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I was wondering if somebody could guide me with this question.

A recent survey examined the working arrangements of married households. It was found that 88% of the households had at least one working member. In 20% of the households with the woman not working, the man also does not work. In 40% of households in which the man does not work, the woman also does not work.

What is the probability the man does not work and the woman works in a randomly selected household?

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Since you don't want the full answer, I'll just make an observation. Say we choose a specific household $H$. I will denote the event "in $H$, the woman is employed and the man is unemployed" by $E_1$ and the event "in $H$, the woman is unemployed and the man is employed" as $E_2$. See that $E_1$ is disjoint from $E_2$. Can you take it from here?

As general advice for problems like these, formalizing things and/or drawing a venn diagram works great! It's probably overkill for this specific problem, but employing this advice for harder problems will certainly keep your thoughts in order.

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Denote an employed or unemployed man by $M$ or $\bar M$ respectively, and an employed or unemployed woman by $W$ or $\bar W$ respectively. From the given probabilities, we have

(1) $P(\bar W,\bar M)=1-0.88=0.12$, and

(2) $P(\bar W|\bar M)=\frac{P(\bar W,\bar M)}{P(\bar M)}=0.4$.

Thus from (2), we have $P(\bar M) = \frac{P(\bar W,\bar M)}{P(\bar W|\bar M)}=0.12/0.4=0.3$.

Also from (2), $P(W|\bar M)=1-P(\bar W|\bar M)=0.6$.

You are asked to compute $P(W, \bar M)$ which can be expressed as

$P(W, \bar M)=P(W|\bar M)P(\bar M)=0.6\times0.3=0.18$.

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