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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 so P(pc
by not-my-account 5y ago
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
so
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!
- ffwacom 5y agoI got the same. When do we start as actuaries?
- TheRealNGenius 5y agoI mean, these probably weren't that hard. I was able to solve all of them with just a standard calculator that could do sqrt.
- ZephyrP 5y agoI 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).
- ummonk 5y agoIt 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).
- not-my-account 5y agoAh clever!
- cortesoft 5y agoI 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%
- grp000 5y agoDid 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.
- cortesoft 5y agoYeah, I couldn’t read the article but saw the first question… I got 33%, too.