Six identical-looking coins are in a box, of which five are unbiased, while the sixth comes up heads with probability $3 \over 4$ and tails with probability $1 \over 4$. Three coins are chosen from the box at random and removed. One of those three is chosen at random and tossed three times, coming up heads every time. Given this information

a). What is the probability that the final coin selected was the biased coin?

b). What is the probability that the biased coin is amongst the three coins removed from the box?

I did this question myself but felt not confident whether I got the right answer or not. So here is my solution:

a). $P={ {5C2 \over 6C3} \times {1 \over 3}}={1 \over 6}$ where the $5C2 \over 6C3$ comes from the probability that the three coins chosen contains the biased coin, and $1 \over 3$ comes from choosing the biased coin from the three coins.

b). $P={1 \over 3}$ as can be seen from a).

Am I getting the correct answer?


For both parts you need to calculate conditional probabilities.

For a) let $B$ denote the event that the biased coin is chosen at the final stage, and let $3H$ denote the event of obtaining three heads.

We require $$p(B|3H)=\frac{p(B\cap3H)}{p(3H)}=\frac{\frac 16\times\frac{27}{64}}{\frac 16\times\frac{27}{64}+\frac 56\times\frac 18}=\frac{27}{67}$$

For b), let C denote the event that the biased coin is amongst those removed from the box.

Now we require $$p(C|3H)=\frac{p(C\cap3H)}{p(3H)}=\frac{\frac 16\times\frac{27}{64}+\frac 13\times\frac 18}{\frac 16\times\frac{27}{64}+\frac 56\times\frac 18}=\frac{43}{67}$$


You have ignored the information from the fact that the coin came up heads all three times. Even so, in your model the answer to b should be $1/2$ because three coins were selected out of six. For a, you have two possibilities: the biased coin was selected (initial probability $1/6$) and of course it came up heads all three times, or some other coin was selected (initial probability $5/6$) and it happened to come up heads all three times (probability $1/8$). Do you know how to combine these?

  • $\begingroup$ I could solve part a) now, yeah I missed the information about three times..... How could I solve part b)? $\endgroup$ – Rescy_ Jul 27 '15 at 3:04
  • $\begingroup$ I don't think the answer to part b is $\frac 12$ $\endgroup$ – David Quinn Jul 27 '15 at 11:17
  • $\begingroup$ @DavidQuinn: It would be if the information from the flips is ignored. That was why I said "in your model" $\endgroup$ – Ross Millikan Jul 27 '15 at 13:52
  • $\begingroup$ I don't know what "flips" are, but I think I get you $\endgroup$ – David Quinn Jul 27 '15 at 14:26
  • $\begingroup$ @DavidQuinn: The flips are the three times the coin is flipped and comes up heads. OP computed the chance the biased coin was chosen without using that information. In that case, choosing one coin of six has probability $\frac 16$, though OP took a little longer route to get there. I was trying to point OP in the direction of your answer. $\endgroup$ – Ross Millikan Jul 27 '15 at 14:36

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