Question

Let A1, A2, A3, and A4, be events from a common sample space. these events are pairwise mitially exclusive, with the exception tha A1 and A2 occur simultaneously with a probability of 0.1. if events A1, A2, and A3 each occur with probability 0.3, what is the largest possible value for the probability of A4?

this is my work:
P(A1) = P(A2) = P(A3) = 0.3
sum of the probabilities of all events must be equal to 1, so we have:
P(A1) + P(A2) + P(A3) + P(A4) = 1
the probability of both A1 and A2 is 0.1
P(A1 and A2) = 0.1
the probability of the intersection of two events is given by P(AnB) = P(A) + P(B) - P(AuB)
So, we can write:
P(A1) + P(A2) - P(A1 and A2) = 0.3 + 0.3 - 0.1 = 0.5
we have:
P(A1) + P(A2) + P(A3) + P(A4) = 1
0.3 + 0.3 + 0.3 + P(A4) = 1
P(A4) = 1 - 0.9 = 0.1

Answer 1

The largest possible value for the probability of A4 is 0.3.

If events A1, A2, A3, and A4 are pairwise mutually exclusive with the exception that A1 and A2 occur simultaneously, then the probability of A1 and A2 occurring together is 0.1. This means that the probability of A1 occurring separately is 0.3 - 0.1 = 0.2, and the same is true for A2.

Since A1, A2, A3, and A4 are pairwise mutually exclusive, the probability that they all occur together is 0. Therefore, the sum of their individual probabilities is also 0. Therefore, the probability of A4 is the remaining probability after accounting for the probability of A1, A2, and A3. The largest possible value for the probability of A4 is the total probability minus the probability of A1, A2, and A3.

So, the largest possible value for the probability of A4 would be:

1 - (0.2 + 0.2 + 0.3) = 1 - 0.7 = 0.3

Therefore, the largest possible value for the probability of A4 is 0.3.