You have six strips of paper. The word 'YES' is written on three on them. The word 'NO' is written on the other three Then all six are folded in a manner that they are identical.
Mr. A draws in the following manner. All six strips are tossed and Mr. A chooses one of them. The chosen strip is opened and the choice recorded. The strip is then folded again and a second draw is done in the same manner.
Mr. B draws in the following manner. All six strips are tossed and Mr. B chooses two of them. The chosen strips are opened and the choices recorded.
Who has the better chance of drawing two 'YES'es?
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Math is for all sigma factorial? Grin
Edit: Sorry i misread the question. I thought that the strips were drawn 3 times, instead of 2.
I disagree with the probability of drawing two YESes. What you found is the probability that the first strip is a YES, and the second strip is a YES, and the third strip is anything.
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I think that Mr. Janardhanan Sivaramakrishnan forgot to mention that "Mr. A"
makes two draws only.
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Did I not write "...and a second draw is done in the same manner." ? OR was that courtesy of the moderators?
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@Janardhanan Sivaramakrishnan – You are right Mr. Janardhanan Sivaramakrishnan.
I must have misread.
My apologies.
Case for Mr A. Since the drawn strip is returned to the lot, the probability of the second draw being a YES is independent of the first one. Hence, probability of 2 YESs = 6 3 6 3 = 4 1
Case of Mr. B Drawing two strips can be regarded as drawing the second strip without returning the first one to the lot. Hence probability of first YES = 2 1 Probability of a second YES (subject to first being YES) = 5 2
Total probability of 2 YES for B = 2 1 5 2 = 5 1
Hence, Mr A has a better chance.
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For the first draw, A and B have the same probability of drawing a YES.
Probability of A drawing two YESes is 6 3 ∗ 6 3 = 4 1 .
Probability of B drawing two YESes is ( 2 6 ) ( 2 3 ) = 5 1 .
Therefore A has a better chance of drawing two YESes