Bunuel wrote:
A box contains 5 radio tubes of which 2 are defective. The tubes are tested one after another until 2 defective tubes are discovered. If the process stopped on the 3rd test, what is the probability that the 1st tube is nondefective?
A. 1/10
B. 3/10
C. 1/3
D. 1/2
E. 2/3
This is a conditional probability problem, which can be solved using:
(The probability that the 1st tube is nondefective when the process stopped on the 3rd test) divided by
(The sum of all probabilities for the event that the process is stopped on the 3rd test)
Let D denote a defective tube and N a non-defective tube. There are two ways the process can be stopped on the 3rd test: NDD and DND. (Note that the third tube must be defective, since the process was stopped after the third test.) The probability of the former is P(NDD) = 3/5 x 2/4 x 1/3 = 1/10 and the probability of the latter is P(DND) = 2/5 x 3/4 x 1/3 = 1/10. Therefore, the probability that the 1st tube is nondefective, given that the process stopped on the 3rd test is:
(1/10) / (1/10 + 1/10) = (1/10) / (2/10) = 1/2
Answer: D
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