Question:
An unfair coin is flipped four times. $P(Heads)=0.6$ and $P(Tails)=0.4$. Every flip is independent of every other flip.
Find the probability of getting exactly $2$ heads and $2$ tails given that the first coin flip was a head.
For some reason I cannot seem to get the right answer. What I have tried:
$$P(A∩B) = 216/625$$ The above was obtained using $(nCx) p^x q^{n-x}$
$$ P(\text {Getting heads on first flip})=P(B) = 0.6 = 3/5$$ $$ P(A|B) = \frac {P(A∩B)}{P(B)}$$
I can't figure out where I am going wrong. Feeling dumb. Please help
Let $X_1,X_2,X_3,X_4$ be i.i.d. random variables with $\mathbb P(X_1=1)=p=1-\mathbb P(X_1=0)$. We compute the conditional probability $$ \mathbb P\left(\sum_{i=1}^4 X_i=2\mid X_1=1\right) = \mathbb P\left(\sum_{i=2}^4 X_i=1\right) = 3 p (1-p)^2. $$ With $p=\frac35$, this reduces to $\frac{36}{125}$.