IF THE PROBABILITY OF A IS 3/5.AND PROBABILITY OF A INTERSECTION B IS 3/10.WHAT IS THE PROBABILITY OF B?
1/30
7/10
I drew a Venn diagram
A is 6 of 10 including intersection
A intersection B is 3/10
That leaves 4 of 10 for B alone
so total B is 4 plus the intersection 3
To find the probability of event B, we need to know the probability of event A and the probability of the intersection of events A and B.
Given:
Probability of event A (P(A)) = 3/5
Probability of the intersection of events A and B (P(A ∩ B)) = 3/10
To find the probability of event B (P(B)), we can use the formula for the probability of the union of two events:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Rearranging the formula, we can solve for P(B):
P(B) = P(A ∪ B) - P(A) + P(A ∩ B)
We know that P(A ∪ B) is equal to 1 (since the union of two events encompasses the entire sample space, hence its probability is 1):
P(B) = 1 - P(A) + P(A ∩ B)
Substituting the values we know:
P(B) = 1 - 3/5 + 3/10
P(B) = 1 - 6/10 + 3/10
P(B) = 7/10
Therefore, the probability of event B is 7/10.