12^n.

The key to solving this problem is the note “When there are multiple pieces of a given color and type (for example, 8 white pawns), we will assume they are indistinguishable from one another.” This means we count each type of chess piece for the colors black and white. There are 6 different types of chess pieces: pawn, bishop, knight, rook, queen, and king. There are 2 possible colors: black and white. This means there are 6 \cdot 2 = 12 unique colored types of chess pieces.

There are 12 options for the first person, there are 12 options for the second person, and so on, for n people, so we can write:

12 \cdot 12 \cdot \dots \cdot 12 = 12^n

We see there are 12^n possible results for this experiment because there are n people in this experiment who can each select 12 unique combinations.

\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.”

{16 \choose 8} \cdot {8 \choose 2} \cdot {6 \choose 2} \cdot {4 \choose 2} \cdot {2 \choose 1} \cdot {1 \choose 1} = \dfrac{16!}{8!2!2!2!}

Think about the problem this way. You have 16 spots on the line and for each subgroup of pieces (pawns, bishops, rooks, etc.), you need to assign spots from the line to place the pieces on that group. For example, we have 16 spots available in the line, and we want to place the pawns first. Well, since we have 8 pawns and 16 unique spots on the line, there are exactly {16 \choose 8} ways to place the pawns differently. Now, once you place the pawns you’ll have only 8 remaining spots on the line, so if you wish to place the 2 bishops now, there are {8 \choose 2} ways to do it. This means that there are {16 \choose 8} \cdot {8 \choose 2} ways of arranging 8 pawns and 2 bishops. Repeat the process for all subgroups and you’ll get:

{16 \choose 8} \cdot {8 \choose 2} \cdot {6 \choose 2} \cdot {4 \choose 2} \cdot {2 \choose 1} \cdot {1 \choose 1} = \dfrac{16!}{8!2!2!2!}

{14 \choose 6} = {14 \choose 8} = \dfrac{14!}{8!6!}

Similarly to 3.1, you know the placement of two pawns, thus you have 14 spots remaining. Then, you only want to place the remaining white pawns. Once you do that, the remaining spaces are for the black pawns. Therefore, the answer is {14 \choose 6}= {14 \choose 8}

\frac{7}{15}

For this question you can use your answer for 3.2. We found that there’s {14 \choose 6} = {14 \choose 8} = \dfrac{14!}{8!6!} lines of pawns with only whitepawns in the ends, by the same reasoning there would be another {14 \choose 8} = \dfrac{14!}{8!6!} lines with black pawns at the ends. Then there is 2 \cdot {14 \choose 8} lines with same color pawns at the end, since we know ther are {16 \choose 8} lines in total then we can calculates the answer by \frac{2 \cdot {14 \choose 8}}{{16 \choose 8}}=\frac{7}{15}

Alternatively, As soon as you know the color of the first pawn, you know that there are 15 remaining pawns, from which only 7 are the same color, since each pawn is equally likely to get placed in the last spot you get a probability of \frac{7}{15}

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 \%.

Bubble 3: Because it uses a dataset where there are many more non-winners than winners.

Each year there is only one winner, but then hundreds of non-winners! This skews the data and makes it harder to figure out if someone won or lost because of their birthday and level.

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.