Suppose that A,B and C are mutually exclusively events such that; P(A) =0.3, P(B)=0.1, and P(C)=0.5. Evaluate;

P(AuBuC)

To evaluate P(AuBuC), we need to find the probability of the union of events A, B, and C.

In probability theory, the union of events can be found by adding their individual probabilities and subtracting the intersection of the events (if any).

In this case, since A, B, and C are mutually exclusive events, it means that they cannot occur at the same time. Therefore, their intersection is empty, and we don't need to subtract anything.

To find P(AuBuC), we simply add the individual probabilities of A, B, and C:

P(AuBuC) = P(A) + P(B) + P(C)

Substituting the given values:

P(AuBuC) = 0.3 + 0.1 + 0.5

Simplifying:

P(AuBuC) = 0.9

Therefore, the probability of the union of events A, B, and C (P(AuBuC)) is 0.9.