1
$\begingroup$

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 $\frac{1}{36}$?

$\endgroup$
  • 1
    $\begingroup$ Because "either one of..." $\not \equiv$ " exactly one of..." $\endgroup$ – Ehsan M. Kermani Jun 9 '13 at 4:31
  • $\begingroup$ You claim that the problem is with $P($either one of the tosses is heads on the two coins$)$. You need to calculate the probability of there being precisely one occurrence of a head in the tossing of the second coin. In your calculation, you subtracted the probability of no heads from one. The remaining probability covers one head or both heads, but you want to exclude the latter. If you make this adjustment, you will get the correct answer. $\endgroup$ – Michael Albanese May 30 '15 at 19:38
2
$\begingroup$

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 $$\binom44\left(\frac12\right)^4\left(1-\frac12\right)^{4-4} \cdot\binom21\left(\frac23\right)^1\left(1-\frac23\right)^{2-1}$$

$$+\binom43\left(\frac12\right)^3\left(1-\frac12\right)^{4-3} \cdot\binom22\left(\frac23\right)^2\left(1-\frac23\right)^{2-2}$$

$$=\frac1{16}\cdot\frac49+\frac14\cdot\frac49=\frac5{36}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.