Question

A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4 and the probability that it contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random one by one, without replacement and tested till all defective articles are found. The probability that the testing procedure ends at the 12th testing is

• A. $\frac{11}{1900}$
• B. $\frac{44}{1900}$
• C. $\frac{66}{1900}$
• D. $\frac{99}{1900}$

Answer 1

The correct option is D $\frac{99}{1900}$

Let A be the event that there are 2 defects, B be the event that there are 3 defects and C be the event that the testing ends on the twelfth trial (i.e. the last defect is found on the twelfth trial).
$P\left(A\right)=0.4$, $P\left(B\right)=0.6$.

$P\left(C|A\right)=\frac{{}_1^{2}C{\times }_{10}^{18}C\times 11!}{{}_{12}^{20}C\times 12!}=\frac{11}{19\times 10}$

$P\left(C|B\right)=\frac{{}_2^{3}C{\times }_9^{17}C\times 11!}{{}_{12}^{20}C\times 12!}=\frac{11}{19\times 12}$
Therefore, required probability = $P\left(C|A\right)\times P\left(A\right)+P\left(C|B\right)\times P\left(B\right)=0.4\times \frac{11}{190}+0.6\times \frac{11}{228}=\frac{99}{1900}$