On: Want to be an actuary? Odds are, you’ll fail the t...

[grp000]

Did you reverse the likelihoods? P(pc) = 2 * P(pd), probability of purchasing C is twice as likely to probability of purchasing D.

[ogogmad → grp000]

In 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 → ogogmad]

Nicely 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 → HighChaparral]

Worth 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.