Coin is tossed five times. Probability of getting head in first toss given that head was observed 3 times? Let A: head observed on first toss
and B: 3 heads observed.
So far I have P(A)= 3/5 and P(B)=0.3125     [  5C3 x (0.5)^3 x (0.5)^2]
I tried to work out P(AnB) and I got 0.375   [ 4C2 x (0.5)^2 x (0.5)^2]
but when using the conditional probability formula I am not getting the right answer. Anyone know what I'm missing out or going wrong on?
 A: Direct answer:  Of the five tosses, three were heads and two were tails.  Since any of the five was equally likely to be the first, the probability that it was heads is 3/5.
A: $P(A)$ is not $\frac35$, it's $\frac12$ since it concerns itself with a single toss. This is not needed for the conditional probability formula, though.
The real mistake is that $P(A\cap B)$ is a factor of $\frac12$ off; it should be $\frac6{32}$ since the first toss is fixed as a head.
A: $P[X_1 = 1 \text{ and } X_2+\cdots +X_5 =2] = {1 \over 2} \cdot \binom{4}{2} {1 \over 2^4}$.
$P[X_1+X_2+\cdots +X_5 = 3] = \binom{5}{3} {1 \over 2^5}$.
Hence the conditional probability is ${ \binom{4}{2} \over \binom{5}{3}} = {3 \over 5}$.
A: Since the other answer responded to the OP's question, I can point out the alternative approach.  Since (presumably) exactly 3 heads out of the 5 tosses were observed, the question is equivalent to the
following:
You have an urn with 5 balls, numbered 1 through 5.  You draw 3 of the 5 balls out, without replacement.  What is the probability that ball number 1 is among the 3 balls drawn.
Here, the numbers on each of the drawn balls can be construed to represent which of the 5 tosses came up heads.
