Suppose a girl throws a dice. 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 dice?

.: Let ${E_1}$ : ‘ 1, 2, 3 or 4 is shown on dice’ i.e., she tosses a coin once.

$\therefore$ Sample elements of ${E_1} = \{ (1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T)\}$

${E_2}$ : ‘5 or 6’ is shown on die' i.e., she tosses coin three times

$\therefore$ Sample points for ${E_2} = \{ (5,HHH),(5,HTT),(5,THT),(5,TTH),(5,HHT)$

$(5,HTH),(5,THH),(5,TTT),(6,HHH)$

$(6,HTT),(6,THT),(6,TTH),(6,HHT),(6,HTH),(6,THH),(6,TTT)\}$

$\Rightarrow$ $P\left( {{E_1}} \right) = \cfrac{4}{6} = \cfrac{2}{3}$ and $P\left( {{E_2}} \right) = \cfrac{2}{6} = \cfrac{1}{3}$

Let $A$ : ‘exactly one head shows up',

then
$P\left( {A|{E_1}} \right) = P$

(head shows up when coin is tossed once) $= \cfrac{1}{2}$ and

$P\left( {A|{E_2}} \right) = P$

(exactly one head shows up when coin is tossed thrice) $= \cfrac{3}{8}$

Hence, the required probability is

$P\left( {{E_1}|A} \right) = \cfrac{{P\left( {A|{E_1}} \right)P\left( {{E_1}} \right)}}{{P\left( {A|{E_1}} \right)P\left( {{E_1}} \right) + P\left( {A|{E_2}} \right)P\left( {{E_2}} \right)}}$

$= \cfrac{{\cfrac{1}{2} \times \cfrac{2}{3}}}{{\cfrac{1}{2} \times \cfrac{2}{3} + \cfrac{3}{8} \times \cfrac{1}{3}}} = \cfrac{8}{{8 + 3}} = \cfrac{8}{{11}}$