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# Suppose a girl throws a die. If she throws a 5 or 6, she tosses a coin 3 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 a die?

$(A)\;\frac{8}{11}$
$(B)\;\frac{7}{11}$
$(C)\;\frac{5}{11}$
$(D)\;\frac{3}{11}$

Let $E_1$ be the event of showing 1, 2, 3 or 4
$E_2$ be the event showing 5 or 6
$P(E_1) = \frac{4}{6} = \frac{2}{3}; P(E_2) = \frac{2}{6} =\frac{1}{3}$
$\therefore P(A/E_1) = \frac{1}{2}$ and $P(A/E_2) = \frac{3}{8}$
\begin{align*}Required\; probability & = \frac{ P( E_1) . P(A/E_1)}{P(E_1). P(A/E_1) + P(E_2). P(A/E_2)} \\ & = \frac{\frac{2}{3} . \frac{1}{2}}{\frac{2}{3} . \frac{1}{2} + \frac{1}{3}. \frac{3}{8}} \\ & = \frac{8}{11}\end{align*}