# A coin is tossed $5$ times. How many possible outcomes contain AT MOST $3$ heads

There are $$32$$ possible outcomes in total when a coin is tossed $$5$$ times. I have found that there are 10 possible outcomes that contain exactly$$3$$ heads by using $$5C3=5!/3!2!$$, but how do I find out how many possibilities contain at most $$3$$ heads? Thanks!

• Add up the number of possibilities that contain 1,2 or 3 heads. – Noe Blassel May 4 at 7:50
• @NoeBlassel what about $0$ heads? – Mohammad Zuhair Khan May 4 at 7:54
• Right, those too – Noe Blassel May 4 at 8:39

This would be modeled with a sum of binomial coefficients. If you want precisely $$k$$ events to occur out of $$n$$ events, where event can only have success or failure, then the number of corresponding outcomes is $$n$$ choose $$k$$, i.e. $$\binom n k$$.

If you want at most some number of events, you can sum over the corresponding $$k$$ values, the $$k$$ values you deem valid. For example, if you want at most $$3$$ successes, you take the sum of $$\binom n k$$ for $$k=0,1,2,3$$.

In your case, $$n=5$$ and thus your result is

$$\sum_{k=0}^3 \binom 5 k$$

It is $$\displaystyle \frac{32}{2}+\binom 5 3$$.

(Half of the possibilities are for getting $$0$$, $$1$$ or $$2$$ heads)

Obviously there are $$5$$ possibilities for exactly $$4$$ heads (as there must occur exactly one tail) and only one for $$5$$ heads. Hence the number of possibilities for at most three heads equals $$32-5-1$$.