Question

The probability that an item produced by a factory is defective is p. From a certain lot, a sample of n items is drawn with replacement. If it contains no defective items, the lot is accepted, while if it has more than two defective items, the lot is rejected. If the sample has one or two defective items, an independent sample of m items is drawn with replacement from the lot and combined with the previous sample. If the combined sample does not contain more than two defective items, the batch is accepted. The probability that the batch is accepted is $(q=1-p)$

• A. $q^n+np{q}^{n-1}\left(q^m+mp{q}^{m-1}\right)+\frac{n\left(n-1\right)}{2}{p}^2{q}^{m+n-2}$
• B. $np{q}^{n-1}\left(q^m+mp{q}^{m-1}\right)$
• C. $q^n+\frac{n\left(n-1\right)}{2}{p}^2{q}^{m+n-2}$
• D. None of these

Answer 1

Answer: (A)

Let $X$ be the number of defective items in the first sample of size $n$, and $Y$ the number of defective items in the sample of size $m$. Then $X\sim B(n,p)$ and $Y\sim B(m,p)$. The batch will be accepted in the following cases :
$(X=0);(X=1,Y\le 1);(X=2,Y=0).$
The probabilities of these events are as follows :
$P(X=0)=q^n;P(X=1,Y\le 1)=P(X=1)P(Y\le 1)$
$=P(X=1)[P(Y=0)+P(Y=1)]$
${=}^n{C}_{1}p{q}^{n-1}\left(q^m{+}^m{C}_{1}p{q}^{m-1}\right)=np{q}^{n-1}\left(q^m+mp{q}^{m-1}\right);$
and $P\left(X=2,Y=0\right)=P\left(X=2\right)P\left(Y=0\right)$
$=\left({}^n{C}_2{p}^2{q}^{n-2}\right)\left({}^m{C}_0{q}^m\right)=\frac{1}{2}n\left(n-1\right)p^2{q}^{m+n-2}$
Adding these three results, we see that the probability thatthe batch is accepted is
$q^n+np{q}^{n-1}\left(q^m+mp{q}^{m-1}\right)+\frac{n\left(n-1\right)}{2}{p}^2{q}^{m+n-2}$