For each problem, write out the number of objects ($n$), the length of the sequence/sample ($l$), if repeats are allowed, and if you are computing permutations or combinations. 1 point for this on each problem
You can answer two ways. First you could imagine choosing yes/no for each day of the week. Then you're choosing a sequence like so: YNYYN. You would view the problem as:
$n = 2\, l = 7$, replacement, permutations
$$ 2^{7} = 128 $$You can also use the rule from lecture about adding up combinations (section 1.3):
$n = 7\, l = 0,\ldots 7$, no replacement, combinations
$$ 2^{7} = 128 $$Simplfiy the following expression and write in words the probability law or definition being used to go from the left-hand-side to the right-hand-side. There should be no intermediate steps. Answer the following questions in Markdown. 2 points each
$\DeclareMathOperator{\P}{P}$ $\DeclareMathOperator{\E}{E}$
2.1 $\sum_{\mathcal{Y}}\P(X = 3|Y = y)\P(Y = y) = \P(X = 3)$. Marginalization of conditional
2.2 $\P(X = 0|Y = 1) = \P(X = 0)$. Definition of independence
2.3 $\sum_{\mathcal{Y}}\P(X = -1 ,Y = y) = \P(X = -1)$. Definition of marginal
2.4 $\sum_{\mathcal{X}}\P(X = x|Y = 1) = 1$, due the the Law of Total Probability.
2.5 $\frac{\P(X = 4, Y = 1)}{\P(Y = 1)} = \P(X = 4|Y = 1)$. Definition of conditional
2.6 $P(X = 0 | Z = 0)$. Definition of conditional independence.
2019-nCoV (武汉肺炎, pronounced Wǔhàn fèiyán) is a new coronavirus first found in the city of Wuhan in the Hubei province of China. A new protocol has been suggested for screening people arriving from the Hubei region. We want to use this test to quarantine probable cases to prevent the spread of nCoV and predict probability of death. People can either not have nCov or have nCov and our test can be positive or negative and the person can be dead or alive from the disease. Let's take the random variable $C$ to indicate if the patient has nCov, $T$ to indicate if the test is positive, and $S$ to be their state (alive = 1) after 7 days. On our first day of testing, 1200 people were tested and 200 had a positive test. Follow-up testing of those 200 showed that 11 had confirmed infections by nCov. We also, to ensure accuracy, did additional tests on the 1,000 who had a negative test result and found that one had nCov. Separately, we know that out of the 560 confirmed cases of nCov, 17 people died from the disease. Using this information, answer the following question:
[1 point] Write out the sample space
[3 points] Write out three probability statements given by the prompt.
[8 points] Make a markdown table showing the probability of every element in the state space. State any assumptions you make.
[3 points] What is the probability of dying and having a positive test?
[3 points] What is the probability of dying if your test was positive? So should you be worried about a positive test?
[3 points] What is the probability of dying if your test was negative?
[5 points] If we decide to use this test and 2,400 people arrive per day from Wuhan, how many will get through with nCov? Use $2,400 \times P(C = 1, T = 0)$ Is this a good test?
There are many ways to arrive at these answers. I'll start by only considering the marginal of C,T first.
| CT | Probability |
|---|---|
| 00 | 999 / 1200 |
| 01 | 189 / 1200 |
| 10 | 1 / 1200 |
| 11 | 11 / 1200 |
I know that the result of the test doesn't matter if I know for sure (conditional independence) if someone actually has the disease. So $P(S = 0 | C = 1, T = ?) = P( S = 0 | C = 1) = P(S = 0, C = 1) / P( C = 1)$. I just need $P(C = 1)$ to solve this equation, which I can get from the table above: $P(C = 1) = (1 + 11) / 1200$
$$ P(S = 0, C = 1) = \frac{17}{560} \times \frac{12}{1200} = \frac{17}{56000} $$Now I can add back the $T$ state
$$ P(S = 0, C = 1, T = 0) = P(T = 0 | S = 0, C = 1) P(S = 0, C = 1) = \frac{1}{12} \times \frac{17}{56000} = 2.53\times 10^{-5} $$You can follow this process to get the rest.
| CTS | Probability |
|---|---|
| 000 | ~ 0 |
| 001 | 0.8330000 |
| 010 | ~ 0 |
| 011 | 0.1575000 |
| 100 | 0.0000253 |
| 101 | 0.0008080 |
| 110 | 0.0002780 |
| 111 | 0.0088900 |
Note that I have assumed the probability of dying if you don't have nCov is 0, at least for the time frame we're considering.