let the probability of Aron going on Saturday be A
A
= {Aaron goes to the gym on Monday},
and the probability of him going on a Sunday be B
B
= {Aaron goes to the gym on Sunday}.
In your post, we have the following pieces of information:
P(A)=0.8
P(A)=0.8
,
P(B∣A)=0.3
P(B∣A)=0.3
,
P(B∣ A)=0.9
P(B∣A¯)=0.9
.
What you are looking for is the probability that he goes only once (either on Saturday or Sunday). That is the probability of the event
(A∩ B¯ )∪( A¯ ∩B)
(A∩B¯)∪(A¯∩B)
. [(A∩ B¯)(A∩B¯)
: if he goes on Saturday he won't go on Sunday
A¯∩B
A¯∩B
: if he did not go on Saturday, he will go on Sunday].
We have:
P
((A ∩ B)∪(A ∩B))
=
P(A∩ B )+
P(A∩B)
=
P( B ∣A)
P(A)+
P(B∣ A)
P( A)
=
(1−P(B∣A))
P(A)
+
P(B∣ A)
(1−P(A))
=(1−0.3)×0.8+0.9×0.2
The prob of Aaron going to the gym on exactly one of the two days is =0.74.
P((A∩B¯)∪(A¯∩B))=P(A∩B¯)+P(A¯∩B)=P(B¯)