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Bunuel
A speaks the truth 3 times out of 4, B speaks the truth 7 times out of 10. They both assert that a white ball is drawn from a bag containing 6 balls, all of different colors. What is the probability of the truth of the assertion.


A. \(\frac{12}{49}\)

B. \(\frac{3}{10}\)

C. \(\frac{21}{40}\)

D. \(\frac{3}{40}\)

E. None of these

Truth speaking rate of A = 3/4
Truth speaking rate of A = 7/10

Probability of truth assertion = (3/4)*(7/10) = 21/40

ANswer: Option C

Hey GMATinsight

Why do we not consider the probability of drawing a white ball (1/6) here?

IMO, answer should be E

Posted from my mobile device


The qquestion is not about it being white ball. Question is about two of them speaking truth simultanuously... hence th eprobability includes on the probability of them speaking truth together,
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The answer is clearly not 21/40. That's easy to see by just changing the question setup:

"A tells the truth 3/4 of the time, and B tells the truth 7/10 of the time. A bag contains one white marble, and a marble is picked from the bag. Both A and B assert that the marble is white. What is the probability they are telling the truth?"

Obviously the answer is 1, not 21/40. The prior likelihood of picking the white marble matters. It also matters what they say when they are not telling the truth. If A always says "that's not a marble, it's an avocado" when he lies, then when he says the marble selected is white, it must be true. If he tells plausible lies, then it becomes less certain. Nor do we even know there's a white marble in the bag (the question doesn't say there is), so it's possible the probability their assertion is true is zero.

The wording of the question is far too unclear for the question to be answerable, and it's not properly thought out if the answer is meant to be C.
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Bunuel

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