Browse Questions

# Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

$\begin{array}{1 1}8/11 \\ 2/11 \\ 3/11 \\ 6/11 \end{array}$

Can you answer this question?

Toolbox:
• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: $\;P(E_i/A)$=$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$
Let $E_1$ be the event that the girl gets 5 or 6 on the roll. P($E_1$) = $\large\frac{2}{6} = \frac{1}{3}$
Let $E_2$ be the event that the girl gets 1,2,3 or 4 on the roll. P($E_1$) = $\large\frac{4}{6} = \frac{2}{3}$
Let A be the event that she obtained exactly one head.
If she tossed a coin 3 times and exactly 1 head showed up, then the total number of favourable outcomes = {(HTT), (THT), (TTH),} = 3.
P (A|$E_1$) = $\large\frac{3}{8}$ (Note: in 3 coin tosses, total sample space = 8)
If she tossed the coin only once and exactly 1 showed up, the total number of favourable outcomes = 1.
P (A|$E_2$) = $\large\frac{1}{2}$
We need to find the probability that she threw 1, 2, 3 or 4 with the die, given that she got exactly one head.
We can use Baye's theorem, according to which $P(E_2|A) = \large\frac{P(E_2)(P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
P($E_2$|A) = $\Large \frac {\frac{1}{2}.\frac{2}{3}}{\frac{1}{2}.\frac{2}{3} + \frac{3}{8}.\frac{1}{3}}$ = $\Large \frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{8}}$ = $\large\frac{8}{11}$
answered Jun 19, 2013