Question

A lot of 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 are tested till all defective articles are found. What is the probability that the testing procedure ends at the twelfth testing?

• A. $\frac{99}{1900}$
• B. $\frac{99}{950}$
• C. $\frac{198}{1900}$
• D. $\frac{99}{1000}$

Answer 1

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

Let $A_1=$ the event that 'the lot contain 2 defective articles'
$A_2=$ the event that 'the lot contains 3 defective articles'
$A=$ the event that ' the testing procedure 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 contain 2 defective articles.
So, $P\left(\frac{A}{A_1}\right)=\frac{{}^2{C}_1{\times }^{18}{C}_{10}}{{}^{20}{C}_{11}}\times \frac{1}{9}$

Similarly, $P\left(\frac{A}{A_2}\right)=\frac{{}^3{C}_2{\times }^{17}{C}_7}{{}^{20}{C}_{11}}\times \frac{1}{9}$

Thus, the required probability
$=P\left(A\right)=P(A\cap A_1)+P(A\cap A_2)=P\left(A_1\right).P\left(\frac{A}{A_1}\right)+P\left(A_2\right).P\left(\frac{A}{A_2}\right)$

$=0.4\times \frac{{}^2{C}_1{\times }^{18}{C}_{10}}{{}^{20}{C}_{11}}\times \frac{1}{9}+0.6\times \frac{{}^3{C}_2{\times }^{17}{C}_7}{{}^{20}{C}_{11}}\times \frac{1}{9}$

$=0.4\times \frac{11}{190}+0.6\times \frac{11}{228}=\frac{44}{1900}+\frac{66}{2280}=\frac{99}{1900}$