Question

Choose an American adult at random. The probability that you choose a woman is $0.52$ The probability that the person you choose has never married is $0.25$ The probability that you choose a woman who has never married is 0.11. The probability that the person you choose is either a woman or has never been married (or both) is therefore about $\left(a\right)0.77.\left(b\right)0.66.\left(c\right)0.44.\left(d\right)0.38.\left(e\right)0.13.$

Short answer

The correct option is $\left(b\right)0.66$

Step-by-step solution

Step 1. Given

The woman has a probability of $0.52$

$0.25$chance of never marrying.

Women who have never married have a probability of $0.11$

Step 2. Concept

The probability of an event=$\frac{numberoffavourableoutcomes}{totalnumberofoutcomes}$

Step 3. Calculation

The likelihood of a randomly selected individual never marrying or being a woman can be computed as follows:$$\begin{gathered} P(Nevermarriedorwoman)=P(Nevermarried)+P\left(woman\right)-P(NevermarriedandWoman) \\ =0.25+0.52-0.11 \\ =0.66 \end{gathered}$$

As a result, $0.66$ is the required probability.

As a result, the correct option is (b).