91Ó°ÊÓ

Three newspapers published are A, B, and C.

The percentage of the families who subscribe to newspapers A, B and C is 60,40 and 30 respectively. Also, the percentage of families that subscribe to both A and B, both A and C and both B and C is 20, 10 and 20 respectively.

5 per cent subscribe to all three newspapers A, B, and C.

Let:

\(A = \)Event that the randomly selected family subscribes to newspaper A is selected

\(B = \)Event that the randomly selected family subscribes to newspaper B is selected

\(C = \) Event that the randomly selected family subscribes to newspaper C is selected

The given information can be summarised as:

\(\begin{array}{l}P\left( A \right) = 0.60\\P\left( B \right) = 0.40\\P\left( C \right) = 0.30\end{array}\)

And,

\(\begin{array}{l}P\left( {A \cap B} \right) = 0.20\\P\left( {A \cap C} \right) = 0.10\\P\left( {B \cap C} \right) = 0.20\\P\left( {A \cap B \cap C} \right) = 0.05\end{array}\)

The conditional probability for an event A given B has already occurred is given by:

\(P\left( {A|B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}};P\left( B \right) > 0\)

Thus, for the case, if a family selected at random from the city subscribes to newspaper A, the probability that the family also subscribes to newspaper B is\(P\left( {B|A} \right)\), and is obtained as:

\(\begin{aligned}{}P\left( {B|A} \right) &= \frac{{P\left( {B \cap A} \right)}}{{P\left( A \right)}}\\ &= \frac{{0.20}}{{0.60}}\\ &= \frac{1}{3}\end{aligned}\)

Therefore, the required probability is \(\frac{1}{3}\) .