Question

A lot contains $20$ articles. The probability that the lot contains exactly $2$ defective articles is $0\cdot 4$ and that lot contains exactly $3$ defective articles is $0\cdot 6$. Articles are drawn from the lot at random one by one without replacement and are tested till all defective articles are found. The probability that testing procedure ends at the twelfth testing is $\frac{11k}{1900}$. Find the value of $k$ ?

Answer 1

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
$=P\left(A\right)=P(A\cap A_1)+P(A\cap A_2)=P\left(A_1\right).P\left(\frac{A}{A_1}\right)+P\left(A_2\right).P\left(\frac{A}{A_2}\right)$
$=0.4\times \frac{{}^2{C}_1{\times }^{18}{C}_{10}}{{}^{20}{C}_{11}}\times \frac{1}{9}+0.6\times \frac{{}^3{C}_2{\times }^{17}{C}_7}{{}^{20}{C}_{11}}\times \frac{1}{9}$
$=0.4\times \frac{11}{190}+0.6\times \frac{11}{228}=\frac{44}{1900}+\frac{66}{2280}=\frac{99}{1900}\therefore k=1$