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I was going through this question on a site. It stated as below,

One more problem that is tying me up..I think because it is so much simpler than what I have been working on. Any help would be great =)

What is the probability of getting exactly $6$ heads and $3$ tails in $9$ coin tosses of an unbiased coin?

Now I were to solve the problem, I would have made a table of all the heads and tails. But this problem size is very big ($2^9$ is large). Is there any other possible way to solve the problem?

Please help me out.


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The number of heads or tails in a coin toss is a binomially distributed random variable. – Srivatsan Sep 20 '11 at 19:02
up vote 3 down vote accepted

Since coin is unbiased all possible 9-tuples have equal probability. Now applying the school definition of probability, $p = \frac{N_f}{N_t}$, i.e. the fraction of the number favorable outcomes over the number of total outcomes.

The total number of outcomes is $2^9$ as you noted. The number of favorable outcomes is given by the number of ways to choose $6$ slots for the head out of 9. This is $\binom{9}{6} = 84$.

Thus the probability

$$ p = \frac{\binom{9}{6}}{2^9} = \frac{84}{2^9} \sim 0.164 $$

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How did you calculated that those 9 tuples? – Fahad Uddin Sep 20 '11 at 19:53
9-tuple is a name for collection of 9 outcomes, what I was saying is that all outcomes of 9 tossings are equally likely. – Sasha Sep 20 '11 at 19:54

Imagine you record the outcome of each coin toss; you will get strings like

  • T T T T T T F F F
  • T F T F F T F T T
  • F F F F F F F F F

Now what is the probability of each string? HINT Remember that each string is equally likely since the coin is unbiased.

Now you count a subset of these strings (or outcomes). Exactly which ones? Those strings that have exactly 6 heads and remaining tails. Do you know how to count the number of such strings?

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