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 until all the defective articles are found. What is the probability that the testing procedure stops at the twelfth testing?

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Solution

Suppose $A$ is the event that the testing procedure ends at the twelfth testing.
$A_{1} = {t h e e v e n t t h a t t h e l o t c o n t a i n s 2 d e f e c t i v e a r t i c l e s}$
$A_{2} = {t h e e v e n t t h a t t h e l o t c o n t a i n s 3 d e f e c t i v e a r t i c l e s}$
$P (A_{1}) = 0.4, P (A_{2}) = 0.6,$
$P (A) = P (A_{1}) P (A / A_{1}) + P (A_{2}) P (A / A_{2})$
$= 0.4 ⎧ ⎨ ⎩ \frac{_{C_{10}}^{28}^{2} C_{1}}{^{20} C_{11}} . \frac{1}{9} ⎫ ⎬ ⎭ + 0.6 {\frac{^{17} C_{9} .^{3} C_{2}}{^{20} C_{11}} . \frac{1}{9}}$
$= \frac{1}{9} {\frac{4}{10} . \frac{18!}{10! 8!} . \frac{11! 9!}{20!} .2 + \frac{6}{10} . \frac{17!}{9! 8!} . \frac{11! 9!}{20!} .3}$
$= \frac{1}{9} {\frac{4}{10} \times \frac{11 \times 9}{19 \times 20} .2 + \frac{6}{10} \times \frac{9 \times 10 \times 11}{18 \times 19 \times 20} .3}$
$= \frac{44}{1900} + \frac{55}{1900} = \frac{99}{1900}$

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