Probability Ques 112
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 are tested till all defective articles are found. What is the probability that the testing procedure ends at the twelfth testing?
$(1986,5 M)$
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Answer:
Correct Answer: 112.$(\frac{99}{1900})$
Solution:
- The testing procedure may terminate at the twelfth testing in two mutually exclusive ways.
I : When lot contains $2$ defective articles.
II : When lot contains $3$ defective articles.
Let $A=$ testing procedure ends at twelth testing
$A _1=$ lot contains $2$ defective articles
$A _2=$ lot contains $3$ defective articles
$\therefore$ Required probability
$ =P\left(A _1\right) \cdot P\left(A / A _1\right)+P\left(A _2\right) \cdot P\left(A / A _2\right) $
Here, $P\left(A / A _1\right)=$ probability that first $11$ draws contain $10$ non-defective and one-defective and twelfth draw contains a defective article.
$ =\frac{{ }^{18} C _{10} \times{ }^{2} C _1}{{ }^{20} C _{11}} \times \frac{1}{9} $
$P\left(A / A _2\right)=$ probability that first $11$ draws contains $9$ non-defective and 2-defective articles and twelfth draw contains defective $=\frac{{ }^{17} C _9 \times{ }^{3} C _2}{{ }^{20} C _{11}} \times \frac{1}{9}$
$\therefore$ Required probability
$ \begin{aligned} & =(0.4) P\left(A / A _1\right)+0.6 P\left(A / A _2\right) \\ & =\frac{0.4 \times{ }^{18} C _{10} \times{ }^{2} C _1}{{ }^{20} C _{11}} \times \frac{1}{9}+\frac{0.6 \times{ }^{17} C _9 \times{ }^{3} C _2}{{ }^{20} C _{11}} \times \frac{1}{9}=\frac{99}{1900} \end{aligned} $
BETA
