12^n

Each quark has 6 \cdot 2 = 12 possibilities because there are 6 possible types and 2 possible states. Since our experiment involves independently picking one of these 12 possibilities for each of our n quarks, the total number of results for the experiment is 12^n.

None of the above (72^n).

For making a meson, we need to choose two quarks. The first quark we choose has 6 \cdot 2 = 12 possibilities because there are 6 possible types and 2 possible states. However, our second quark cannot have the same state as our first; so the second quark has 6 \cdot 1 = 6 possibilities because there are still 6 possible types, but only 1 possible state.

This means a single meson has 12 \cdot 6 = 72 possibilities. (If the order of the quark and antiquark did not matter, it would instead have \frac{72}{2} = 36 possibilities). So, an experiment where we select n mesons uniformly at random has 72^n unique results.

{54 \choose 4} = 316251

Since order does not matter, we would use a combination instead of a permutation here. From 54 unique cards in our poker deck, we can choose 4 at random to get a card hand.

An alternative way to think about it: there are 54 options for the first card we pick, 53 for the second card, 52 for the third card, etc. For four cards, we have 54 \cdot 53 \cdot 52 \cdot 51 hands. However this will count the same hand in different card orders as different hands, which we don’t want. To correct this, we divide by 4! (4 \cdot 3 \cdot 2 \cdot 1), because each unique hand will appear 24 times but in different orders.

\frac{54 \cdot 53 \cdot 52 \cdot 51}{4 \cdot 3 \cdot 2 \cdot 1} = {54 \choose 4} = 316251

650

Since the first 4 cards are four-of-a-kind (same number), there are 13 ways to select a four-of-a-kind. For the 5th card, there are 50 total choices (12 remaining values \cdot 4 suits + 2 wildcards). So there are a total 13 \cdot 50 = 650 options.

P(\text{Four of a kind | Bomb}) = \frac{1}{103}

The number of hands containing two jokers (the other two cards are arbitrary) is given by: \begin{align*} \underbrace{{2 \choose 2}}_{\text{Two Jokers}} \cdot \underbrace{{52 \choose 2}}_{\text{Other Two Cards}} = \frac{52 \cdot 51}{2} = 1326. \end{align*} The number of hands that form four of a kind is 13, one for each of the 13 unique values in a deck.

For a bomb to occur, the four cards either contain two jokers or form four of a kind. Hence, \begin{align*} P(\text{Four of a kind | Bomb}) = \frac{13}{13 + 1326} = \frac{1}{103} \end{align*}

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*}