3 ms·
Did you reverse the likelihoods? P(pc) = 2 * P(pd), probability of purchasing C is twice as likely to probability of purchasing D.
by grp000 5y ago
Did you reverse the likelihoods? P(pc) = 2 * P(pd), probability of purchasing C is twice as likely to probability of purchasing D.
- ogogmad 5y agoIn 3 steps: I. P(C) = 2 P(D) II. 0.15 = P(C and D) = P(C) P(D) {by independence} = 2 P(D)^2 {by I} implies P(D) = 0.27 III. P(~C and ~D) = P(~C) P(~D) {by independence} = (1 - P(C))(1 - P(D)) = 0.33 {by I and II}
- HighChaparral 5y agoNicely laid out. I goofed on the last bit by assuming the final answer was: 1 - P(C) - P(D) which gives 0.178 ~ 0.18 or answer (A)
- ogogmad 5y agoWorth remembering that: I. P(A and B) = P(A) P(B) {when independent} II. P(A or B) = P(A) + P(B) {when mutually exclusive} But in general: III. P(A and B) = P(A) P(B|A) IIII. P(A or B) = P(A) + P(B) - P(A and B) The only way that I and II can both be true is that either P(A) = 0 or P(B) = 0. This does not imply though that A and B are mutual exclusive and independent, but it is a necessary condition. The conclusion is that if you're using both I and II, you're almost certainly doing something wrong.