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I'm not sure how to reason about this problem.

Say we toss 12 coins in a row. What is the probability that 7 of those tosses were heads, and five were tails?

I've tried thinking of it as the number of ways to pick 7 things out of 12 options: $ \frac{12}{7!\ (12-7)!} $. I thought this was most reasonable, but it's wrong.

I also considered thinking of it as the number of permutations of 7 things with 2 choices divided by the number of permutations of 12 things: $ \frac{2^7}{2^{12}} $.

I'm not very sure where to go from here. Thank you for the help!

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This is a binomial distribution. The probability of getting 7 heads and five tails is as follows: $\binom{12}{7} (p_{head})^{7} * (1 - p_{head})^{5}$. So you choose your heads, then multiply by the probability of exactly $7$ heads out of $12$. If this is a fair coin, $p_{head} = 1 - p_{head} = \frac{1}{2}$.

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  • $\begingroup$ Thank you, this was correct. I see how it works now! $\endgroup$
    – Ben
    Apr 7, 2014 at 4:55
  • $\begingroup$ Glad I could help! :-) $\endgroup$
    – ml0105
    Apr 7, 2014 at 4:59

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