\dfrac{{12 \choose 8}\cdot8!}{12^8} = \dfrac{12}{12}\cdot\dfrac{11}{12}\cdot\dfrac{10}{12}\cdot\dots\cdot\dfrac{5}{12}

There’s two ways you can think of this problem.

The numerator {12 \choose 8}\cdot8! is the number of total possible permutations of 8 distinct values; {12 \choose 8} possible sets of 8 distinct values times 8! orderings, because the die could roll the same set of distinct values in 8! different ways. The denominator 12^8 is the number of total possible outcomes. Altogether, dividing the numerator by the denominator gives us the probability of rolling 8 distinct values.

We can also solve this problem differently by realizing that the rolls are independent. In the first roll the probability of getting a new number is 1 (since any number would be new), now in the second roll the probability of getting a new number is \frac{11}{12} since we’ve only rolled one number so rolling any of the other 11 numbers would be rolling a new number.If you keep doing this for each roll, and multiply the independent probabilities of getting a new number in each of the rolls you’ll get \dfrac{12}{12}\cdot\dfrac{11}{12}\cdot\dfrac{10}{12}\cdot\dots\cdot\dfrac{5}{12}

\dfrac{{8 \choose 2}\cdot{12 \choose 7}\cdot7!}{12^8}

Again, I’m going to show you two different ways you can think of this problem. We can get the total number of outcomes that get you exactly 7 distinct values by doing the following, first choose 7 from the 12 possible numbers (there’s {12 \choose 7} ways to do it), then choose a number to duplicate (there’s {7 \choose 1} ways to do that), lastly you need to order them. From the 8 possible places you have availible you need to choose 2 for the repeated value (there’s {8 \choose 2} ways to do that), then once you chose a placement for your reapeted value you need to choose one for each of the remaing 6 values(there’s 6! ways to do that). Finally just multiply the different ways you can perform each of those actions and you’ll get the number of ways you can roll 7 different values \begin{align*} {12 \choose 7}\cdot{7 \choose 1}\cdot{8 \choose 2}\cdot6! = {8 \choose 2}\cdot{12 \choose 7}\cdot7! \end{align*}

lastly, just divide by the total number of possible outcomes and you get \dfrac{{8 \choose 2}\cdot{12 \choose 7}\cdot7!}{12^8}

You could also find the number of rolls with exactly 7 different values by doing the following. First, select an ordered string of size 7 from the 12 possible values (there’s {12 \choose 7}\cdot7! ways to do that), then take the first value of the string as the duplicated value and keep the others as they are, now you only need to place 2 copies of the repeated value somewhere on your string of 6 numbers (there’s {8 \choose 2} ways to do that), multiply this the ways you can do these two actions and divide by total number of outcomes and you’ll get \dfrac{{8 \choose 2}\cdot{12 \choose 7}\cdot7!}{12^8} This solution is faster but less intuitive.

P(A|(B \cap C)) and P((A \cap C)|(B \cap C))

Recall the formula for conditional probability is P(A|B) = \frac{P(A \cap B)}{P(B)}. We are going to use this fact to figure out which of the following are the same as P((A\cap B)|(B \cap C)).

We should first expand P((A\cap B)|(B \cap C)) by doing: P((A\cap B)|(B \cap C)) = \frac{P(A \cap B) \cap P(B \cap C)}{P(B \cap C)}. We can further simplify the numerator to be (A \cap B) \cap (B \cap C) \rightarrow A \cap B \cap C. This means we get P((A\cap B)|(B \cap C)) by doing: P((A\cap B)|(B \cap C)) = \frac{P(A \cap B \cap C)}{P(B \cap C)}.

Now we can go through each of the options and figure out if, when put inside the formula for conditional probability, we get \frac{P(A \cap B \cap C)}{P(B \cap C)}.

P(A|(B \cap C)) P(A|(B \cap C)) = \frac{P(A)P(B \cap C)}{P(B \cap C)}= \frac{P(A \cap (B \cap C))}{P(B \cap C)} = \frac{P(A \cap B \cap C)}{P(B \cap C)} This matches \frac{P(A \cap B \cap C)}{P(B \cap C)}!

P(A \cap B \cap C)

We can see that there is no way for this to equal \frac{P(A \cap B \cap C)}{P(B \cap C)} because it is missing the denominator.

P((B \cap C)|(A \cap B))

P((B \cap C)|(A \cap B)) = \frac{P(B \cap C)P(A \cap B)}{P(A \cap B)}= \frac{P(A \cap B \cap C)}{P(A \cap B)} This does not match \frac{P(A \cap B \cap C)}{P(B \cap C)}.

P((A \cap C)|(B \cap C)) P((A \cap C)|(B \cap C)) = \frac{P(A \cap C)P(B \cap C)}{P(B \cap C)} = \frac{P(A \cap B \cap C)}{P(B \cap C)} This matches \frac{P(A \cap B \cap C)}{P(B \cap C)}!

Since options 1 and 4 are correct the answer cannot be None of the Above!

B, C

We are going to rewrite the following in the hopes of being able to simplify things later. To do this we will once again use the formula for conditional probability: P(A|B) = \frac{P(A \cap B)}{P(B)}

Let’s look at the left side of P((A \cap B)|C) = P(A|(B \cap C))*P(B).

P((A \cap B)|C) = \frac{P(A \cap B)P(C)}{P(C)} = \frac{P(A \cap B \cap C)}{P(C)}

Let’s now look at the right side of P((A \cap B)|C) = P(A|(B \cap C))*P(B).

P(A|(B \cap C))*P(B) = \frac{P(A)P(B \cap C)}{P(B \cap C)} * P(B) = \frac{P(A \cap B \cap C)}{P(B \cap C)}* P(B) = \frac{P(A \cap B \cap C) * P(B)}{P(B \cap C)}

Now we can look at them together! \frac{P(A \cap B \cap C)}{P(C)} = \frac{P(A \cap B \cap C) * P(B)}{P(B \cap C)}. We can cross multiply to clear the fractions: P(A \cap B \cap C) * P(B \cap C) = P(A \cap B \cap C) * P(B) * P(C). Assuming that P(A \cap B \cap C) \neq 0 we can divide both sides by P(A \cap B \cap C). We end up with P(B \cap C) = P(B) * P(C), which demonstrates to us that B and C are independent of one another.

280

This problem requires us to the the principle of inclusion-exclusion for three sets. For this problem lets define the following sets:

• Let H be the set of people who included “hall” on their list
• Let C be the set of people who included “chill” on their list
• Let L be the set of people who included “logical” on their list

We are told:

• |H| = 200
• |C| = 100
• |L| = 25
• |H \cap C| = 80
• |C \cap L| = 15
• |H \cap L| = 20
• |H \cap C \cap L| = 10
• |H' \cap C' \cap L'| = 60
  • H' means “not H”

Recall the principle of inclusion-exclusion for three sets follows: |H \cup C \cup L| = |H| + |C| + |L| - |H \cap C| - |C \cap L| - |H \cap L| + |H \cap C \cap L|

We can simply plug and chug into the equation: |H \cup C \cup L| = 200 + 100 + 25 - 80 - 15 - 20 + 10 = 220. This means the number of data science majors who submited words “hall,”chill,” and “logical” is 220. But this is not all data science majors!

We need to add in the number of data science majors who did not submit any of those words in their list, which we are told is 60 students. This means the number of data science majors is 220 + 60 = 280.

\dfrac{8}{16}\cdot\dfrac{7}{15} = \dfrac{7}{30}

We get \dfrac{8}{16} from the fact there are 8 white pawns and 16 white chess pieces. So the chance of first getting a pawn is \dfrac{8}{16}.

Now we assume that we have a white pawn, meaning there are 8-1=7 white pawns left and 16 - 1 = 15 white chess pieces, which means we have a \dfrac{7}{15} chance for getting another white pawn.

From here we simplify the answer by writing:

\dfrac{8 \cdot 7}{16 \cdot 15} = \dfrac{56}{240} = \dfrac{7}{30}

Another way to solve this is by using combinatorics: Note that {8 \choose 2} means “8 choose 2.”

\dfrac{{8 \choose 2}}{16 \choose 2} = \dfrac{\frac{8}{32}\cdot\frac{7}{31}}{\frac{16}{32}\cdot\frac{15}{31}} = \dfrac{7}{30}

False.

We know that two probabilities are independent of each other if P(A \cap B) = P(A) \cdot P(B).

• Let P(A) be the probability of getting a white piece twice in a row.
• Let P(B) be the probability of getting a pawn twice in a row.

We can calculate the probability of getting a white piece twice in a row P(A) = \frac{16}{32} \cdot \frac{15}{31} = \frac{15}{62}. We get \frac{16}{32} from the fact there are 16 white pieces out of the 32 total pieces. We get \frac{15}{31} from the fact there are now 15 white pieces out of the 31 pieces left after taking out a white piece.

We can also calculate the probability of getting a pawn twice in a row. Note that the color of the pawn does not matter! P(B) = \frac{16}{32} \cdot \frac{15}{31} = \frac{15}{62}. We get \frac{16}{32} from the fact there are 16 pawns (8 black pawns plus 8 white pawns) out of the 32 total pieces. We get \frac{15}{31} from the fact there are now 15 pawns out of the 31 pieces left after taking out a pawn.

We can now calculate P(A) \cdot P(B): \frac{15}{62} \cdot \frac{15}{62} = \frac{225}{3844}

Recall we calculate P(A \cap B) = P(A) \cdot P(B|A).

We solved for P(B|A) in part A of the problem: \frac{7}{30}. We also already solved for P(A) above: \frac{15}{62}.

\frac{15}{62} \cdot \frac{7}{30} = \frac{7}{124}

We can see that P(A \cap B) \neq P(A) \cdot P(B) because \frac{7}{124} \neq \frac{225}{3844}, so the answer is “False.”

67 \%

We know the percentages of players that beat Avi (meaning Avi loses) and the percentages of different types of players inside the Avocado Cup.

We can use the percentage of players that beat Avi to figure out the percentage of times Avi wins by writing 1 - p.

Avi wins against beginners with a likelihood of 1 - 0.1 = 0.9, against intermediate players with a likelihood of 1 - 0.4 = 0.6, and against advanced players with a likelihood of 1 - 0.8 = 0.2.

Now we can multiply the probabilities of Avi winning with their respective probabilities of different players’ types to find the probability Avi that will win in the first round! We can do this because you can imagine we are multiplying the probability Avi wins with the amount of people in the tournament of that level.

\begin{align*} 0.9 \cdot 0.5 &= 0.45 \\ 0.4 \cdot 0.5 &= 0.2 \\ 0.2 \cdot 0.1 &= 0.02 \\ 0.45 + 0.2 + 0.02 &= 0.67 \end{align*}

25 \%

• Let P(A) be the probability of Avi playing against an advanced player
• Let P(B) be the probability of Avi losing a match
• Let P(C) be the probability of Avi playing against an intermediate player
• Let P(D) be the probability of Avi playing against a beginner player

We can use Bayes’ theorem to help us find the probability that Avi lost the match to an advanced player (P(A|B)). Recall Bayes’ theorem is:

P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}

We are given the probability that Avi loses the match against an advanced player P(B|A) = 0.8 and the probability of Avi playing against an advanced player P(A) = 0.1.

We can calculate the probability of Avi losing a match P(B) with the law of total probability: P(B) = P(B|A) \cdot P(A) + P(B|C) \cdot P(C) + P(B|D) \cdot P(D)

We know the probability of Avi losing to an intermediate player: P(B|C) = 0.5, the probability of Avi losing to a beginner player: P(B|D) = 0.1, the probability of an intermediate player being the opponent: P(C) = 0.4, and the probability of a beginner player being the opponent: P(D) = 0.5

Plugging in what we know into the law of total probability equation we get: \begin{align*} P(B) &= 0.8 \cdot 0.2 + 0.5 \cdot 0.4 + 0.1 \cdot 0.5 \\ &=0.08 + 0.2 + 0.05 \\ &= 0.33 \end{align*}

Or we can calculate P(B) by using what we know in part A (67 \%). 1 - 0.67 = 0.33.

Back to Bayes’ theorem we have: \begin{align*} P(A|B) &= \frac{0.8 \cdot 0.1}{0.33} \\ &= \frac{0.08}{0.33}\\ &= 0.24 \end{align*}

0.24 is closest to 0.25, so 25 \%.

Only bubble 2: A and B are conditionally independent given C.

We know that A and B are not independent because winning or losing the first match could affect if Avi wins the tournament.

It is helpful to think about “if C happens does it affect my knowledge of A and B happening”? If C happens we know that A cannot happen. If A cannot happen then B is independent of A. This means A and B are conditionally independent given C.

Recall a partition is a collection of non-empty, non-overlapping subsets of a sample space. We know that A affects B, which means that this statement cannot be true.

Since the third option is true the answer cannot be “None of the above.”

P(\overline{A}|(A \cup B)) = \dfrac{P(B)}{P(A)+P(B)}

There are several correct approaches to this problem. The simplest one comes from using the definition of conditional probability to write:

P(\overline{A}|(A \cup B)) = \dfrac{P(\overline{A}\cap (A \cup B))}{P(A \cup B)} To evaluate the numerator, notice that since A and B are disjoint, any outcome that is not in A but is in the union of A and B (A or B) must be in B. The Venn diagram shown below illustrates this. Therefore, we can simplify this as = \dfrac{P(B)}{P(A \cup B)} and because we know A and B are disjoint, we can use the addition rule to expand the denominator as = \dfrac{P(B)}{P(A)+P(B)}

This is the desired final expression because it uses P(A) and P(B) only.

In the Venn Diagram below, \overline{A} is shaded in yellow, and A \cup B is shaded in blue. Their overlap, \overline{A}\cap (A \cup B), is shaded in green, which is just B.

Orange stripes.

\bar{A} is the area with green stripes and orange stripes. \bar{B} is the area with red stripes and orange stripes.

Therefore \bar{A}\cap\bar{B} is the area with orange stripes.

We can write out the inclusion-exclusion principle. From there, we use the law of total probability to split up event B as the regions A\cap B and \bar{A}\cap B and continue from there:

\begin{align*} P(A\cup B)&=P({A}) + P({B}) - P(A\cap B)\\ &=P({A}) + \left[ P(A\cap B)+P(\bar{A}\cap B) \right]- P(A\cap B)\\ &=P({A}) + P(\bar{A}\cap B) \\ &=\left[ 1-P(\bar{A}) \right] + P(\bar{A}\cap B) \\ &=1- \left[P(\bar{A}\cap B) + P(\bar{A}\cap \bar{B}) \right] + P(\bar{A}\cap B)\\ &=1-P(\bar{A}\cap\bar{B}) \end{align*}

Therefore P(\bar{A}\cap\bar{B})=1-P(A\cup B).

12^n

To answer this question we must determine how many different pieces each person can choose. In a complete chess set we know that there’s pawns, knights, bishops, rooks, queens, and kings. Then we also know that each of these pieces can be aither black or white, therefore there’s a totoal of 6 \cdot 2 = 12 different pieces. Notice that a complete chess set has 32 total pieces, but there’s multiple duplicates in those 32 pieces that is why we say there’s only 12 different pieces on a chess set.

Now since every person has their own chess set, we know that each of them can get any of the 12 pieces, therefore if we do this with one person there would be 12 possible outcomes, now if we do it with two persons the first person can get any of the 12 pieces, but then for each piece that person 1 can choose person 2 can choose from their 12 pieces, which means that in this case there’s 12 \cdot 12 outcomes. if we keep doing this for n persons we’ll get that there’s 12 \cdot \dots \cdot 12 = 12^n.

\frac{1}{4}

First, let’s calculate the probability of one person choosing a black pawn. We know that the complete chess set has 32 pieces, and out of those 32 there are 8 black pawns, which gives us a probability of \frac{8}{32} = \frac{1}{4} of selecting a black pawn. Now, think about how we interpret probabilities; for example, if we say that the probability of getting heads when flipping a coin is \frac{1}{2} that just means that if we flip a coin over and over, we expect to see heads \frac{1}{2} of the time. Similarly, when we say that the probability of choosing a black pawn is \frac{1}{4}, that means that if we were to choose a chess piece over and over, we expect to choose a black pawn \frac{1}{4} of the time.

As n becomes large, we choose chess pieces over and over n times. Therefore, you still expect to see the black pawn \frac{1}{4} of the time.

False.

We know that two probabilities are independent of each other if P(A \cap B) = P(A) \cdot P(B).

• Let P(A) be the probability of getting a white piece twice in a row.
• Let P(B) be the probability of getting a pawn twice in a row.

We can calculate the probability of getting a white piece twice in a row P(A) = \frac{16}{32} \cdot \frac{15}{31} = \frac{15}{62}. We get \frac{16}{32} from the fact there are 16 white pieces out of the 32 total pieces. We get \frac{15}{31} from the fact there are now 15 white pieces out of the 31 pieces left after taking out a white piece.

We can also calculate the probability of getting a pawn twice in a row. Note that the color of the pawn does not matter! P(B) = \frac{16}{32} \cdot \frac{15}{31} = \frac{15}{62}. We get \frac{16}{32} from the fact there are 16 pawns (8 black pawns plus 8 white pawns) out of the 32 total pieces. We get \frac{15}{31} from the fact there are now 15 pawns out of the 31 pieces left after taking out a pawn.

We can now calculate P(A) \cdot P(B): \frac{15}{62} \cdot \frac{15}{62} = \frac{225}{3844}

Recall we calculate P(A \cap B) = P(A) \cdot P(B|A).

To find P(B|A), we assume that both pieces are white and imagine the probabilities of getting two pawns from that batch of white pieces. To get the first pawn, the probability is \frac{8}{16} (8 pawns out of 16 white pieces). To get the second pawn, the probability is: \frac{7}{15} (7 remaining pawns out of 15 remaining white pieces). So, the probabilitiy of getting two pawns given we have two white pieces is P(B|A) = \frac{8}{16} \cdot \frac{7}{15} = \frac{7}{30}. We also already solved for P(A) above: \frac{15}{62}.

\frac{15}{62} \cdot \frac{7}{30} = \frac{7}{124}

We can see that P(A \cap B) \neq P(A) \cdot P(B) because \frac{7}{124} \neq \frac{225}{3844}, so the answer is “False.”

P(\text{Decay|Dead}) = \frac{18}{26}

The probability we are looking for is P(\text{Decay|Dead}). Using Bayes’ theorem, we can write:

\begin{align*} P(\text{Decay|Dead}) = \frac{P(\text{Dead|Decay})P(\text{Decay})}{P(\text{Dead})} \end{align*}

Using the Law of Total Probability, we can rewrite the denominator:

\begin{align*} P(\text{Decay|Dead}) = \frac{P(\text{Dead|Decay})P(\text{Decay})}{P(\text{Dead|Decay})P(\text{Decay}) + P(\text{Dead|Not Decay})P(\text{Not Decay})} \end{align*}

We know that:

\begin{align*} P(\text{Dead|Decay}) &= 90\% = 0.9\\ P(\text{Dead|Not Decay}) &= 10\% = 0.1\\ P(\text{Decay}) &= 20\% = 0.2 \end{align*}

Plugging these numbers, we have:

\begin{align*} P(\text{Decay|Dead}) &= \frac{0.9\cdot 0.2}{0.9 \cdot 0.2 + 0.1 \cdot (1-0.2)}\\ &=\frac{18}{26} \end{align*}

Statements 1 and 2 are true. Statements 3, 4, and 5 are false.

When two events are independent, the following statement is true: P(\text{Event 1}) = P(\text{Event 1} | \text{Event 2}). So for statement 1, we need to show: P(\text{Decay}) = P(\text{Decay} | \text{Cat Dead})

• P(\text{Decay}) is the fraction of the total Venn Diagram that is green.
• P(\text{Decay}|\text{Cat Dead}) is the fraction of the “Cat Dead” region that is green.

These two fractions look the same. So that means the independence equation holds, and statement 1 is true.

We can do the same thing but instead with the “Cat Alive” region to also show that statement 2 is true.

The “Cat Alive” and “Cat Dead” regions intersect, implying those two events can occur at the same time. If two events can occur at the same time, they are not mutually exclusive, so statement 3 is false.

A partition of a sample is a perfect division the sample space into a bunch of little pieces, without any overlaps or empty space. For this problem, it would mean finding a way to perfectly divide the Venn Diagram. You cannot divide the Venn Diagram into the regions “Cat Dead”, “Cat Alive”, and “Decay”; go ahead and try, no matter how you slice it the overlaps will mess up the division. This means that “Cat Dead”, “Cat Alive”, and “Decay” do not form a partition of the sample, so statement 4 is false.

Since statements 1 and 2 were true, the answer cannot be none of the above, so statement 5 is false.

P(\text{Alive} \cap \text{Dead}) = \frac{1}{8}

Since we know that the decay is independent to cat’s state, we have: \begin{align*} P(\text{Dead} \cup \text{Decay}) &= P(\text{Dead}) + P(\text{Decay}) - P(\text{Dead} \cap \text{Decay})\\ &=P(\text{Dead}) + P(\text{Decay}) - P(\text{Dead})\cdot P(\text{Decay}) = \frac{4}{5} \end{align*}

Similarly, we have: \begin{align*} P(\text{Alive} \cup \text{Decay}) &= P(\text{Alive}) + P(\text{Decay}) - P(\text{Alive} \cap \text{Decay})\\ &=P(\text{Alive}) + P(\text{Decay}) - P(\text{Alive})\cdot P(\text{Decay}) = \frac{1}{2} \end{align*}

We are given P(\text{Decay}) = \frac{1}{5}, so we can plug it into the equations we made for P(\text{Dead} \cup \text{Decay}) and P(\text{Alive} \cup \text{Decay}): \begin{align*} &P(\text{Dead} \cup \text{Decay}) = P(\text{Dead}) + \frac{1}{5} - (\frac{1}{5} \cdot P(\text{Dead})) = \frac{4}{5}\\ \\ &P(\text{Alive} \cup \text{Decay}) = P(\text{Alive}) + \frac{1}{5} - (\frac{1}{5} \cdot P(\text{Alive})) = \frac{1}{2}\\ \\ \end{align*}

Solving these two equations, we have: \begin{align*} &\frac{4}{5} = P(\text{Dead}) + \frac{1}{5} - (\frac{1}{5} \cdot P(\text{Dead})) \\ &(\frac{4}{5}) \cdot 5 = (P(\text{Dead}) + \frac{1}{5} - (\frac{1}{5} \cdot P(\text{Dead}))) \cdot 5 \\ &4 = 5 \cdot (P(\text{Dead})) + 1 - P(\text{Dead}) \\ &3 = 4 \cdot (P(\text{Dead})) \\ &P(\text{Dead}) = \frac{3}{4} \end{align*} and \begin{align*} &\frac{4}{5} = P(\text{Alive}) + \frac{1}{5} - (\frac{1}{5} \cdot P(\text{Alive})) \\ &(\frac{1}{2}) \cdot 10 = (P(\text{Alive}) + \frac{1}{5} - (\frac{1}{5} \cdot P(\text{Alive}))) \cdot 10 \\ &5 = 10 \cdot (P(\text{Alive})) + 2 - 2 \cdot P(\text{Alive}) \\ &3 = 8 \cdot (P(\text{Alive})) \\ &P(\text{Alive}) = \frac{3}{8} \end{align*}

The probability of a superposition state is P(\text{Alive} \cap \text{Dead}). Since we know that P(\text{Alive} \cup \text{Dead}) = 1 we have:

\begin{align*} P(\text{Alive} \cup \text{Dead}) &= P(\text{Alive}) + P(\text{Dead}) - P(\text{Alive} \cap \text{Dead})\\ \end{align*}

Rearranging terms, we have: \begin{align*} P(\text{Alive} \cap \text{Dead}) &= P(\text{Alive}) + P(\text{Dead}) - P(\text{Alive} \cup \text{Dead})\\ &= \frac{3}{4} + \frac{3}{8} - 1\\ \\ &= \frac{6}{8} + \frac{3}{8} - 1\\ \\ &=\frac{1}{8} \end{align*}

32.

There are a few ways to approach the problem, all of which involve a similar sequence of calculations – specifically, the principle of inclusion-exclusion for 3 sets.

The key is remembering that there are two unknowns: the number of people overall, n, and the number of people who have been to both Atlanta and Detroit, which we’ll call k.

Many students drew Venn Diagrams, but ultimately they don’t help answer the question directly; one needs to use the independence between Atlanta and Detroit.

One way to proceed is:

1 = \frac{20}{n} + \frac{16}{n} + \frac{10}{n} + \frac{1}{n} - \frac{5}{n} - \frac{k}{n},

where \frac{k}{n} = \frac{20}{n} \cdot \frac{16}{n}. This is the key step!

Simplifying gives 1 = \frac{42}{n} - \frac{320}{n^2}, which is equivalent to n^2 - 42n + 320 = 0.

This factors into (n - 32)(n - 10) = 0, so n = 32.

28.

Without any assumptions on the value of k, the number of people who have been to Atlanta and Detroit, it can be shown that:

n = 42 - k

Using the conditional independence assumption given in this part, we have that \frac{20-k}{n - 16} \cdot \frac{10}{n - 16} = \frac{5}{n - 16}, which simplifies to 2(20 - k) = n - 16 \implies n = 56 - 2k.

We now have a system of equations: \begin{cases} n = 42 - k \\ n = 56 - 2k \end{cases}

Solving this system of equations for n gives n = 28.

Option 3 is correct (A \cap D   D^c   A^c \cap D \cap M^c).