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