# A fair coin is tossed $5$ times. It is known that there are more than $2$ heads in the $5$ tosses. What is the probability…

A fair coin is tossed $5$ times. It is known that there are more than $2$ heads in the $5$ tosses. What is the probability that there are exactly $3$ heads in the $5$ tosses?

I believe the answer is $\binom{5}{3}$, but I'm confused by the second sentence. How does this change the outcome?

• Let $H$ be the number of heads in the $5$ tosses. You are looking for $P(H=3|H\ge 2)$. Are you familiar with Bayes rule? – mzp Apr 23 '18 at 2:51
• The number of favorable cases is $\binom{5}{3}$, but this cannot be the answer for a probability question since it is larger than $1$. – N. F. Taussig Apr 23 '18 at 3:00

$$P(H=3|H> 2)= \frac{P([H=3] \cap [H>2])}{P(H>2)}=\frac{P(H=3)}{P(H>2)}$$
• Thanks for your answer. How were you able to deduce that $$\frac{P([H=3] \cap [H\ge2])}{P(H\ge2)}=\frac{P(H=3)}{P(H\ge2)}?$$ – beepboopbeepboop Apr 23 '18 at 3:26
• There are two event that should occur simultaneously. First is the getting 3 heads. Second is getting $\ge 2$ heads. Intersection is the situation when both these occur simultaneously. So how many heads can you have so that both are satisfied?\ – Sonal_sqrt Apr 23 '18 at 3:31
• $P(H\ge 2)=1-P(H=1)-P(H=0)$ check. – Sonal_sqrt Apr 23 '18 at 3:37