A box contains 5 radio tubes of which 2 are defective. The tubes are

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

Hi, I do not understand why the OA is D.

If the test stopped after inspecting the third tube and we need 2 defective tubes to stop the process that means that in order for the 1st tube to be non-defective, the next two tubes have to be defective:
\(\frac{3}{5}*\frac{2}{4}*\frac{1}{3} = \frac{1}{10}\)

The test stopped at the third tells us for sure that the 3rd was DEF. which means we are considering now only the 1st and 2nd = 50:50 = 1/2

Testing was stopped after 3rd draw which means the 2nd defective tube was drawn on the 3rd draw which means the first defective tube had to have been picked on either the 1st or 2nd draw. Therefore 1/2 --> D

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Question says, 2 defectives were found by 3rd test, so there must be 2 defectives(D) and 1 non-defective(N)
so let us find total possibilities of (2 Ds and 1 N)

total arrangements = (3!/2!) = 3
but we should NOT consider this combination, D D N, if it were the case, test would have stopped by 2nd test itself, as the question says, test stops as soon as 2 defectives are found,

so total arrangements = 2
(N D D) , (D N D)
out of above 2, only 1 combination has first test finding out to be defective, so probability is 1/2 (D)

Let's not overcomplicate this question: it's half logic, half math.

Since the process ends after the third step, we can infer two important things: 1) the 3rd step must be defective, and 2) at least one of the other steps must be defective.

This leaves us with only two possibilities: DND and NDD. In the first case, the probability of choosing D is 100%. In the second case, the probability of choosing D is 0%.

The average of 0% and 100% is 50%, or 1/2.­