SOLUTION: For the events A and B:P(A) = 0.7,P(AUB)=0.9,P(A and B)=0.3 find P(-B and A)

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Question 953339: For the events A and B:P(A) = 0.7,P(AUB)=0.9,P(A and B)=0.3 find P(-B and A)
Answer by Theo(13342) (Show Source):
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p(a and not b) is equal to .4.

you are given that:

p(a) = .7
p(a or b) = .9
p(a and b) = .3

the formula to use is:

p(a or b) = p(a) + p(b) - p(a and b)

replace p(a or b) with .9 and replace p(a) with .7 and replace p(a and b) with .3 and the formula becomes:

.9 = .7 + p(b) - .3

solve for p(b) to get p(b) = .5

since p(a) includes p(a and b), you have to remove p(a and b) to get p(a) only.

this results in p(a only) = .7 - .3 = .4

similarly, since p(b) includes p(a and b), you have to remove p(a and b) to get p(b) only.

this results in p(b only) = .5 - .3 = .2

your solution is that p(a and not b) is equal to p(a only) which is equal to .4

the following diagram show the relationship in pictures.

in this diagram, the set of c is defined as the set that contains all of the elements that are not in the set of a and also not in the set of b.

the probability of that occurring is defined as:

p(not(a or b)) = 1 - p(a or b) which becomes 1 - .9 which is equal to .1.

that's why the set of c has a probability of .1 associated with it.

from the diagram, you can see that the set of a intersects with the set of b and intersects with nothing else.

the set of (a and b) is defined as the intersection of the sets a and b.

those are the elements that are in set a and b at the same time.

take those away and you have the set of elements that are in set a and not in b.