Chapter Two Answers

Question 3

According to the CDC (Centers for Disease Control and Prevention), men who smoke are 23 times more likely to develop lung cancer than men who don’t smoke, Also according to the CDC, 21.6% of men in the U.S. smoke, What is the probability that a man in the U.S. is a smoker, given that he develops lung cancer?

Question

$$ P(S|C)=\dfrac{\frac{23}{24} \cdot .216}{.239\overline{6}}\approx.864 $$

Answer

$$P(S|C)=\dfrac{P(C|S)P(S)}{P(C)}$$ $$P(C) = P(C|S)P(S) + P(C|S^c)P(S^c)$$ plugging in our values from above, we get $$P(C) = \frac{23}{24}\cdot .216 + \frac{1}{24} \cdot .784 = .239\overline{6} $$ our final equation is $$ P(S|C)=\dfrac{23/24 \cdot .216}{.239\overline{6}}\approx.864$$ Notice that we do not actually know what the rate of cancer is. Knowing that smokers are 23 times more likely to get cancer is enough for us to answer this question.

Explaination

Question 4

Fred is answering a multiple-choice problem on an exam, and has to choose one of n options (exactly one of which is correct). Let K be the event that he knows the answer, and R be the event that he gets the problem right (either through knowledge or through luck). Suppose that if he knows the right answer he will definitely get the problem right, but if he does not know then he will guess completely randomly. Let P(K) = p.

(a) Find P(K|R) (in terms of p and n).
(b) Show that P(K|R) ≥ p, and explain why this makes sense intuitively. When (if ever) does P(K|R) equal p?

Question

A) $$ P(K|R) = \dfrac{p}{p + (1-p)/n} $$ B) Intuitivly, $P(K|R)$ will always be greater than or equal to $p$ because getting the answer right is usually evidence in favor of knowlege.
When $p=1$ or $p=0$, then $P(K|R)=p$

Answer

This equation might not look like Bayes because $P(R|K) = 1$, and therefore we can implicitly multiply by 1.

Explanation

Question 6

Question

$$ P(D|H) = \dfrac{.01}{(.01)+(.99 \cdot .5^7)} \approx 0.564 $$

Answer

First let's define our events $ D = \text{double-sided coin selected} $ $ H = seven heads in a row $ We are told that $P(D) = .01$ and $P(H|D) = 1$. With a bit of common sense we can finish building this equation $$ P(D|H) = \dfrac{P(H) \cdot P(H|D)}{P(D) \cdot P(H|D) + P(D^c) \cdot .5^7} $$ $.5^7$ is the probability of flipping a fair coin 7 times and getting heads heach time.

Explanation

Question 7

Question

A) $$ P(D|H) = \dfrac{\dfrac{1}{2} \left(\dfrac{99}{100} \cdot \left(\dfrac{1}{2}\right)^7 + \dfrac{1}{100}\right)} {\dfrac{1}{2} \left(\dfrac{99}{100} \cdot \left(\dfrac{1}{2}\right)^7 + \dfrac{1}{100}\right) + \left(\dfrac{1}{2}\right)^8 } \approx .694 $$ B) $$ P(D|H) = \dfrac{.005}{.005 + (.995 \cdot .5^7)} \approx .391 $$

Answer

Explanation

Question 8

The screens used for a certain type of cell phone are manufactured by 3 companies, A, B, and C. The proportions of screens supplied by A, B, and C are 0.5, 0.3, and 0.2, respectively, and their screens are defective with probabilities 0.01, 0.02, and 0.03, respectively. Given that the screen on such a phone is defective, what is the probability that Company A manufactured it?

Question

$$ P(A|D) = \dfrac{\frac{1}{200} }{\frac{1}{200} + 2(\frac{3}{500}) } = \dfrac{5}{17} $$

Answer

The numerator is the probability that a screen made by company A is defective times the probability that a screen is made by company A, which is $.5 \cdot .01$ or $\dfrac{1}{200}$
The denominator is the probability that any screen is defective. We find this by adding up the probability of each company making a screen and probability that a screen by that company is defective.

Explanation

Chapter 9

(a) Show that if events A1 and A2 have the same prior probability P (A1) = P (A2), A1 implies B, and A2 implies B, then A1 and A2 have the same posterior probability P (A1|B) = P (A2|B) if it is observed that B occurred.
(b) Explain why (a) makes sense intuitively, and give a concrete example.

Question

A)
If two events have the same prior probability, that means they are equally likely to happen. If they both imply the same event and that event happens, that just means the probability of each event stays the same.
B)
If there is a 10 percent chance of rain and a 10 percent chance of thunder, and both events imply clouds, then the fact that it is cloudy changes nothing.

Answer

Question 10

Fred is working on a major project. In planning the project, two milestones are set up, with dates by which they should be accomplished. This serves as a way to track Fred’s progress. Let A1 be the event that Fred completes the first milestone on time, A2 be the event that he completes the second milestone on time, and A3 be the event that he completes the project on time. Suppose that $P(A_{j+1}|A_j )=0.8$ but $P(Aj+1|A^c_j )=0.3$ for j = 1, 2, 3. If Fred falls behind on his schedule it will be hard for him to get caught up. Also, assume that the second milestone supersedes the first, in the sense that once we know whether he is on time in completing the second milestone, it no longer matters what happened with the first milestone. We can express this by saying that $A_1$ and $A_3$ are conditionally independent given $A_2$ and they’re also conditionally independent given $A^c_2$.

(a) Find the probability that Fred will finish the project on time, given that he completes the first milestone on time. Also find the probability that Fred will finish the project on time, given that he is late for the first milestone.
(b) Suppose that P(A1)=0.75. Find the probability that Fred will finish the project on time.

Question

A) $$ P(A_3|A_1) = .8^2 + .2(.3) = .7 $$ $$ P(A_3|A_1^c) = .3(.8) + .7(.3) = .45 $$ B) $$ P(A_3) = 0.75 \cdot 0.8 \cdot 0.8 + 0.75 \cdot 0.2 \cdot 0.3 + 0.25 \cdot 0.3 \cdot 0.8 + 0.25 \cdot 0.7 \cdot 0.3 = .6375 $$

Answer

Explanation

Question 11

An exit poll in an election is a survey taken of voters just after they have voted. One major use of exit polls has been so that news organizations can try to figure out as soon as possible who won the election, before the votes are officially counted. This has been notoriously inaccurate in various elections, sometimes because of selection bias: the sample of people who are invited to and agree to participate in the survey may not be similar enough to the overall population of voters. Consider an election with two candidates, Candidate A and Candidate B. Every voter is invited to participate in an exit poll, where they are asked whom they voted for; some accept and some refuse. For a randomly selected voter, let A be the event that they voted for A, and W be the event that they are willing to participate in the exit poll. Suppose that P(W|A)=0.7 but P(W|A c )=0.3. In the exit poll, 60% of the respondents say they voted for A (assume that they are all honest), suggesting a comfortable victory for A. Find P(A), the true proportion of people who voted for A.

Question

$$ P(A) \approx .391 $$

Answer

To find $P(A)$, we find the equation for $P(A|W)$ and then solve for $P(A)$. $$ P(A|W) = \dfrac{.7x}{.7x + .3(1-x)} = .6 $$ Noticed I replaced $P(A)$ with $x$ for clarity.
Now solve for $x$ to get the answer.

Explanation

Question 12

Alice is trying to communicate with Bob, by sending a message (encoded in binary) across a channel.

(a) Suppose for this part that she sends only one bit (a 0 or 1), with equal probabilities. If she sends a 0, there is a 5% chance of an error occurring, resulting in Bob receiving a 1; if she sends a 1, there is a 10% chance of an error occurring, resulting in Bob receiving a 0. Given that Bob receives a 1, what is the probability that Alice actually sent a 1?
(b) To reduce the chance of miscommunication, Alice and Bob decide to use a repetition code. Again Alice wants to convey a 0 or a 1, but this time she repeats it two more times, so that she sends 000 to convey 0 and 111 to convey 1. Bob will decode the message by going with what the majority of the bits were. Assume that the error probabilities are as in (a), with error events for different bits independent of each other. Given that Bob receives 110, what is the probability that Alice intended to convey a 1?

Question

A) $$ P(S_1|R_1) = \dfrac{.5 \cdot .9}{(\frac{.9}{2})+(\frac{.05}{2})} \approx .947 $$ B) $$ P(S_1|R_{110}) = \dfrac{\frac{.081}{2}}{\frac{.081}{2}+\frac{.0023}{2}} \approx .972 $$

Answer

Explanation

Question 13

Company A has just developed a diagnostic test for a certain disease. The disease afflicts 1% of the population. As defined in Example 2.3.9, the sensitivity of the test is the probability of someone testing positive, given that they have the disease, and the specificity of the test is the probability that of someone testing negative, given that they don’t have the disease. Assume that, as in Example 2.3.9, the sensitivity and specificity are both 0.95. Company B, which is a rival of Company A, offers a competing test for the disease. Company B claims that their test is faster and less expensive to perform than Company A’s test, is less painful (Company A’s test requires an incision), and yet has a higher overall success rate, where overall success rate is defined as the probability that a random person gets diagnosed correctly.

(a) It turns out that Company B’s test can be described and performed very simply: no matter who the patient is, diagnose that they do not have the disease. Check whether Company B’s claim about overall success rates is true.
(b) Explain why Company A’s test may still be useful.
(c) Company A wants to develop a new test such that the overall success rate is higher than that of Company B’s test. If the sensitivity and specificity are equal, how high does the sensitivity have to be to achieve their goal? If (amazingly) they can get the sensitivity equal to 1, how high does the specificity have to be to achieve their goal? If (amazingly) they can get the specificity equal to 1, how high does the sensitivity have to be to achieve their goal?

Question

A) first we define our events $$ D = \text{has disease} $$ $$ T_p = \text{tests positive} $$ $$ T_n = \text{tests negative} $$ With those events defined, we can set up an equation to see how likely a person is to actually have a disease given that company A's test gave them a positive result. $$ P(D|T_p) = \dfrac{.01 \times .95}{(.01 \times .95) + (.99 \times .05)} \approx .161 $$ And an equation for how likely someone is to not be diseased given a negative result. $$ P(D^c|T_n) = \dfrac{.99 \times .95}{(.99 \times .95) + (.01 \times .05)} \approx .998 $$ This means that using company A's test, there is a $1-.161$ false positive and basically no false negative.
Company B has a false negative of .01 and a false positive of 0, so in the general population, company B performs much better.

B)
This may lead some to believe that company A's test is useless or even fake! This is misguided though. Company A's test could still be usefull if there is a population where the rate of disease is higher than the general population.
You may crunch the numbers and find out what the precision of both tests are when the prior probability of having the disease is $.5$. Also, If company A gives multiple tests to each patient and uses the majority to decide, accuracy will be improved.

lastly, company A's test can allow us to update our probability that a patient has the disease, while company B's test does not do this at all. Notice that the accuracy of a test is not the only important thing about a test!

C)
If sensitivity and specificity must be the same, then company A's test will outperform company B's test when both are greater than .99
iF sensitivity is 1, then specificity must be really close to .99 or better
when specificity is 1, then sensitivity must be better than 0.
Because most people do not have the disease, it is more important not to get false positives than to avoid false negatives.

Answer

Question 14

Consider the following scenario, from Tversky and Kahneman:

Let A be the event that before the end of next year, Peter will have installed a burglar alarm system in his home. Let B denote the event that Peter’s home will be burglarized before the end of next year. (a) Intuitively, which do you think is bigger, P (A|B) or P (A|Bc)? Explain your intuition.
(b) Intuitively, which do you think is bigger, P (B|A) or P (B|Ac)? Explain your intuition.
(c) Show that for any events A and B (with probabilities not equal to 0 or 1), P (A|B) > P (A|Bc) is equivalent to P (B|A) > P (B|Ac).
(d) Tversky and Kahneman report that 131 out of 162 people whom they posed (a) and (b) to said that P (A|B) > P (A|Bc) and P (B|A) < P (B|Ac). What is a plausible explanation for why this was such a popular opinion despite (c) showing that it is impossible for these inequalities both to hold?

Question

A)
Intuitively, I would think that $P(A|B)$ is higher than $P(A|B^c)$ because if his house gets broaken into, he is more likely to install the burgaler alarm.
B)
Here my intuition says $P(B|A^c)$ should be greater than $P(B|A)$, but this cannot be the case if the above condition holds.
C)
We use the definition of dependent probability to demostrate why $P(A|B) \gt P(A|B^c)$ is equivelant to $P(B|A) \tg P(B|A^c)$ as long as both $P(A)$ and $P(B)$ are not 0 or 1.
$$ \dfrac{P(B\cap A}{P(A)} \gt \dfrac{P(B\cap A^c)}{A^c} $$ and $$ \dfrac{P(A\cap B)}{P(B)} \gt \dfrac{P(A \cap B^c}{P(B^c)} $$ plug in any number otherthan 0 or 1 for both probabilities and see if they hold, this should give you a better intuition about why this is true

Answer

Question 15

Question

$$ A $$

Answer

of the options, A is the least likely to have occured, therefor it is best to see that event occure if we want both A and B to occure.
seeing the event $A\cup B$ does not give us as much information because it is the most probable.

Explanation

Question 16

Show that P (A|B) $\le$ P (A) implies P(A|B^c) $\ge$ P(A), and give an intuitive explanation of why this makes sense.

Question

Intuitively, we can view the events as circles in a ven diagram. When we say $P(A|B) \le P(A)$, we are saying that at most half of $P(A|B)$ is inside of $P(A)$.
This means that half or more of $P(A|B)$ lies outside $P(A)$.

Answer

Question 17

In deterministic logic, the statement “A implies B” is equivalent to its contrapositive, “not B implies not A”. In this problem we will consider analogous statements in prob- ability, the logic of uncertainty. Let A and B be events with probabilities not equal to 0 or 1.

(a) Show that if P $(B|A) = 1$, then P $(A^c|B^c) = 1$.
Hint: Apply Bayes’ rule and LOTP. (b) Show however that the result in (a) does not hold in general if = is replaced by $\approx$. In particular, find an example where $P (B|A)$ is very close to 1 but $P (A^c|B^c)$ is very close to 0.

Question

A) If B is guarenteed to occure then A occures, that means A is a subset of B. Therefore if B does not occure, then A also cannot occure.
We can denonstrate this using Bayes theorem. $$ P(B|A) = \dfrac{P(B) \cdot P(A|B)}{ P(B) P(A|B) + P(B^c) P(A|B^c) } = 1 $$ For the denomniator, we know that $P(B^c) P(A|B^c)$ must be zero for this equation to equal 1 according to LOTP.
Therefore, $P(A^c|B^c)$ must equal 1.
B) if $P(B|A) = .99$, then A is not a subset of B, therefore it is no longer a guarentee that $P(A^c|B^c) \approx 1$.
If your sample space is not much larger than $A \cup B$, then you can have a situation where $P(B|A) \approx 1$ but $P(A^c|B^c) \approx 0$
For example, let A be the probability that someone has a very common disease and let B be the probability that someone tests positive for the disease.
Assuming that almost everyone with the disease tests positive, with only a few false negatives. Let B = test positive and A = has disease. If only a few people have the disease and test negative but even fewer people test positve, then $P(A^c|B^c) \approx 0$

A simpler answer I arrived at via the hint is that A and B can occure with probability close to 1 as independent events.

Answer

Question 18

Question

When $P(A) = 1$, then bayes theorem looks like this: $$ P(A|B) = \dfrac{P(B|A) }{P(B|A)} = 1 $$ Intuitively, If you believe something is guarenteed to happen, no data can change that. If you are sure you have cancer, then a test result saying you do not have cancer will be totally unconvincing.

Answer

Question 20

The Jack of Spades (with cider), Jack of Hearts (with tarts), Queen of Spades (with a wink), and Queen of Hearts (without tarts) are taken from a deck of cards. These four cards are shuffled, and then two are dealt.

(a) Find the probability that both of these two cards are queens, given that the first card dealt is a queen.
(b) Find the probability that both are queens, given that at least one is a queen.
(c) Find the probability that both are queens, given that one is the Queen of Hearts.

Question

A) $$ P(\text{both}| \text{first}) = \dfrac{ 1 }{ {3 \choose 1} } = \dfrac{1}{3} $$ B) $$ P(\test{both}| \test{either}) = \dfrac{1}{ {5 \choose 1} -1 } = \dfrac{1}{5} $$ C) $$ P(\text{both}|\text{queen of hearts}) = \dfrac{1}{3} $$

Answer

These are easiest to understand if we simply write out all possible events meeting the first condition then count the events where the first two cards are queens.

A) there are 3 events where a queen is first and 1 where the first two cards are both queens
$$ QQJJ \cdot $$ $$ QJQJ $$ $$ QJJQ $$ B) There are 5 events where at least 1 of the first two is a queen and 1 where both are queens
$$ QQJJ \cdot $$ $$ QJQJ $$ $$ QJJQ $$ $$ JQQJ $$ $$ JQJQ $$ C) For this one, we must use labeled cards, so I will use $q$ and $Q$. There are 2 where both are queens and 6 in total. Uppercase $Q$ will represent queen of hearts
$$ qQJJ \cdot $$ $$ QqJJ \cdot $$ $$ QJqJ $$ $$ QJJq $$ $$ JQqJ $$ $$ JQJq $$

Explanation

Question 21

A fair coin is flipped 3 times. The toss results are recorded on separate slips of paper (writing “H” if Heads and “T” if Tails), and the 3 slips of paper are thrown into a hat.

(a) Find the probability that all 3 tosses landed Heads, given that at least 2 were Heads.
(b) Two of the slips of paper are randomly drawn from the hat, and both show the letter H. Given this information, what is the probability that all 3 tosses landed Heads?

Question

A) $$ P(3 heads | 2 heads) = \dfrac{1}{4} $$ B) $$ P(3 heads | 2 random) = \dfrac{1}{2} $$

Answer

What! How could these possibly be different? It seems like having at least two heads is the same as pulling 2 random H's out of a hat, but these are indeed different.

A) In the case of 2 or more heads, we add up all the ways we can have either 2 or 3 heads. There are 3 ways to have 2 heads and 1 way to have 3.
B) When randomly pullying 2 H's, there are 3 ways to do that when there are 3 heads and 3 ways when there are 2 heads.

Explanation

Question 22

A bag contains one marble which is either green or blue, with equal probabilities. A green marble is put in the bag (so there are 2 marbles now), and then a random marble is taken out. The marble taken out is green. What is the probability that the remaining marble is also green?

Question

$$ P(GreenRemains | GreenRemoved) = \dfrac{2}{3} $$

Answer

There are 2 equally likely combinations of marbles after the green one is added, those are GG and GB. If a green marble is taken out, then we are left with B,G, and G again. Each being equally likely.

Explanation

Question 23

Let G be the event that a certain individual is guilty of a certain robbery. In gathering evidence, it is learned that an event $E_1$ occurred, and a little later it is also learned that another event $E_2$ also occurred. Is it possible that individually, these pieces of evidence increase the chance of guilt (so $P (G|E_1) > P (G)$ and $P (G|E_2) > P$ (G)), but together they decrease the chance of guilt (so $P (G|E_1, E_2) < P$ (G))?

Question

This is possible.
Maybe we know that The suspect was at the location 5 hours before it was robbed, and the suspect also was there 2 hour after it was robbed. Either one of these make him more likely to rob the store, but together it just means he had a 7 hour shift.

Answer

Question 24

Is it possible to have events A1, A2, B, C with P (A1|B) > P (A1|C) and P (A2|B) > P (A2|C), yet P (A1 $\cup$ A2|B) $\lt$ P (A1 $\cup$ A2|C)? If so, find an example (with a “story” interpreting the events, as well as giving specific numbers); otherwise, show that it is impossible for this phenomenon to happen

Question

I have two answers for this question:

Answer 1

There are 2 types of kidney stone treatments, we will call them treatments B and C. There are also two types of kidney stones, small and large. $$ A_1 = successful treatment of small stone $$ $$ A_2 = successful treatment of large stone. $$ $$ P(A_1|B) = .93 $$ $$ P(A_1|C) = .87 $$ $$ P(A_2|B) = .73 $$ $$ P(A_2|C) = .69 $$ however, $$ P(A_1 \cup A_2 | B) = .78 $$ $$ P(A_1 \cup A_2 | C) = .83 $$ How is this possible? it is only possible if treatment B is used more often for treating smaller stones. For treatment B, there was a huge sample of small stones and a small sample of large stones. As it turns out, small stones are more likely to be treated successfully. This means that even though treatment B is actually better, treatment C is more likely to be successful because it is more commonly used on smaller stones.

Answer 2

Suppose we have two decks of cards, deck B and deck C. each deck has 6 cards and each card has a number written on it. The numbers in each deck are as follows:

Deck B: 2 4 6 101 103 105

Deck C: 1 3 4 6 102 104

Let the event $A_1$ be the event where you draw an odd number from a deck and let event $A_2$ be the event where you draw a number higher than 100 and let events $B$ and $C$ be the events where you draw from decks B and C respectively.
We can clearly see that $$ P(A_1 | B) = \dfrac{1}{2} $$ $$ P(A_2 | B) = \dfrac{1}{2} $$ and $$ P(A_1 | C) = \dfrac{1}{3} $$ $$ P(A_2 | C) = \dfrac{1}{3} $$ However, $$ P(A_1 \cup A_2 | B) = \dfrac{1}{2} $$ $$ P(A_1 \cup A_2 | C) = \dfrac{2}{3} $$

Answer

Question 27

Suppose that there are 5 blood types in the population, named type 1 through type 5, with probabilities p1, p2, . . . , p5. A crime was committed by two individuals. A suspect, who has blood type 1, has prior probability p of being guilty. At the crime scene blood evidence is collected, which shows that one of the criminals has type 1 and the other has type 2. Find the posterior probability that the suspect is guilty, given the evidence. Does the evidence make it more likely or less likely that the suspect is guilty, or does this depend on the values of the parameters p, p1, . . . , p5? If it depends, give a simple criterion for when the evidence makes it more likely that the suspect is guilty.

Question

$$ P(G|E) = \dfrac{p}{p + \frac{p_1}{p_1+p_2}} $$ Finding this evidence does not always increase the probability that he is guilty! If his prior probability of guilt is $0$ or $1$ then none of this matters (think about Oliver Cromwell).

Answer

For this question, we assume that blood tests at a crime scene are 100% accurate. So there is a probability of $0$ that our suspect is guilty given his bloodtype is not found at the crimescene and a probability of $1$ that you will find his blood at the crime scene given he is guilty.

Explanation

Question 30

A family has 3 children, creatively named A, B, and C.

(a) Discuss intuitively (but clearly) whether the event “A is older than B” is independent of the event “A is older than C”.
(b) Find the probability that A is older than B, given that A is older than C.

Question

A) If “A is older than B” is independent of "A is older than C", that would mean that finding out that A is older than B tells us nothing about whether A is older than C.
Intuitively, this cannot be true. If A is the youngest then A is not older than either and if A is oldest then A is older than both. Finding out that A is older than B makes is more likely that A is the oldest child.
b) There are 6 possible birth orders:
ABC A older than C
ACB A older than C
BAC A older than C
BCA
CAB
CBA
We clearly see 3 orderings where A is born before C (and therefore older) and of those, 2 where A is older than B, leaving us with the answer of: $$ P(A>B|A>C) = \dfrac{2}{3} $$

Answer

Question 31

Is it possible that an event is independent of itself? If so, when is this the case?

Question

when $P(A) = 1$ or when $P(A) = 0$, the event A is independant of itself.

Answer

This was a weird question for me to think about... hurt my brain a little :-)
First, let's refresh our understanding of independance:
Events A and B are independant if:
$$ P(A|B) = P(B|A) $$ and $$ P(A \CAP B) = P(A)P(B) $$ If event A happens, then the probability that it happened is 1, if it did not happen, the probability it did not happen is 0. So unless $P(A) = 0$ or $P(A) = 1$, there is no way for it to be independant of itself.

Explanation