Question

A survey was recently done in a certain town to determine readership of newspapers available. 50% of the resident read Daily Nation, 60% read the standard and 20% read both newspapers. Determine the probability that a resident selected does not read any newpaper.

Answer 1

Answer on Question #67010 – Math – Statistics and Probability

Question

A survey was recently done in a certain town to determine readership of newspapers available. 50% of the resident read Daily Nation, 60% read the standard and 20% read both newspapers. Determine the probability that a resident selected does not read any newspaper.

Solution

We denote the event ‘a resident reads Daily Nation’ by $DN$, the event ‘a resident reads Standard’ by $S$ and the event ‘a resident does not read any newspaper’ by $A$.

Then

$$P(DN) = 0.5,$$ $$P(S) = 0.6,$$ $$P(DN \text{ and } S) = 0.2.$$

By the addition law of probability,

$$P(DN \text{ or } S) = P(DN) + P(S) - P(DN \text{ and } S) = 0.5 + 0.6 - 0.2 = 0.9.$$

The event $A$ can be represented as a complementary event to $DN$ or $S$.

Therefore,

$$P(A) = P((DN \text{ or } S)^c) = 1 - P(DN \text{ or } S) = 1 - 0.9 = 0.1$$

or 10% of the resident selected does not read any newspaper.

Answer: 0.1.

Answer provided by https://www.AssignmentExpert.com