Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.

Question

Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze. They all toss each of their coins once and record the results. Two results are the same if two gold coins land on the same face, two silver coins land on the same face, and two bronze coins land on the same face. What is the probability that Paul's result is the same as at least one of the other two results?

A. 15/32
B. 15/64
C. 31/64
D. 3/8
E. 1/2

Show Answer

Official Answer

B

A. 15/32
B. 15/64
C. 31/64
D. 3/8
E. 1/2

IanStewart · Accepted Answer

Regardless of what Paul gets, there will be a 1/2 chance Quinn's gold coin matches Paul's, a 1/2 chance Quinn's silver matches Paul's, and a 1/2 chance Quinn's bronze matches Paul's, so a (1/2)(1/2)(1/2) = 1/8 probability that all three of Quinn's coins match Paul's. Similarly there is a 1/8 probability all three of Richard's coins match Paul's. If you wanted to now solve directly, you might notice that the answer will be very close to 1/8 + 1/8, from which you can already pick the right answer (only B is plausible), but we're double-counting the case when both Quinn and Richard's coins match Paul's, which will happen (1/8)(1/8) = 1/64 of the time, and the answer is 1/4 - 1/64 = 15/64.

But in questions like this, where we need at least one thing to happen, it's usually easiest to reverse the problem (work out the probability we do  not  get the result the question asks about), since that will leave us with only one case, and we won't need to worry about the double-counting issue. The probability is 7/8 that Quinn's coins do not match Paul's, and 7/8 that Richard's do not match Paul's, and thus the probability is (7/8)(7/8) = 49/64 that neither of their coins match Paul's, and 1 - 49/64 = 15/64 that at least one does match.

hadimadi · Answer

Hi,

the question is looking for the probability of:
Only Paul Richard Same (OPRS)+OPQS(Only Paul Quinn Same)+PQRS(...)=Result.

OPRS=PRS and PQ NOT same. We have 2 possibilities per coin toss, meaning for a 3 combination, there are 8 unique values. This means that Paul and Richard have the same in 8/64 cases. However, we now have to make sure that Paul and Quinn DON'T have the same. Now, for every of the 8 cases which were valid for PRS, we can combine 7 out of the 8 possible coin tosses of Quinn (because we can't take same results in!). This means that we are left with 56/512 possibilities.

OPQS= ... -> This case is just the same as above, we have the same result

PQRS: There are only 8 in 8^3 cases in which all tosses are the same. In total we get

2*(56/512)+(8/512)=120/512=60/256=...=15/64

Bunuel  ?

thakurarun85 · Answer

How do you get 8/64.

hadimadi · Answer

thakurarun85

Each can toss 8 different results (2^3). In total, it means that combining both, we can have 2^3*2^3=64 results. But only 8 of these results are the same. For example, if Richard has HHH, then Paul needs HHH. Or for Richard HTH, Paul needs HTH. And there are 8 such cases

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Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.

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Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze.

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