p sample, july 15

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

See here.

Cheating a little bit here: on two problems I got the right answer, but I wrote the wrong letter down on my paper. Super annoying. I don't think I'm going to do that when I take my exam on Monday.

135. How much did the student know?

A student takes an examination consisting of 20 true-false questions. The student knows the answer to N of the questions, which are answered correctly, and guesses the answers to the rest. The conditional probability that the student knows the answer to a question, given that the student answered it correctly, is 0.824..

Calculate N.

I got the conditional probability wrong twice here, so I started plugging in answers. With $K$ for "known answer" and $C$ for "correct answer," I found

But that's a clumsy way to do it, which relies on knowing some possible correct answers.

I tried conditional probability here:

$$\begin{aligned} P(K|C) &= \frac{P(C|K) P(K)}{P(C)} \\&= \frac{ 1 \times \frac{N}{20}}{ N + \frac{20-N}{2}} \\&= \frac{ N }{200 + 10N} \\ ({200 + 10N})\times P(K|C) &= N \\ N &= \frac{200P}{1-10P} \end{aligned}$$

But that's negative. What did I miss?

Solution manual

We have

$$\begin{aligned} P(C|\bar K) &= 0.5 \\ P(C| K) &= 1 \\ P(K|C) &= 0.824 \\ P(K) &= N/20 \\ \end{aligned}$$

Oooooh, I should have said something like $P(\bar K) = \frac{20-N}{20}$. That was my problem.

31. How many excess high-risk drivers?

A large pool of adults earning their first driver’s license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% high-risk drivers. Because these drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool.

This month, the insurance company writes four new policies for adults earning their first driver’s license.

Calculate the probability that these four will contain at least two more high-risk drivers than low-risk drivers.

I wrote that the only option was (LHHH) and its permutations, with probability 0.016. This wasn't an option.

I didn't consider the cases with zero low-risk drivers. I should have counted (HHHH), (MHHH), (MMHH), and (LHHH).

33. Calculate the probability that a machine part fails early.

The lifetime of a machine part has a continuous distribution on the interval $(0, 40)$ with probability density function $f(x)$, where $f(x)$ is proportional to $(10 + x)^{-2}$ on the interval.

Calculate the probability that the lifetime of the machine part is less than 6.

I found the normalization,

$$\begin{aligned} \frac{1}{A} &= \int_0^{40} \frac{\mathrm dx}{(10 + x)^2} \\&= \left.\frac{-1}{10+x}\right|_0^{40} \\&= \frac{1}{10} - \frac{1}{50} = \frac{2}{25} A &= \frac{25}{2} \end{aligned}$$

But then, when I did the probability integral, I forgot the normalization, finding $P =\frac{1}{10} - \frac{1}{16}$ rather than $P =\frac{25}{2}\left(\frac{1}{10} - \frac{1}{16}\right)$.

Oops.

533. Six employees are in department A, and four are in department B.

I did the arithmetic right, saw that the number I wanted was at the end of the five-letter list, and wrote down "B" because the question was about how many injured employees are in group B. I'm not concerned about repeating that on a computer-based test.