Suppose you have a die, with numbers 1-6.
You roll it twice. Independent events.
What's the chance you roll a multiple of 2 or a multiple of 3?
P(*3) = P(3 or 6) = 2/6
P(*2) = P(2 or 4 or 6) = 3/6
P(*2 & *3)) = P(6) = 1/6
P(*2 or *3) = P(2 or 3 or 4 or 6) = 4/6
P(either) = P(*2) + P(*3) - P(*2 & *3)
if the event sets are not disjoint, some events are counted twice.
If events A and B are independent and
P(A) and P(B) are nonzero, explain why P(A U B) does not equal P(A)+ P(B)
im not sure how to answer this
1 answer