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Suppose Player A and Player B are flipping a coin. Player A flips the coin 20 times, and Player B 40 times. What would be the probability that Player A gets 10 heads from his 20 flips, given that 20 heads were found in total.

I know we can use the binomial distribution to find the denominator(20 heads in 60 flips), but I'm wondering what exactly would b in the numerator. Would I use the binomial distribution as well?

This is my denominator:

$ 60 \choose 20 $$ * 0.5^{20} * (1 -.5)^{40}$

Thanks in advance!

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    $\begingroup$ Yes. Assuming you mean "exactly" in each case (so the total is exactly $20$ Heads and so on) then the numerator is the probability that $A$ throws exactly $10$ and $B$ also throws exactly $10$. $\endgroup$ – lulu Oct 15 '17 at 17:26
  • $\begingroup$ @lulu Thanks! You can put your comment as an answer so I can check it. $\endgroup$ – idude Oct 15 '17 at 17:29
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Trusting that you mean "exactly" in each instance (so "exactly" $20$ heads tossed, etc.) then the numerator is the joint probability that $A$ throws exactly $10$ Heads AND $B$ throws exactly $10$. By independence, this joint probability is just a product and the two factors can easily be computed from the binomial distribution.

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