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There is a problem I try to solve:

Arthur and Dutch are planned together to go out the next Saturday which is predicted to be sunny. There is a probability of 0.4 that Dutch wears sunglasses the day they are supposed to meet. The probability that Arthur wears sunglasses on the same day is 0.7 if Dutch wears sunglasses and 0.35 if he does not.

Suppose that Micah, a friend of both Dutch and Arthur, is going to join them on Friday. The probability that Micah wears sunglasses is 0.55 if both Dutch and Arthur wear sunglasses and 0.25 if exactly one of them wears sunglasses.

What is the probability that Micah and Arthur wear sunglasses but Dutch does not?

I want to calculate $P(MA\bar{D})$.

The question says: $$P(M| A \oplus D) = 0.25$$

In other words, I have: $$P(M| A\bar{D} \cup \bar{A}D) = 0.25$$

I calculated: $$P(A \oplus D) = P(A \cap \bar{D}) + P(D \cap \bar{A})= P(A \cup D)-P(A \cap D) = 0.33$$

I also have calculated $P(A\bar{D}) = P(A|\bar{D})P(\bar{D}) = 0.35*0.6=0.21$

But, I still can't relate these calculations to find $P(MA\bar{D})$, and my main problem is that I can't break or simplify $$P(M| A\bar{D} \cup \bar{A}D)$$

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  • $\begingroup$ Is it important that Saturday and Friday are different days? $\endgroup$ – Henry Mar 12 at 8:07
  • $\begingroup$ @Henry I guess that must be a typo!!! $\endgroup$ – Ahmad Mar 13 at 17:43
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Hint:

$$P(MA\bar{D}) = P(M \mid A\bar{D})\,P(A\bar{D}) $$ $$P(A\bar{D}) = P(A\mid \bar{D})\,P( \bar{D})$$

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  • $\begingroup$ my problem is $P(M \mid A\bar{D})$, $P(A\bar{D}) = 0.21$ $\endgroup$ – Ahmad Mar 13 at 17:56
  • $\begingroup$ @Ahmad The question says "The probability that Micah wears sunglasses is ... 0.25 if exactly one of [Dutch and Arthur] wears sunglasses." $\endgroup$ – Henry Mar 13 at 18:05
  • $\begingroup$ As I wrote, to me it means $P(M| A\bar{D} \cup \bar{A}D) = 0.25$, it seems you suggest $P(M | A\bar{D}) = 0.25$, if yes, could you please explain why my understanding is not correct? $\endgroup$ – Ahmad Mar 13 at 18:13
  • $\begingroup$ My reading is that it means both $P(M \mid A\bar{D} ) =0.25$ and $P(M \mid \bar{A}D) =0.25$, and indeed $P(M \mid A\bar{D} \cup \bar{A}D) =0.25$ too. My interpretation provides an answer to the question, which is a point in its favour $\endgroup$ – Henry Mar 13 at 18:24
  • $\begingroup$ You mean from the probability point of view all three can be held at the same time? Also, regarding the question title, is $P(A | B \cup C) = P(A |B) +P(A|C) - P(A| B \cap C)$ $\endgroup$ – Ahmad Mar 13 at 20:28

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