Foundations of Probability

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This page contains all problems about Foundations of Probability.


Problem 1

Source: Spring 2023 Final Part 2, Problem 2

You roll a 12-sided die 8 times.


Problem 1.1

What is the probability you roll 8 distinct values?

\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}


Problem 1.2

What is the probability you roll exactly 7 distinct values?

\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.



Problem 2

Source: Spring 2023 Final Part 2, Problem 3

Consider three events A, B, and C in the same sample space.


Problem 2.1

Which of the following is equivalent to P((A\cap B)|(B \cap C))? Select all that apply.

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!


Problem 2.2

Suppose P((A \cap B)|C) = P(A|(B \cap C))*P(B). Which of the following pairs of events must be independent?

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.



Problem 3

Source: Spring 2023 Final Part 2, Problem 6

Suppose UCSD implements a new graduation requirement for data science majors. All data science majors must submit a list of English words they can make using the letters of “Halicioglu”. When these lists are collected, it is determined that:

How many data science majors submitted a list?

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.


Problem 4

Source: Spring 2023 Midterm 2, Problem 2

A set of chess pieces has 32 pieces. 16 of these are black and 16 of these are white. In each color, the 16 pieces are:

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.

Suppose you randomly select 2 pieces from a set of 32 chess pieces, without replacement.


Problem 4.1

You glance at the pieces just long enough to see that both pieces are white. What is the probability that you have 2 pawns?

\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}


Problem 4.2

True or False: Having two pawns is independent of having two white pieces.

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.”



Problem 5

Source: Spring 2023 Midterm 2, Problem 4

Suppose that there are three possible experience levels in chess (beginner, intermediate, advanced). Only 10% of beginner players beat Avi at a game of chess, while 50% of intermediate players and 80% of advanced players do.

Avi signs up to participate in a certain chess tournament called the Avocado Cup. Aside from Avi, 50% of the players in the tournament are beginners, 40% are intermediate, and 10% are advanced.

The tournament proceeds in rounds. In the first round of the tournament, players are randomly paired up for a game of chess.


Problem 5.1

What is the probability that Avi wins in the first round of the tournament?

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*}


Problem 5.2

It turns out that, sadly, Avi loses to his opponent in the first round. What is the probability that Avi’s opponent is an advanced player? Choose the closest answer among the choices listed.

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 \%.



Problem 6

Source: Spring 2023 Midterm 2, Problem 6

The Avocado Cup is organized into rounds. In each round, players who win advance to the next round, and players who lose are eliminated. Rounds continue on like this until there is a single tournament winner.

Define the following events in the sample space of possible outcomes of the Avocado Cup:


Problem 6.1

Which of the following statements is true? Select all that apply.

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.”



Problem 6.2

The events A and B are mutually exclusive, or disjoint. More generally, for any two disjoint events A and B, show how to express P(\overline{A}|(A \cup B)) in terms of P(A) and P(B) only. For this problem only, show your work and justify each step.

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.



Problem 7

Source: Winter 2023 Final, Problem 6


Problem 7.1

In the above figure, S is the sample space. Two events are denoted by two circles: A (Red + Blue) and B (Green + Blue). Which area denotes the event \bar{A}\cap\bar{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.



Problem 7.2

Use the law of total probability to prove De Morgan’s law: P(\bar{A}\cap\bar{B})=1-P(A\cup B). You may use the figure above to visualize the statement, but your proof must use the law of total probability.

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).



Problem 8

Source: Winter 2024 Final Part 2, Problem 1

Consider an experiment where each of n people selects one piece from their own set of chess pieces, uniformly at random. A chess set has 32 pieces. 16 of these are black and 16 of these are white. In each color, the 16 pieces are:

The result of the experiment is a description of the colors and types of the pieces each person selected. For example, if n=3, one possible result is:


Problem 8.1

How many results are possible for this experiment with n people?

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.


Problem 8.2

As n becomes large, what fraction of n people has selected a black pawn? Choose the answer that’s closest to the actual fraction.

\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.



Problem 8.3

True or False: Having two pawns is independent of having two white pieces.

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.”



Problem 9

Source: Winter 2024 Midterm 2, Problem 3

Schrödinger’s cat is a famous thought experiment in quantum mechanics proposed by physicist Erwin Schrödinger in 1935. The experiment is described below:

“Imagine there’s a hypothetical cat in a closed box with a toxic radioactive element that might decay. If it decays, the cat dies; if it doesn’t, the cat lives.”

Please note that Schrödinger’s Cat is a purely theoretical concept, so a thought experiment. It has never been executed in the real world, and no cats have ever been harmed as a result of it.


Problem 9.1

Suppose the cat has 90 \% of the chance to die if the decay happens. The cat also has 10 \% chance to die even if the decay does not happen. Suppose the decay happens with a 20 \% probability. After you open the box, you find the cat dead. What is the probability that the decay happened?

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*}



Problem 9.2

Quantum superposition is a mind-bending and counter-intuitive concept in physics. In the Schrödinger’s cat scenario, when the box is closed, the cat can be both dead and alive simultaneously. However, once we open the box and observe, the superposition collapses, and the cat must be either dead or alive, not both.

The Venn Diagram below depicts the Schrödinger’s cat scenario with the box closed. In this diagram, the black dotted line is the event of cat dead, the red dashed line denotes the event of cat being alive, and the shaded region is the event of decay.


Based on this Venn diagram, which of the following is true? Select all that apply:

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.


Problem 9.3

You’re given the following probabilities:

Using the Venn diagram in Problem 3.2, calculate the probability for Schrodinger’s cat to be in the superposition state (i.e. both dead and alive):

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*}



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