Flip a fair coin 5 times. $A$ is the event that the first flip is heads, and $B$ is the event that you get at least three heads. What is $P(A|B)$?

Question

I'm trying to use the conditional probability formula, $P(A|B)= \frac{P(A \cap B)}{P(B)}$. I've determined that $P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{2^5}(\binom{5}{3} + \binom{5}{4} + \binom{5}{5}) = \frac{1}{2}$, but I can't figure out how to solve for $P(A \cap B)$.

Answer 1

Given the event $B$, that you get at least three heads in five flips, there are $\binom{5}{3} + \binom{5}{4} + \binom{5}{5}$ equiprobable such outcomes. How many of these have heads in the first flip? If there were exactly $3$ heads flipped and the first flip is heads, then there are $\binom{4}{2}$ ways to distribute the remaining $2$ heads and $2$ tails; if there are exactly $4$ heads flipped, there are $\binom{4}{1}$ ways to distribute the remaining $3$ heads and $1$ tail; and if $5$ heads are flipped there is only $1$ outcome. So the desired probability is $$\frac{\binom{4}{2} + \binom{4}{1} + \binom{4}{0}}{\binom{5}{3} + \binom{5}{4} + \binom{5}{5}} = \frac{6+4+1}{10+5+1} = \frac{11}{16}.$$

---

The above is a pure enumeration approach. We can consider a more general approach in which the probability of heads is some $p \in (0,1)$ rather than the fair case $p = 1/2$. This is done by Bayes' theorem: $$\Pr[A \mid B] = \frac{\Pr[B \mid A]\Pr[A]}{\Pr[B]}.$$ The unconditional number of heads is a binomial random variable, $$X \sim \operatorname{Binomial}(n = 5, p), \\ \Pr[X = x] = \binom{5}{x} p^x (1-p)^{5-x},$$ hence event $B$ is equivalent to $X \ge 3$, and $$\Pr[B] = \Pr[X \ge 3] = \sum_{x=3}^5 \binom{5}{x} p^x (1-p)^{5-x}.$$ The conditional random variable $B \mid A$ corresponds to the outcome $Y \ge 2$ where $Y$ is binomial but with $n = 4$, because flips are independent; i.e. $$Y \sim \operatorname{Binomial}(n = 4, p).$$ And of course, $\Pr[A] = p$. So it follows that $$\Pr[A \mid B] = \frac{\Pr[Y \ge 2]p}{\Pr[X \ge 3]} = \frac{p \sum_{y=2}^4 \binom{4}{y} p^y (1-p)^{4-y}}{\sum_{x=3}^5 \binom{5}{x} p^x (1-p)^{5-x}} = \frac{6-8p+3p^2}{10-15p+6p^2}.$$ Note when $p = 1/2$ we recover the original answer $11/16$.

Answer 2

Flip a fair coin 5 times. A is the event that the first flip is heads, and B is the event that you get at least three heads. What is P(A|B) ?

For the event $(A\cap B)$ you need at least $2$ heads in the next $4$ tosses, thus

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

$=\dfrac{\binom 4 2 + \binom 4 3 + \binom 4 4}{\binom 5 3 + \binom 5 4 + \binom 5 5} = \dfrac {11}{16}$

PS

If you want to use probabilities, $P(A\cap B) = P(A)\cdot P(B|A)$, but you can further simplify the computation by dealing with number of ways rather than probabilities as shown above