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  • 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 fr

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 are found. The probability that the testing procedure stops at the twelfth testing, is
 

Option: 1

\frac{44}{1900}
 


Option: 2

\frac{55}{1900}
 


Option: 3

\frac{77}{1900}
 


Option: 4

\frac{99}{1900}


Answers (1)

best_answer

Suppose \mathrm{A} is the event that the testing procedure ends at the twelfh testing, \mathrm{A_1} is the event that the lot contains 2 defective and \mathrm{A_2} is the event that the lot contains 3 defective.
Then, \mathrm{P\left(A_1\right)=0.4 \: and \: P\left(A_2\right)=0.6}

\mathrm{P\left(A_1\right)=0.4\: and \: P\left(A_2\right)=0.6}

             \mathrm{=0.4\left(\frac{{ }^{18} C_{10} \cdot{ }^2 C_1}{{ }^{20} C_{11}} \cdot \frac{1}{9}\right)+0.6\left(\frac{{ }^{17} C_9 \cdot{ }^3 C_2}{{ }^{20} C_{11}} \cdot \frac{1}{9}\right) }

             \mathrm{=\frac{1}{9}\left[\frac{4}{10} \cdot \frac{18 !}{10 ! 18 !} \cdot \frac{11 ! 9 !}{20 !} \cdot 2+\frac{6}{10} \cdot \frac{7 !}{9 ! 18 !} \cdot \frac{11 ! 9 !}{20 !} \cdot 3\right] }

            \mathrm{=\frac{1}{9}\left[\frac{4}{10} \times \frac{11 \times 9}{19 \times 20} \times 2+\frac{6}{10} \times \frac{9 \times 10 \times 11}{18 \times 19 \times 20} \times 3\right] }

            \mathrm{ =\frac{44}{1900}+\frac{55}{1900}=\frac{99}{1900}}

Hence option 4 is correct.






 

Posted by

Divya Prakash Singh

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