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.

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 \%.

P(\text{long}) = \frac{7}{15}

We are given that: P(B_1)=P(B_2)=\frac{1}{2} P(\text{long}|B_1) = \frac{60}{60 + 40} = \frac{3}{5} P(\text{long}|B_2) = \frac{10}{10 + 20} = \frac{1}{3}

Using the law of total probability:

P(\text{long})=P(\text{long}|B_1) \cdot P(B_1)+P(\text{long}|B_2) \cdot P(B_2) = \frac{3}{5} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2} = \frac{7}{15}

There are 3! \cdot 3! permutations.

We can treat the group of A, B, and C (in any order) as a single block of letters which we can call X. This will gurantee that any permutation we find will satisfy our condition of having A, B, and C being all together.

Our problem now has 3! = 6 permutations of our new letter bank: X, D, and E.

XDE XED DXE EXD DEX EDX

Let’s call these 6 permutations “structures”. For each of these structures, there are 3! = 6 rearrangements of the letters A, B, and C inside of X.

So, with 3! different arrangements of D, E and our block of letters (X), and with each of those arrangements individually having 3! ways to re-order the group of ABC, the overall number of permutations of ABCDE where ABC are all together, in any order, is 3! \cdot 3!.

There are 20 ways the series can be won.

The series ends when the last game in the series is the third win by either player. Overall, there must be at least 3 games to end a series (all wins by the same player) and at most 5 games to end a series (each player has two wins, and then a final win by either player) before either player wins 3 games. Let’s assume player A always wins, and then we’ll multiply the number of options by two to account for the possibilities where player B wins instead.

Breaking it down:

• For a series that ends in 3 games - there is a single option (all wins by player A).

• For a series that ends in 4 games, we’ve assumed the last game is a player A win to end the series, therefore the 3 preceding games have {3 \choose 2} options for player A winning 2 of the 3 preceding games.

• For a series that ends in 5 games, we’ve assumed the last game is a player A win to end the series, therefore the 4 preceding games have {4 \choose 2} options for player A winning 2 of the 4 preceding games.

Overall, the number of ways the series can occur is: 2 \cdot (1 + {3 \choose 2} + {4 \choose 2}) = 20

The false positive rate should be \approx 0.001 to satisfy this condition.

Let A be the event a patient has a genetic defect. Let B be the event a patient got a positive test result.

We know:

• P(A) = 0.01

• The true positive rate (probability of a positive test result given someone has the genetic defect) can be modeled by P(B|A)=0.9

• The false positive rate (probability of positive test result given someone doesn’t have the genetic defect) can be modeled by P(B|\bar{A}) = p

We want:

• The probability that a patient having the genetic defect given a positive result (P(A|B)) is 90\%

Let’s set up a relationship between what we know and what we want by using Bayes’ Theorem:

\begin{aligned} P(A|B) &= \frac{ P(B|A)P(A)}{P(B)} \\ &=\frac{ P(B|A)P(A)}{P(B|A)P(A)+ P(B|\bar{A})P(\bar{A})}\\ &=\frac{0.9 \cdot 0.01}{0.9 \cdot 0.01 + p \cdot 0.99} =0.9 \end{aligned}

Rearranging the terms:

\begin{aligned} 0.9 \cdot 0.01 & = 0.9 \cdot (0.9 \cdot 0.01 + p \cdot 0.99)\\ 0.1 \cdot 0.9 \cdot 0.01 & = 0.9 \cdot p \cdot 0.99 \\ p = P(B|\bar{A}) &=\frac{0.1 \cdot 0.01}{0.99} \approx 0.001 \end{aligned}

We’ve found that the probability of a false positive (positive result for someone who doesn’t have the genetic defect) needs to be \approx 0.001 to satisfy our condition.

The probability of a true positive needs to be 0.99 to satisfy our condition.

Let A be the event a patient has a genetic defect. Let B be the event a patient got a positive test result.

We know:

• P(A) = 0.01

• The true positive rate (probability of positive test result if someone has the genetic defect) can be modeled by P(B|A) = p

• The false positive rate (probability of positive test result if someone doesn’t have the genetic defect) can be modeled by P(B|\bar{A}) = 0.01

We can set up Bayes’ Theorem again to gain more information:

\begin{aligned} P(A|B) &= \frac{ P(B|A)P(A)}{P(B|A)P(A)+ P(B|\bar{A})P(\bar{A})}\\ &=\frac{p \cdot 0.01}{p \cdot 0.01 + 0.01 \cdot 0.99} \\ &=\frac{p}{p + 0.99} = 0.5 \end{aligned}

Rearranging the terms: \begin{aligned} p & = 0.5 \cdot (p + 0.99)\\ 0.5 \cdot p & = 0.5 \cdot 0.99 \\ p = P(B|{A}) &=0.99 \end{aligned}

We find that the probability of a true positive needs to be 0.99 to satisfy our condition.

Let A be the event a patient has a genetic defect. Let B be the event a patient got a positive test result.

We know:

• P(A) = 0.01

• The true positive rate (probability of positive test result if someone has the genetic defect) can be modeled by P(B|A) = p

• The false positive rate (probability of positive test result if someone doesn’t have the genetic defect) can be modeled by P(B|\bar{A}) = 0.01

Let’s try solving as before:

\begin{aligned} P(A|B) &=\frac{ P(B|A)P(A)}{P(B|A)P(A)+ P(B|\bar{A})P(\bar{A})}\\ &=\frac{p \cdot 0.01}{p \cdot 0.01 + 0.01 \cdot 0.99} \\ &=\frac{p}{p + 0.99} = 0.9 \end{aligned}

Rearranging the terms: \begin{aligned} p & = 0.9 \cdot (p + 0.99)\\ 0.1 \cdot p & = 0.9 \cdot 0.99 \\ p = P(B|{A}) &= \frac{0.99 \cdot 0.9}{0.1} = 8.91 > 1 \end{aligned}

We have found that p, a probability value, must be greater than 1, which is impossible!

Given the rate of false positives, we cannot find a true positive rate such that the probability of a patient having the genetic defect given a positive result is 0.9.

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.

Bubble 1: A and B are independent. A and B are conditionally dependent given C.

We can test if A and B are independent by using the equation: \mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B).

• We know \mathbb{P}(A) = \frac{6}{26} because there are six vowels the Stringle string can start with out of 26 letters in the alphabet.
• We know \mathbb{P}(B) = \frac{13}{26} = \frac{1}{2} because the letters A through M is numbered 1 to 13, so there are 13 possible letters out of 26.
• We know \mathbb{P}(A \cap B) = \frac{3}{26} because the letters A, E, and I are between A and M and are also vowels out of all 26 letters.

When plugging these values into the equation we get: \begin{align*} \mathbb{P}(A \cap B) &= \mathbb{P}(A) \cdot \mathbb{P}(B) \\ \frac{3}{26} &= \frac{6}{26} \cdot \frac{1}{2} \\ \frac{3}{26} &= \frac{3}{26} \end{align*}

This proves that A and B are independent.

Recall we can test if something is conditionally independent by doing \mathbb{P}(A \cap B|C) = \mathbb{P}(A|C) \cdot \mathbb{P}(B|C).

• We know there are 25 letters in the alphabet that are not equal to Z.
• We can calculate \mathbb{P}(A|C) = \frac{6}{25} because there are six vowels out of the 25 non-Z letters.
• We can calculate \mathbb{P}(B|C) = \frac{13}{25} because there are 13 letters out of the 25 non-Z letters.
• We can calculate \mathbb{P}(A \cap B|C) = \frac{3}{25} because the letters A, E, and I are between A and M and are also vowels out of 25 non-Z letters.

When plugging these values into the equation we get: \begin{align*} \mathbb{P}(A \cap B|C) &= \mathbb{P}(A|C) \cdot \mathbb{P}(B|C) \\ \frac{3}{25} &= \frac{6}{25} \cdot \frac{13}{25} \\ \frac{3}{25} &\neq \frac{78}{625} \end{align*}

Since \mathbb{P}(A \cap B|C) \neq \mathbb{P}(A|C) \cdot \mathbb{P}(B|C) this means that A and B are conditionally dependent given C.

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.

k=3

According to the Naive Bayes formula we will be solving these equations: \begin{align*} &P(\text{successful}|\text{urban, winter, late night})\\ &= P(\text{successful}) * P(\text{urban}|\text{successful}) * P(\text{winter}|\text{successful}) * P(\text{late night}|\text{successful}) \end{align*} and \begin{align*} &P(\text{unsuccessful}|\text{urban, winter, late night}) \\ &= P(\text{unsuccessful}) * P(\text{urban}|\text{unsuccessful}) * P(\text{winter}|\text{unsuccessful}) * P(\text{late night}|\text{unsuccessful}) \end{align*}

This means we need to calculate our prior probabilities P(\text{successful}), P(\text{unsucessful}), P(\text{urban}|\text{successful}), P(\text{urban}|\text{usuccessful}), P(\text{winter}|\text{successful}), P(\text{winter}|\text{unsuccessful}), P(\text{late night}|\text{successful}), and P(\text{late night}|\text{unsuccessful}).

We can calculate P(\text{successful}) by looking at the number of successful outcomes from all possible outcomes. We find P(\text{successful}) = \frac{4}{12} = \frac{1}{3}.

Similarily we can calculate P(\text{unsucessful}). P(\text{unsucessful}) = \frac{8}{12} = \frac{2}{3}.

We can now calculate P(\text{urban}|\text{successful}) by looking at how many settings are urban when the outcome is successful. P(\text{urban}|\text{successful}) = \frac{1}{4}.

Similarily we can calculate P(\text{urban}|\text{unsuccessful}). P(\text{urban}|\text{unsuccessful}) = \frac{3}{8}.

Following this same pattern we will find:

P(\text{winter}|\text{successful}) = \frac{3}{4}

P(\text{winter}|\text{unsuccessful}) = \frac{4}{8} = \frac{1}{2}

P(\text{late night}|\text{successful}) = \frac{2}{4} = \frac{1}{2}

P(\text{late night}|\text{unsuccessful}) = \frac{6}{8} = \frac{3}{4}

From here all we need to do is plug our prior probnabilities into the equations we made earlier and solve!

\begin{align*} & P(\text{successful}|\text{urban, winter, late night})\\&= P(\text{successful}) * P(\text{urban}|\text{successful}) * P(\text{winter}|\text{successful}) * P(\text{late night}|\text{successful})\\ &= \frac{1}{3} * \frac{1}{4} * \frac{3}{4} * \frac{1}{2}\\ &= \frac{1}{32} \end{align*}

and

\begin{align*} & P(\text{unsuccessful}|\text{urban, winter, late night}) \\ &= P(\text{unsuccessful}) * P(\text{urban}|\text{unsuccessful}) * P(\text{winter}|\text{unsuccessful}) * P(\text{late night}|\text{unsuccessful})\\ &= \frac{2}{3} * \frac{3}{8} * \frac{1}{2} * \frac{3}{4}\\ &= \frac{3}{32} \end{align*}

We are told in the problem that “the probability you are unsuccessful is k times the probability you are successful.”

This means k * \frac{1}{32} = \frac{3}{32}, which we can easily see shows k = 3.

The predicted class is Healthy(-).

If p_+ and p_- are the numerators of the Naive Bayes comparison that represent a Sick(+) prediction and a Healthy(-) prediction, respectively, we have to compare: \begin{aligned} p_+=P(N,\bar C,\bar F|+)P(+)\quad \text{and}\quad p_-=P(N,\bar C,\bar F|-)P(-). \end{aligned}

We can find: \begin{aligned} &P(+)=\frac35\\ &P(-)=\frac25 \\ &P(N,\bar C,\bar F|+)=P(N|+)P(\bar C|+)P(\bar F|+)=\frac23\times \frac13\times\frac13=\frac{2}{27}\\ &P(N,\bar C,\bar F|-)=P(N|-)P(\bar C|-)P(\bar F|-)=\frac12\times \frac12\times 1=\frac{1}{4}. \end{aligned}

Then we can build up our comparison: \begin{aligned} p_+&= P(+)P(N,\bar C,\bar F|+) = \frac35\times\frac{2}{27}=\frac{2}{45}\quad \\ p_-&= P(-)P(N,\bar C,\bar F|-) = \frac25\times\frac{1}{4}=\frac{1}{10}\\ p_-&>p_+ \end{aligned}

So, the predicted class is Healthy(-).

The predicted class is Sick(+).

From the dataset we see P(F|-)=0, so we know: P(N,\bar C, F|-)=P(N|-)P(\bar C|-)P(F|-)=0

This will make p_- = P(N,\bar C, F|-)P(-) = 0

Contrast this against P(F|+), P(N|+), and P(\bar C|+), which are all nonzero values, making P(N,\bar C, F|+), (and therefore p_+) nonzero.

So p_+>p_-. This means our predicted class is Sick(+).

With smoothing, the result is \dfrac{1}{16}.

To apply smoothing, we add 1 to the numerator, and add the number of possible categories of the given event, healthiness (which would be 2 possible options in this case: sick, or healthy) to the denominator of our conditional probabilities. That way, other probabilities avoid being multiplied by zero. Let’s smooth our three conditional probabilities:

P(N|-)= \frac{1}{2} \to \frac{1 + 1}{2 + 2} = \frac{2}{4} P(\bar C|-)= \frac{1}{2} \to \frac{1 + 1}{2 + 2} = \frac{2}{4}

P(F|-)=0 is a zero probability that also needs to be smoothed. The smoothed version becomes: P(F|-)= \frac{0}{2} \to \frac{0 + 1}{2 + 2} = \frac{1}{4}

We carry on evaluating P(N,\bar C, F|-), just as the problem asks.

\begin{aligned} P(N,\bar C, F|-)=P(N|-)P(\bar C|-)P(F|-)=\frac24\times \frac24\times \frac14=\frac{1}{16}. \end{aligned}

\text{Odds of being sick}=\frac27

Using previous information, we have that the probability of a person being sick with a running nose but no coughing and no fever is:

\begin{aligned} P(+|N,\bar C,\bar F)=\frac{p_+}{P(N,\bar C,\bar F)}=\frac{\frac{2}{45}}{\frac{1}{5}}=\frac29. \end{aligned}

\begin{aligned} \text{Odds of being sick}=\frac{P(+|N,\bar C,\bar F)}{1-P(+|N,\bar C,\bar F)}=\frac27. \end{aligned}

One interpretation is as follows:

“If x is increased by 1, then the odds of y=+1 increases by a factor of \exp(1)=2.718.”

k = 3

Following the equation for Naive Bayes we will create the folllowing two formulas: \begin{align*} &P(\text{formed hadron}|\text{up, particle, blue})\\ &=P(\text{formed hadron}) * P(\text{up}|\text{formed hadron}) * P(\text{particle}|\text{formed hadron}) * P(\text{blue}|\text{formed hadron}) \end{align*} and \begin{align*} &P(\text{not formed hadron}|\text{up, particle, blue})\\ &=P(\text{not formed hadron}) * P(\text{up}|\text{not formed hadron}) * P(\text{particle}|\text{not formed hadron}) * P(\text{blue}|\text{not formed hadron}) \end{align*}

Now all we have to do is calculate the prior probabilities!

We can calculate P(\text{formed hadron}) by looking at the number of times a formed hadron happens out of the total number of outcomes. We can see this probability is P(\text{formed hadron}) = \frac{5}{10} = \frac{1}{2}.

Similarily we can calculate for a not formed hadron P(\text{not formed hadron}) = \frac{5}{10} = \frac{1}{2}.

We can calculate P(\text{up}|\text{formed hadron}) by looking at the number of times up appears out of all formed hadrons. This will give us something like: P(\text{up}|\text{formed hadron}) = \frac{2}{5}.

We can use this same method to calculate P(\text{up}|\text{not formed hadron}) by looking at the number of times up appears out of all not formed hadrons. P(\text{up}|\text{not formed hadron}) = \frac{1}{5}

If you continue finding the conditional probabilities you will find:

• P(\text{particle}|\text{formed hadron}) = \frac{3}{5}
• P(\text{particle}|\text{not formed hadron}) = \frac{2}{5}
• P(\text{blue}|\text{formed hadron}) = \frac{2}{5}
• P(\text{blue}|\text{not formed hadron}) = \frac{2}{5}

Now we simply plug and chug using the equations we had before! \begin{align*} &P(\text{formed hadron}|\text{up, particle, blue})\\ &= P(\text{formed hadron}) * P(\text{up}|\text{formed hadron}) * P(\text{particle}|\text{formed hadron}) * P(\text{blue}|\text{formed hadron})\\ &= \frac{1}{2} * \frac{2}{5} * \frac{3}{5} * \frac{2}{5}\\ &= \frac{6}{125} \end{align*} and \begin{align*} &P(\text{not formed hadron}|\text{up, particle, blue})\\ &= P(\text{not formed hadron}) * P(\text{up}|\text{not formed hadron}) * P(\text{particle}|\text{not formed hadron}) * P(\text{blue}|\text{not formed hadron})\\ &= \frac{1}{2} * \frac{1}{5} * \frac{2}{5} * \frac{2}{5}\\ &= \frac{2}{125} \end{align*}

Now the only thing left to do is calculate k. We can do this by solving the equation k * \frac{2}{125} = \frac{6}{125}. You should find k=3.

k = 3

All we have to do for this problem, after completing the first part, is to see how P(\text{top}|\text{formed hadron}) and P(\text{top}|\text{not formed hadron}) changes our previous equations.

We calculate these two probabilites by looking at the number of formed/not formed hadrons respectively and counting how many of those are in the top position. You will find P(\text{top}|\text{formed hadron}) = \frac{2}{5} and P(\text{top}|\text{not formed hadron}) = \frac{1}{5}. These are the same as our conditional probabilities when in the up position! Which means k=3 again.

If you want to go through the entire calculation again it can be found below: \begin{align*} &P(\text{formed hadron}|\text{top, particle, blue})\\ &= P(\text{formed hadron}) * P(\text{top}|\text{formed hadron}) * P(\text{particle}|\text{formed hadron}) * P(\text{blue}|\text{formed hadron})\\ &= \frac{1}{2} * \frac{2}{5} * \frac{3}{5} * \frac{2}{5}\\ &= \frac{6}{125} \end{align*} and \begin{align*} &P(\text{not formed hadron}|\text{top, particle, blue})\\ &= P(\text{not formed hadron}) * P(\text{top}|\text{not formed hadron}) * P(\text{particle}|\text{not formed hadron}) * P(\text{blue}|\text{not formed hadron})\\ &= \frac{1}{2} * \frac{1}{5} * \frac{2}{5} * \frac{2}{5}\\ &= \frac{2}{125} \end{align*}

We can do this by solving the equation k * \frac{2}{125} = \frac{6}{125}. You should find k=3.