Csdrhrt

The members of a wine tasting club are married couples. For any married couples in the club, the probability that "husband is retired " $= 0.7$ , "wife is retired" $= 0.4$ , the probability that "the husband retires given his wife retires" $=0.8$ .

Two married couples are chosen at random. Find the probability that only one of the two husbands and only one of the two wives is retired.

My work:

$P$ (Both of them are retired) = $0.32 $
$P$ (Only the husband is retired) = $0.38 $
$P$ (Only the wife is retired) = $0.08 $
$P$ (Only one of them is retired) = $0.46 $

So
$P$ (Only one of two husbands and only one of the two wives is retired) = $$2 times 0.38 times 0.08 = 0.0608$$

But the answer says $0.2016$ . Why is it so? I tried $P$ (Only one of them retired)$^2$ but this equals $0.2116$ and does not make sense since it also accounts for having both husbands retired or wives retired from two couples -- not one each.

Many thanks in advance!

The members of a wine tasting club are married couples. For any married couples in the club, the probability husband is retired = 0.7, wife retired = 0.4, the probability that the husband retire is 0.8.

This appears to be a miscopy. The second of those "probability that the husband retires" is apparently conditioned on when the wife is retired.   Let $H_k$ be the event that husband of couple $k$ is retired, and $W_k$ that of the wife being so.   Then $mathsf P(H_k)=0.7, mathsf P(W_k)=0.4, mathsf P(H_kmid W_k)=0.8$ which does mean (as you obtained):

$mathsf P(H_kcap W_k) ~=~ 0.4cdot 0.8 ~=~ 0.32
\ mathsf P(H_kcap W_k^complement) ~=~ 0.7-0.32 ~=~ 0.38
\ mathsf P(W_kcap H_k^complement) ~=~ 0.4-0.32 ~=~ 0.08
\ mathsf P(H_ktriangle W_k) ~=~ 0.38+0.08 ~=~ 0.46 $

This is okay. $color{green}checkmark$

We also have $mathsf P(H_1^complementcap W_k^complement)=1-(0.7+0.08)= 0.22$

Then you attempted: $ 2 mathsf P((H_1cap W_2^complement)cap(W_1cap H_2^complement)) = 2cdot 0.38cdot 0.08 = 0.0608$, making use of symmetry, and that is okay for the "probability for only one of each from each couple is retired."   However that is not what you needed, so it is not okay. $require{cancel}color{red}{xcancelcdot}$

tl;dr

You were actually asked to find "the probability that only one of the two husbands and only one of the two wives is retired", which is $mathsf P((H_1triangle H_2)cap (W_1triangle W_2))$   This includes the event that both do come from the same couple.

$$begin{align} =~& 2mathsf P(H_1cap W_1)mathsf P(H_2^complementcap W_2^complement)+2mathsf P(H_1cap W_1^complement)mathsf P(H_2^complementcap W_2) \ =~& 2cdot 0.32cdot 0.22 + 0.0608 \ =~& 0.2016end{align}$$

You seem to be using the probability the husband is retired is $0.8$ and assuming independence between the husband being retired and the wife being retired. As you write the question you might be expected to use $0.7$ for the probability the husband is retired. Under your reading, the probability that only the husband is retired is $0.8(1-0.4)=0.48$ so the probability of exactly one retired is $0.56$ Then in your final calculation where you ask the probability that exactly one husband and one wife is retired, you assume the retired people come from separate couples. It could be that both people of one couple are retired and neither of the other couple.

Given the correct question to find...

...the probability that only one of the two husbands and only one of the two wives is retired,

this is how I understood it.

Who are the couples in this scenario? Husband 1 and Wife 1 ($H_1W_1$) and Husband 2 and Wife 2 ($H_2W_2$). These couples may be thought of as choices in a decision tree, where we multiply probabilities "across" and sum "down".

Given the condition that only one husband and one wife can be retired, we have the following possibilities:

$$(H_1W_1)(H_2'W_2')implies P(Hcap W)P(H'cap W')$$
$$(H_1W_1')(H_2'W_2)implies P(Hcap W')P(H'cap W)$$
$$(H_1'W_1)(H_2W_2')implies P(H'cap W)P(Hcap W')$$
$$(H_1'W_1')(H_2W_2)implies P(H'cap W')P(Hcap W)$$

Hence we can sum "down" and collect terms to obtain

$$2P(Hcap W)P(H'cap W')+2P(Hcap W')P(H'cap W)=0.2016.$$