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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 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.

Grade:10

## 1 Answers

Jitender Pal
askIITians Faculty 365 Points
6 years ago
Hello Student,
Please find the answer to your question
Let A1 be the event that the lot contains 2 defective articles and A2 the event that the lot contains 3 defective articles. Also let A be the event that the testing procedure ends at the twelfth testing. Then according to the question :
P (A1) = 0.4 and P (A2) = 0.6
Since 0 < P (A1) <1, 0 < P (A2) < 1, and P (A1) + P (A2) = 1
∴ The events A1, A2 form a partition of the sample space. Hence by the theorem of total probability for compound events, we have
NOTE THIS STEP:
P (A) = P (A1) P (AI A1) + P (A) P (AI A2) . . . . . . . . . . . . . . . . . . . . (1)
Here P (AI A1) is the probability of the event the testing procedure ends at the twelfth testing when the lot contains 2 defective articles. This is possible when out of 20 articles; first 11 draws must contain 10 non defective and 1 defective articles and 12th draw must give a defective article.
∴ P (AI A1) = 18C­10 x 2C1/20 C11 x 1/9 = 11/190
Similarly, P (AI A2) = 17 C9 x 3 C1/20 C11 x 1/9 = 11/228
Now substituting the values of P (AI A1) and P (AI A2) in eq. (1), we get
P (A) = 0.4 x 11/190 + 0.6 x 11/228 = 11/475 + 11/380 = 99/1900

Thanks
Jitender Pal
askIITians Faculty

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