I am stuck on this question.
A coin with $P(H) = \frac{1}{2}$ is flipped $4$ times and then a coin with $P(H) = \frac{2}{3}$ is tossed twice. What is the probability that a total of $5$ heads occurs?
I keep getting $\frac{1}{6}$ but the answer is $\frac{5}{36}$.
Attempt: $P($all heads on the four coins$)P($either one of the tosses is heads on the two coins$)+P(3$ heads on the four coins$)P($both coins are heads$)$
$P($all heads on the four coins$) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$.
$P($either one of the tosses is heads on the two coins$) = 1-P($no heads on both tosses$) = 1-\frac{1}{3}\cdot\frac{1}{3} = \frac{8}{9}$.
$P($exactly $3$ heads on the four tosses$) = \frac{1}{4}$.
$P($both coins are heads$) = \frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}$.
Final Equation: $\frac{1}{16}\cdot\frac{8}{9}+\frac{1}{4}\cdot\frac{4}{9} = \frac{1}{6}$.
Why am I off by $\frac{1}{36}$?
$\endgroup$ 21 Answer
$\begingroup$The required probability will be
$P($exactly $4 $ heads from the $4$ flips of $1$st coin$)\cdot P($exactly $1 $ head from the $2$ flips of $2$nd coin $)+$
$P($exactly $3 $ heads from the $4$ flips of $1$st coin$)\cdot P($exactly $2 $ heads from the $2$ flips of $2$nd coin $)$
Using Binomial Distribution, the required probability $$\binom44\left(\frac12\right)^4\left(1-\frac12\right)^{4-4} \cdot\binom21\left(\frac23\right)^1\left(1-\frac23\right)^{2-1}$$
$$+\binom43\left(\frac12\right)^3\left(1-\frac12\right)^{4-3} \cdot\binom22\left(\frac23\right)^2\left(1-\frac23\right)^{2-2}$$
$$=\frac1{16}\cdot\frac49+\frac14\cdot\frac49=\frac5{36}$$
$\endgroup$
