Question

In a certain city two newspapers A and B are published, it is known that 25% of the city population reads A and 20% reads B while 8% reads both A and B. It is also known that 30% of those who read A but not B look into advertisements and 40% of those who read B but not A look into advertisements while 50% of those who read both A and B look into advertisements. What is the percentage of the population that reads an advertisement?

Answer 1

If E1, E2, E3 . . . . . . En are mutually exclusive and exhaustive events and E is an event which can take place in conjunction with any one of E1 then

P(E) = P (E1) P (EI E1) + P(E2) P(EIE2) + . . . . . . . . + P (En) P(EIEn)

Let P(A) denote the prob. of people reading newspaper A and P(B) that of people reading newspaper B

Then, P(A) = 25/100 = 0.25

P(B) = 20/100 = 0.20, P(AB) = 8/100 = 0.08

Prob. of people reading the newspaper A but not B = P (ABc)

= P (A) – P(AB) = 0.25 – 0.08 = 0.17

Similarly,

P(AcB) = P(B) – P (AB) = 0.20 – 0.08 = 0.12

Let E be the event that a person reads an advertisement.

ATQ, P (EI ABc) = 30/100; P(EI AcB) = 40/100

P(EI AB) = 50/100

∴ By total prob. theorem (as ABc , AcB and AB are mutually exclusive)

P(E) = P(EI ABc) P (ABc) + P(EI Ac B) P (AcB) + P (EI AB). P (AB)

= 30/100 x 0.17 + 40/100 x 0.12 + 50/100 x 0.08

= 0.051 + 0.048 + 0.04

Thus the population that reads an advertisements is 13.9%