Question
$$A,\,B$$ and $$C$$ are contesting the election for the post of secretary of a club which does not allow ladies to become members. The probabilities of $$A,\,B$$ and $$C$$ winning the election are $$\frac{1}{3},\,\frac{2}{9}$$ and $$\frac{4}{9}$$ respectively. The probabilities of introducing the clause of admitting lady members to the club by $$A,\,B,$$ and $$C$$ are $$0.6,\,0.7$$ and $$0.5$$ respectively. The probability that ladies will be taken as members in the club after the election is :
A.
$$\frac{{26}}{{45}}$$
B.
$$\frac{5}{9}$$
C.
$$\frac{{19}}{{45}}$$
D.
none of these
Answer :
$$\frac{{26}}{{45}}$$
Solution :
Let $${E_A} = $$ the event of $$A$$ becoming secretary. Similarly, $${E_B}$$ and $${E_C}.$$
$$\,\,\,\,\,\,\,\,\,\,\,{E_L} = $$ the event of admitting lady members.
Here, $$P\left( {{E_A}} \right) = \frac{1}{3},\,\,P\left( {{E_B}} \right) = \frac{2}{9},\,\,P\left( {{E_C}} \right) = \frac{4}{9}$$
Clearly, $${E_A},\,{E_B},\,{E_C}$$ are mutually exclusive and exhaustive.
Also, $$P\left( {\frac{{{E_L}}}{{{E_A}}}} \right) = 0.6,\,\,\,P\left( {\frac{{{E_L}}}{{{E_B}}}} \right) = 0.7,\,\,\,\,P\left( {\frac{{{E_L}}}{{{E_C}}}} \right) = 0.5$$
$$\therefore $$ the required probability
$$\eqalign{
& = P\left( {{E_A}} \right).P\left( {\frac{{{E_L}}}{{{E_A}}}} \right) + P\left( {{E_B}} \right).P\left( {\frac{{{E_L}}}{{{E_B}}}} \right) + P\left( {{E_C}} \right).P\left( {\frac{{{E_L}}}{{{E_C}}}} \right) \cr
& = \frac{1}{3} \times \frac{3}{5} + \frac{2}{9} \times \frac{7}{{10}} + \frac{4}{9} \times \frac{5}{{10}} \cr
& = \frac{{26}}{{45}} \cr} $$