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not-my-account

Is the answer to the question at the top 0.33?

let pc = purchased_collision, pd = purchased_disability

we have:

2 * P(pc) = P(pd) P(pc & pd) = 0.15

P(pc) * P(pd) = 0.15 2 P(pc)^2 = 0.15 --> P(pc) = 0.274, P(pd) = 0.548

So the probability of neither pc nor pd is

!P(pc | pd) = 1 - P(pc) - P(pd) + P(pc & pd) = 0.328 ~= 0.33?

The addition of P(pc & pd) was to take account of the double counting of pc and pd.

Please let me know if I've made a mistake!

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It is yes.

Definitely need a calculator to be able to do sqrt(0.075) and calculate the result.

Though I was able to figure it out in my head without a calculator by eliminating choice A and choice C (I tried pd=0.3 and pd=0.2 to prove the answer had to be >0.28 and <0.48).

Ah clever!

I did trial and error with some possible numbers until I got really close with .55 and .275… then I used those and got .32625, which was really close to 33%

I got the same result as you, but in my mind the expression "twice as likely to purchase" could just as easily have meant P(C) = 2 * P(D) / 1 + P(D) (yielding 0.35824) rather than 2 * P(C) (yielding 0.32841).

I got the same. When do we start as actuaries?

I mean, these probably weren't that hard. I was able to solve all of them with just a standard calculator that could do sqrt.

Yeah, I couldn’t read the article but saw the first question… I got 33%, too.

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

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}

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)

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.

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