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. Articles are drawn from the lot at random one by one without replacement and tested till all the defective articles are found. The probability that the testing procedure ends at the twelfth testing $\frac{100-k}{1900}$. Find the value of $k$.

Answer 1

Let $A_1$ be the event that the lot contains $2$ defective articles and $A_2$ the event the lot contains $3$ defective articles. Also let $A$ be the event that the testing procedure ends at the twelth testing. Then according to the question:
$P\left(A_1\right)=0.4$ and $P\left(A_2\right)=0.6$
Since $0<P\left(A_1\right)<1,0<P\left(A_2\right)<1,$ and $P\left(A_1\right)+P\left(A_2\right)=1$
the events $A_1,A_2$ form a partition of the sample space. Hence by the theorem of total probability for compound events, we have
$P\left(A\right)=P\left(A_1\right)P\left(A/A_1\right)+P\left(A_2\right)P\left(A/A_2\right)$ ...(1)But $P\left(A/A_1\right)=\left(\frac{{}^{18}{C}_{10}{\times }^2{C}_1}{{}^{20}{C}_{11}}\right)\times \left(\frac{1}{9}\right)$ ...(2)

$=\frac{2}{9}\times \frac{18!}{10!8!}\times \frac{11!\times 9!}{20!}=\frac{11}{190}.$
[Note that out of $20$ articles, first $11$ draws must contain $10$ non-defective and $1$ defective article and the ${12}^{th}$ draw must give a defective article. Probabilities of these two events are shown in separate parantheses in (2)].

Similarly, $P\left(A/A_2\right)=\frac{{}^{17}{C}_9{\times }^3{C}_1}{{}^{20}{C}_{11}}\times \frac{1}{9}$

$=\frac{1}{3}\times \frac{17!}{9!8!}\times \frac{11!.9!}{20!}=\frac{11}{228}.$

Now substituting the values of
$P\left(A_1\right),P\left(A_2\right),P\left(A/A_1\right)$ and $P\left(A/A_2\right)$ in (1), we get

$P\left(A\right)=0.4\times \frac{11}{190}+0.6\times \frac{11}{228}$

$=\frac{11}{475}+\frac{11}{380}=\frac{99}{1900}.$

Hence, value of $k$ is $1$.

Alternate Method:-

Let $A$ denote the event that the lot contains $2$ defective articles and $B$ the event that the lot contains $3$ defective articles.
$\therefore P\left(A\right)=0.4,P\left(B\right)=0.6$
Suppose C denotes the event that all the defective articles are found by the twelfth test, i.e., all but one defective article must be found in the first eleven tests and the last defective article must be found at the twelfth test.

$\therefore P(C\mid A)=\frac{\left({}^2{C}_1\right)\left({}^{18}{C}_{10}\right)}{{}^{20}{C}_{11}}\times \frac{1}{9}=\frac{11}{190}$

$P(C\mid B)=\frac{\left({}^3{C}_2\right)\left({}^{17}{C}_9\right)}{{}^{20}{C}_{11}}\times \frac{1}{9}=\frac{11}{228}$

$\therefore P\left(C\right)=P(C\cap A)+P(C\cap B)$

$=P\left(A\right)P(C\mid A)+P\left(B\right)P(C\mid B)$
$=0.4\times \frac{11}{190}+0.6\times \frac{11}{228}=\frac{99}{1900}.$