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 ?
My Doubt?
Let E1 be the event that the outcome on the die is 5 or 6 and E2 be the event that the outcome on the die is 1,2,3 or 4. A - exactly one head.
I am not able to understand why the probability of 4 / 6 in {1,2,3,4} not getting multiplied with the probability of exactly one head i.e 1/2 to get P (A/E2)?
The probability you mentioned is
$$P[E_2 \cap A] = P[E_2] \cdot P[A|E_2] = \left(\frac{2}{3}\right) \left(\frac{1}{2}\right) = \frac{1}{3}$$
However, the question is asking for $P[E_2|A]$, which, by Bayes' Theorem, is
$$\begin{align} P[E_2|A] &= \frac{P[A|E_2] \cdot P[E_2]}{P[A]} \\ \\ &= \frac{P[A|E_2] \cdot P[E_2]}{P[A|E_1] \cdot P[E_1] + P[A|E_2] \cdot P[E_2] } \\ \\ &= \frac{\left(\frac{2}{3}\right) \left(\frac{1}{2}\right)}{\left(\frac{1}{3}\right) \left(\frac{3}{8}\right) + \left(\frac{2}{3}\right) \left(\frac{1}{2}\right)} \\ \\ &= \frac{8}{11} \end{align}$$