Prove (3.1) for a nonuniform sample space. Hints: Remember that the probability of an event is the sum of the probabilities of the sample points favorable to it. Using Figure 3.1, let the points in A but not in AB have probabilities p₁, p₂, ... p_n, the points in have probabilities p_n+1, p_n+2, .... + p_n+k, and the points in B but not in AB have probabilities p_n+k+1, p_n+k+2, ....p_n+k+l. Find each of the probabilities in (3.1) in terms of the ’s and show that you then have an identity.

Let p₁, p₂, ... p_n be the probabilities of points in A but not in AB.

Assume that p_n+1, p_n+2, ....p_n+k be the probabilities of points in AB and p_n+k+1, p_n+k+2, ....p_n+k+lbe the probabilities of points in B but not in AB.

Therefore, the sum of probabilities of points in is given by,

p(AB) = p_n+1+ ....p_n+k.

Also, the sum of probabilities of points in A but not in AB is given by,

p(A) = p₁ + .... + p_n + p_n+1+ .... + p_n+k

We can write from the above two equations,

p_A(B) = p_n+1 + ....p_n+k/p₁ + .... + p_n + p_n+1 + .... + p_n+k

.

Since, it given that A happened (considered only the points in A) and taking the sum of probabilities of points in B (which is equal to AB since A happened ).

Hence, we get,

P(AB) = P(A). P_A(B)

Hence, the proof of result (3.1) is done.