# Probability - Biased Coin Given First Flip Is Heads

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 $$(nCx) p^x q^{n-x}$$
$$P(\text {Getting heads on first flip})=P(B) = 0.6 = 3/5$$ $$P(A|B) = \frac {P(A∩B)}{P(B)}$$

• Use q^{n-x} to get $q^{n-x}$. Commented Aug 1, 2022 at 9:41
• Given the initial $H$, the question comes down to "what's the probability of getting exactly one $H$ out of $3$?" which is a straight binomial computation.
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 $$\frac{36}{125}$$.