I am trying to figure out the problem below.

The number of ways a coin can be tossed 6 times so that there is exactly 3 heads and no two heads occur in a row is?

At first it seemed like it would just be .5•.5•.5•.5•.5•.5 which would be 0.015625, since none of the outcomes are dependent on each other. Then as I was reading the question more carefully, I thought maybe the answer is 2? Since there are only 2 outcomes for each toss?

All help greatly appreciated.

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    $\begingroup$ "The number of ways" will result in an integer, not a number between $0$ and $1$ (exclusive). "The probability" will result in a number between $0$ and $1$ (inclusive), but never a number larger than $1$. The phrasing of the question you posted is "The number of ways." $\endgroup$ – JMoravitz Jun 6 '15 at 21:07
  • $\begingroup$ I see, that makes sense. Can you by chance tell me how to calculate it? Not sure where to go with this one. $\endgroup$ – Omar N Jun 6 '15 at 21:09

The question asks for "The number of ways" that you can flip six coins with exactly 3 heads and 3 tails such that no two heads appear next to one another.

Use the tails as a barrier:


The three heads may go into those four available spaces with at most one head per space. Choose which three spaces are occupied.

Four spaces, you want to choose three of them, there are then $\binom{4}{3}=4$ different such sequences. The sequences are specifically HTHTHT, HTHTTH, HTTHTH, THTHTH

As an aside, the probability of this happening is: The number of ways it could happen / The total number of ways of flipping coins regardless of this happening.

$\frac{4}{2^6} = \frac{1}{2^4} = \frac{1}{16}$

  • $\begingroup$ Gotcha! I see where I went wrong. Thanks so much for the help. $\endgroup$ – Omar N Jun 6 '15 at 21:14

It is actually 4:

$$(T, H, T, H, T, H)$$ $$(H, T, T, H, T, H)$$ $$(H, T, H, T, T, H)$$ $$(H, T, H, T, H, T)$$

  • $\begingroup$ Thank you! This is an excellent way to visualize it $\endgroup$ – Omar N Jun 6 '15 at 21:14





Number of WAYS = 4

Probability of this occurring =$\frac{4}{2^6}$ = $\frac{1}{16}$


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