From a lot of 20 baskets, which includes 6 defective baskets, a sample of 2 baskets is drawn at random one…
From a lot of 20 baskets, which includes 6 defective baskets, a sample of 2 baskets is drawn at random one by one without replacement. The expected value of number of defective basket is
$0.6$
$0.06$
$0.006$
$1.07$
Solution
Let X denotes the number of defective baskets.
\(\therefore\) Possible values of X are \(0,1,2\).
Now, selection of baskets is done without replacement, we get
\(\begin{aligned}
& P(X=0)=\frac{14}{20} \times \frac{13}{19}=\frac{182}{380} \\
& P(X=1)=\frac{14}{20} \times \frac{6}{19}+\frac{6}{20} \times \frac{14}{19}=\frac{168}{380} \\
& P(X=2)=\frac{6}{20} \times \frac{5}{19}=\frac{30}{380}
\end{aligned}\)
\(\therefore \quad\) Required Expected Value
\(\begin{aligned}
& =0 \times \mathrm{P}(\mathrm{X}=0)+1 \times \mathrm{P}(\mathrm{X}=1)+2 \times \mathrm{P}(\mathrm{X}=2) \\
& =\frac{168+60}{380}=\frac{228}{380}=0.6
\end{aligned}\)