Question

A lot contains $20$ articles. The probability that the lot contains exactly $2$ defective articles is $0.4$ and the probability that the lot contains exactly $3$ defective articles is $0.6$. Article are drawn from the lot at random one by one without replacement and are tested till all defective articles are found. What is the probability that the testing procedure ends at the twelfth testing?

Answer 1

Let $A_1$ denotes that the lot contains 2 defective articles
$A_2$ denotes that the lot contains 2 defective articles
$A$ the event that testing procudure ends with the twelfth testing
Now $P\left(A_1\right)=0.4$ and $P\left(A_2\right)=0.6$
The testing procedure ending at the twelfth testing means that, one defective article must be found in the first eleven testing and the remaining one must be found at the twelfth testing, in case the lot contains 2 defective articles
So, $P\left(A|A_1\right)=\frac{{}^2{C}_1{.}^{18}{C}_{10}}{{}^{20}{C}_{11}}\times \frac{1}{9}$
Similarly $P\left(A|A_2\right)=\frac{{}^3{C}_2{.}^{17}{C}_9}{{}^{20}{C}_{11}}\times \frac{1}{9}$
Thus the required probability
$P\left(A\right)=P(A\cap A_1)+P(A\cap A_2)=P\left(A_1\right)P\left(A|A_1\right)+P\left(A_2\right)P\left(A|A_2\right)=0.4\frac{{}^2{C}_1{.}^{18}{C}_{10}}{{}^{20}{C}_{11}}\times \frac{1}{9}+0.6\frac{{}^3{C}_2{.}^{17}{C}_9}{{}^{20}{C}_{11}}\times \frac{1}{9}=\frac{44}{1900}+\frac{66}{2280}=\frac{99}{1900}=0.0521$