Choose an American adult at random. The probability that you choose a woman is 0.52 . 0.52. The probability that the person you choose has never married is 0.26 . 0.26. 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:

(a) 0.77.
(b) 0.66.
(c) 0.44.
(d) 0.38.
(e) 0.13.

The probability that the person you choose is either a woman or has never been married (or both) is therefore about 0.66

Step-by-step explanation:

Given Data:

Probability of choosing a women = P(W) = 0.52

Probability that chosen person has never married = P(M) = 0.26

Probability of choosing a women that has never married = P(W and M) = 0.11

We need to find the probability that the person you choose is either a woman or has never been married (or both). The addition rule of probabilities of two events is given as:

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) is the probability of occurrence of either A or B or both. P(A and B) is the probability of occurrence of A and B at the same time.

Re-writing the addition formula for given case, we get:

P(W or M) = P(W) + P(M) - P(W and M)

Substituting the given values results in:

P(W or M) = 0.52 + 0.26 - 0.11

P(W or M) = 0.67

Therefore, the probability that the person you choose is either a woman or has never been married (or both) is 0.67.

Note: There is either typo in given options or the questions. The question I found on Google has 0.25 as the probability that the person you choose has never married ( i.e. P(M) = 0.25 ).

So, in this case the correct answer would be option (b) 0.66

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step-by-step explanation: