p sample, july 1

$ \gdef\and{\cap} \gdef\or{\cup} \gdef\bar#1{\overline{#1}} $

See here.

The moral of the story today is that conditional probabilities are really hard for me.

81: two life policies

Two life insurance policies, each with a death benefit of 10,000 and a one-time premium of 500, are sold to a married couple, one for each person. The policies will expire at the end of the tenth year. The probability that only the wife will survive at least ten years is 0.025, the probability that only the husband will survive at least ten years is 0.01, and the probability that both of them will survive at least ten years is 0.96.

Calculate the expected excess of premiums over claims, given that the husband survives at least ten years.

Clearly a conditional probability problem, with $$ \begin{aligned} P(\text{both}) &=& P(H\and W) &= 0.96 \\ P(\text{wife only}) &=& P(\bar H\and W) &= 0.025 \\ P(\text{husband only}) &=& P(H \and \bar W) &= 0.01 \\ P(\text{neither survives}) &=& P(\bar H\and\bar W) &= 0.005 \end{aligned} $$

Given that the husband is alive, the profit on his policy is 100%, so we expect a correct answer above $\$500$. So we need to extract that probability of a payment on the wife's policy, given that the husband survives, which is $ P(\bar W | H)$, to find the policy's expected payout.

The probability should clearly be approximately $P(H\and \bar W) = 0.01$, giving a $\$400$ expected profit on the wife's policy, but none of the options are nice round numbers. There's some conditional correction.

The total probability that the husband survives should be $$ \begin{aligned} P(H) &= P(H\and W) + P(H \and \bar W) \\ &= 0.96 + 0.01 = 0.97 \end{aligned} $$

I tried $$ \begin{aligned} P(\bar W | H) &= \frac{P(H\and \bar W)}{P(H)} \\ &= \frac{0.01}{0.97} \overset{???}{=} 0.0155 \end{aligned} $$

to find the expected payout $$ \begin{aligned} E(\text{payout} | H) & = \$0\times P(W|H) + \$10\mathrm k\times P(\bar W|H) \end{aligned} $$

That's the right computation. But for some reason I used $\frac{0.015}{0.97}$ instead. I don't really have any idea why.

Anyway, I was close enough to choose the correct $\$897$ instead of the incorrect $\$870$.

274: customers insure different subsets of cars

I felt good about this, but apparently I shouldn't have.

An insurance company examines its pool of auto insurance customers and gathers the following information:

1. All customers insure at least one car.
2. 62% of the customers insure more than one car.
3. 15% of the customers insure a sports car.
4. Of those customers who insure more than one car, 20% insure a sports car.

Calculate the probability that a randomly selected customer insures exactly one car, and that the car is not a sports car.

I wrote that $20\%\times62\% = 12.4\%$ of all customers insure a sports car and something else, which means that $2.6\%$ of all customers insure only a sports car.

The solution says,

Consider the following events about a randomly selected auto insurance customer:

• A = customer insures more than one car
• B = customer insures a sports car

We want to find the probability of the complement of A intersecting the complement of B (exactly one car, non-sports). $$ \begin{aligned} P(A^c\and B^c) &= 1- P(A\or B) \\ &= 1-\left( P(A) + P(B) - P(A\and B) \right) \end{aligned} $$ Using $$ \begin{aligned} P(A\and B) &= P(B|A) P(A) = 0.124, \end{aligned} $$ we find $$ \begin{aligned} P(A^c \and B^c) &= 1 - (0.62 + 0.15 - 0.124) = 0.354. \end{aligned} $$

Oh, I read the fucking thing wrong. I wanted $38\% - 2.6\% = 35.4\%.$

Okay, not serious.

329: left-handed children

A geneticist compiled the following information:

1. 1/2 of children who have two left-handed parents are left-handed.
2. 1/6 of children who have exactly one left-handed parent are left-handed.
3. 1/16 of children who have no left-handed parents are left-handed.
4. 1/50 of children have two left-handed parents.
5. 1/5 of children have exactly one left-handed parent.

Calculate the probability that a randomly selected left-handed child has no left-handed parents.

I got this one right, but it took me a long time and I wrote profanity on my paper nine minutes into the problem.

I always end up solving these conditional-probability problems by finding the population of every set and every subset

These are very hard for me to do algebraically. I need to come up with a graphical way that's consistent. Perhaps I can spend some time on that later today.