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

Question

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

A. 1/10
B. 3/10
C. 1/3
D. 1/2
E. 2/3

chetan2u · Accepted Answer

Hi..

The process of picking stops when both defective bulbs are picked.

The process stops after the third pick, so the total events are not 5 balls but 3 balls that are picked up.
The next inference will be that third bulb picked was defective as the process stopped then, so the third becomes a defective bulb for sure.

After we know that the third is defective, we are left with 1st pick and 2nd pick, one of which is defective and other non-defective. We are to find whether the first pick was defective..
So our events are 2 - 1st pick and 2nd pick, which can be either DN(Defective-NONdefective) or ND (NONDefective-Defective)..

Finally what the question has become is probability of ND between two cases ND and DN, thus probability is 1 event out of 2 events or  \frac{1}{2}

Archit3110 · Answer

P of non defective and defective tube; 1/2
IMO D

NinetyFour · Answer

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}

Rafymono · Answer

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

nmccull · Answer

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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ScottTargetTestPrep · Answer

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

hellosanthosh2k2 · Answer

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)

mcelroytutoring · Answer

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.

bumpbot · Answer

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

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

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