# The Probability of 4 heads given the first toss is a head

The Question

Alice tosses a fair coin seven times. Find the probability that she tosses 4 heads given her first toss is a head. Then, find the probability that she tosses 4 heads given her first and last tosses are heads.

My Work

Part A

$A =$ Alice gets 4 heads $B =$ Her first toss is a head

Part a asks us to find $P(A|B)$

$P(A|B) = \frac{P(A\cap B)}{P(B)}$

We know that $P(B) = \frac{1}{2}$

$P(A\cap B) = P(A)P(B|A)$

$P(A) = \frac{\binom{6}{4}}{2^7}$

$P(B|A) = \frac{P(B\cap A)}{P(A)}$

Now I'm back to where I started trying to find $P(B \cap A)$ can anyone give any hints on how to take a different approach to this problem to get the correct solution?

• @VladimirVargas why is that? – Dunka Jan 20 '15 at 3:43
• I didn't read "seven times", I'm sorry. – Vladimir Vargas Jan 20 '15 at 3:43
• In your calculation (which is not necessary) the probability of $B\cap A$ is $(1/2)\binom{6}{3}(1/2)^6$. – André Nicolas Jan 20 '15 at 3:49