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

For a randomly chosen married couple who are members of the club , find the probability that

a) both of them are retired ,

b) only one of them is retired ,

c) neither of them is retired

Two married couples are chosen at random .

d)Find the probability that only one of the two husbands and only one of the two wives is retired

My turn : a) $0.4 \times 0.8 = 0.32$

b) Using Venn-diagram i got the probability would be $0.08 + 0.38 = 0.46$

c) Also using the same Venn-diagram i got $0.22$

d)I do not understand this part ?

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    $\begingroup$ You are correct for parts a,b,c. For part d, you just have to list the possibilities. Either in one couple both are retired, and in the other neither is, or in one couple only the man is retired, and in the other one the woman is. In each case, the statuses of the two couples are presumably independent. $\endgroup$ – saulspatz Mar 12 '20 at 20:36
  • $\begingroup$ There's not enough information to answer the question. Does each of the married couples consist of a husband and a wife? $\endgroup$ – joriki Mar 12 '20 at 21:04
  • $\begingroup$ I do not know how can i list the possibilities @saulspatz $\endgroup$ – Hussien Mohamed Mar 12 '20 at 21:20
  • $\begingroup$ I listed the possibilities in my prior comment. I don't understand what the difficulty is. There are $4$ possibilities: $2$ choices for which of the men is retired, and $2$ choices for which of the women is retired. $\endgroup$ – saulspatz Mar 12 '20 at 21:25
  • $\begingroup$ Excuse me , would you tell me how can i find the probability of these choices ? @saulspatz $\endgroup$ – Hussien Mohamed Mar 12 '20 at 22:52
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For part d) there are two possibilities. Either in one couple both partners are retired, or in one couple only the husband is retired, and in the other only the wife is retired.

You have shown in parts a) and c) the the probability that both are retired is $.32$ and the probability that neither is retired is $.22$. Also in part b), you showed that the probability that only the wife is retired is $.08$ and that the probability that only the husband is retired is $.38$.

Since the statuses of the two couples are independent, we compute the probability that both events occur by multiplication. Thus the probability that both partners in one marriage are retired and that neither partner in the other marriage is retired is $$2\cdot.32\cdot.22=.1408,$$ where the $2$ comes from the fact that we have $2$ ways to choose which of the two couples is retired.

Can you see how to finish the problem now?

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  • $\begingroup$ Why do we consider the case that the two husbands are retried ? Is not it correct to consider the case of a husband is retired and a wife is not or a wife is retired and a husband is not ? @saulspatz $\endgroup$ – Hussien Mohamed Mar 12 '20 at 23:25
  • $\begingroup$ @HussienMohamed We don't consider the case that two husbands are retired. There is one husband and one wife who are retired; they may be married to each other or not. read what I wrote again. $\endgroup$ – saulspatz Mar 12 '20 at 23:45
  • $\begingroup$ You have written the probability of both the husband and the wife are retired multiplied by the probability that no one of then is retired and o do not understand why did you do so @saulspatz $\endgroup$ – Hussien Mohamed Mar 12 '20 at 23:58
  • $\begingroup$ @HussienMohamed Perhaps you are interpreting the problem differently. If so, please tell me what you think it means instead of just saying you can't understand what I'm writing. I don't know how to make it any plainer. $\endgroup$ – saulspatz Mar 13 '20 at 0:28

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