p sample, july 14

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

See here.

Well, that was interesting.

First off, about half of these were repeats. But I'm not sure that I really did better on the repeats. In fact, the one that I missed today was a repeat that I'm pretty sure I got right the last time I did it.

190. Life insurance auditor.

A life insurance company has found there is a 3% probability that a randomly selected application contains an error. Assume applications are mutually independent in this respect.

An auditor randomly selects 100 applications.

Calculate the probability that 95% or less of the selected applications are error-free.

Two errors would be a 94% correct rate, so we want the probability that the number of errors is zero or one. This is a binomial problem, with

$$\begin{aligned} P(0) &= {100\choose 0} 0.03^0 0.97^{100} &&= 0.04755 \\ P(1) &= {100\choose 1} 0.03^1 0.97^{100-1} &&= 0.14707 \\ P(2) &= {100\choose 2} 0.03^2 0.97^{100-2} &&= 0.2251 \\ & \vdots \end{aligned}$$

I add these first two to get 0.1946. But the options are $$ 0.08\quad 0.10\quad 0.13\quad 0.15\quad 0.18 $$ What gives?

Oh, I'm being ridiculous. A 95% correct rate is four or fewer incorrect.

from scipy.special import comb

def binom_pmf(p, n):
    k = np.arange(0,n+1)
    return comb(n,k) * p**k * (1-p)**(n-k)
def binom_cdf(p, n, k=None):
    return np.cumsum(binom_pmf(p,n))
for k, F in enumerate(binom_cdf(0.03,100)):
    print(f"{k=} {F=:0.4f} {1-F=:0.4f}")
    if k > 4: break

k=0 F=0.0476 1-F=0.9524
k=1 F=0.1946 1-F=0.8054
k=2 F=0.4198 1-F=0.5802
k=3 F=0.6472 1-F=0.3528
k=4 F=0.8179 1-F=0.1821

Gosh, it's a real coincidence that I was close at all.

458. Diversify to minimize total variance.

An investor wants to purchase a total of ten units of two assets, A and B, with annual payoffs per unit purchased of X and Y, respectively. Each asset has the same purchase price per unit. The payoffs are independent random variables with equal expected values and with Var(X) = 30 and Var(Y) = 20.

Calculate the number of units of asset A the investor should purchase to minimize the variance of the total payoff.

I forgot here that $\text{Var}(aX) = a^2\text{Var}(X)$, and I make the problem inear.

I'm annoyed because I've actually solved this one before, and I distinctly remember getting the answer somewhere in the middle of the table.

459. Average claims over many groups.

For its group life policies, an insurer models the number of claims per group as independent Poisson random variables with common mean 16.

The insurer randomly selects 64 of its groups.

Calculate the probability that the average number of claims per group is between 15 and 18.

Same issue here: trouble with the difference between the variance versus the standard deviation, and which one is squared.

Blugh.

265. Find the resident.

A health insurer sells policies to residents of territory X and territory Y. Past claims experience indicates the following:

1. 20% of the total policyholders from territory X and territory Y combined filed no claims.

2. 15% of the policyholders from territory X filed no claims.

3. 40% of the policyholders from territory Y filed no claims.

Calculate the probability that a randomly selected policyholder was a resident of territory X, given that the policyholder filed no claims.

Honestly it looks like I just made an algebra mistake here, and caught it just as I was about to move on.