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The events A, B, C have the probabilities: P(A|B)=0.25, P(C|B)=0.5, P(A∩C|B)=0.10. Given that B has happened, find the following probabilities:

a) That only C has happened 

b) That only C or only A has happened, but not both of them

c) That C or A has happened

I have answered the three questions, but want to know if the logic is correct (A' = A compliment):

a) P(A'∩C|B) = P(C|B) - P(A∩C|B) = 0.5 - 0.1 = 0.4

On the RHS I simply take the probability that C happened given B, and subtract from it the probability that A and C happened. The LHS is the only way I can find of representing only C without including C∩A, though I'm not sure if this is the correct way of representing that.

b) P(C∪A|B) = P(C|B) + P(A|B) - 2*P(C∩A|B) = 0.5 + 0.25 - 0.2 = 0.55

Addition rule of probabilities. Subtract twice the intersection, once for repeat, and second because we don't want to include it at all.

c) P(C∪A|B) = P(C|B) + P(A|B) - P(C∩A|B) = 0.5 + 0.25 - 0.1 = 0.65

Same as above, except only subtract intersection once.

Is my logic above correct?

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  • $\begingroup$ a) only C has happened among the three or among A and C only? $\endgroup$ – msm Sep 29 '16 at 4:16
  • $\begingroup$ I'm not sure, since the question does not clarify. But does it really matter? If we know B happened, then we're inside the sample space of B already, so technically speaking wouldn't it be among the three anyways? $\endgroup$ – Leo Denni Sep 29 '16 at 4:24
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Sep 29 '16 at 4:31
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a) if by "only C has happened" it means among the two, then you are correct. However, if it means among the three, then the probability is zero $$P(A'\cap B' \cap C|B)=0$$

b) $$\begin{align}P(\text{only A or C}|B)&=P((A\cap C')\cup(A'\cap C)|B)\\ &=P(A\cap C'|B)+P(A'\cap C|B)-P((A\cap C')\cap(A'\cap C)|B)\\ &=P(A|B)-P(A\cap C|B)+P(C|B)-P(A\cap C|B)-0 \end{align}$$

c) is correct

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  • $\begingroup$ Isn't b) the same as I wrote? $\endgroup$ – Leo Denni Sep 29 '16 at 4:32
  • $\begingroup$ Yes but you have written $P(C∪A|B)$ at the beginning (which is actually the same as c) and is wrong. $\endgroup$ – msm Sep 29 '16 at 4:34
  • $\begingroup$ Oh right, I didn't consider the LHS was wrong. It wouldn't make sense for it to be the same as c). Thanks! $\endgroup$ – Leo Denni Sep 29 '16 at 4:35

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