Question

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kor an event the noiation ร A represcits ihe number of outcomes in A. Rccall iliai if the ouicomcs m a sample space S are equally likely, then the probability of an event A is given by n(A) n(S) # of outcomes in A # of outcomes in S P(A) = Show that this definition of probability satisfies the three axioms of a probability measure. In other words, show that for this definition of P the following statements hold (a) P(S) 1 (b) P(A) 20, for any event A (c) P(A U B) = P(A) + P(B), for any disjoint events A and B

Answer 1

A) P(S)=1

As all the outcomes in S are equally likely.

P(A) = n(A)/n(S)

P(~A) = (n(S) - n(A))/n(S)

P(S) = P(A) + P(~A) = 1

B) for any event A , n(A) >=0 and always n (S) >=0

So P(A) = n(A)/n(S) >=0

C) P(B) = n(B)/n(S)

If A and B are disjoint than elements in A are not in B..

P(A union B) = n(A union B)/n(S) = (n(A) + n(B))/n(S) = n(A)/n(S) + n(B)/n(S) = P(A) + P(B)