} \\
& P(X=1)=\frac{{ }^{10} C_2 \times{ }^2 C_1}{{ }^{12} C_3}=\frac{9}{22} \\
& P(X=2)=\frac{{ }^{10} C_1 \times{ }^2 C_2}{{ }^{12} C_3}=\frac{1}{22}
\end{aligned}
$$
$$
\begin{aligned}
& \bar{x}=\Sigma x \cdot P_i=\frac{1}{2} \\
& \text { Variance }=\Sigma x^2 P_i-(\bar{x})^2=\frac{13}{22}-\frac{1}{4}=\frac{15}{44}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \quad X=0,1,2 \\
& P(X=0)=P(\text { all } 3 \text { good eggs })=\frac{{ }^{10} C_3}{{ }^{12} C_3}=\frac{12}{22} \\
& P(X=1)=\frac{{ }^{10} C_2 \times{ }^2 C_1}{{ }^{12} C_3}=\frac{9}{22} \\
& P(X=2)=\frac{{ }^{10} C_1 \times{ }^2 C_2}{{ }^{12} C_3}=\frac{1}{22}
\end{aligned}
$$
$$
\begin{aligned}
& \bar{x}=\Sigma x \cdot P_i=\frac{1}{2} \\
& \text { Variance }=\Sigma x^2 P_i-(\bar{x})^2=\frac{13}{22}-\frac{1}{4}=\frac{15}{44}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.