1. You have k slots (of the same size) to drop n tokens into. How many ways

can you do this if the tokens are marked? Unmarked? (Hint: try with 3 tokens
and 7 slots first)
2. What is the probability of getting them all in the unique “winning” slot?
(Do you use the marked or unmarked count from #1 to compute this?)
3. What is the probability of getting at least two in the same slot (n≥2)?
4. What is the probability of getting at most one in each slot? At least one in
each slot, n≥k?
5. Suppose we let k approach in�finity, what happens in #4? Take a moment to
think about how slot size is important. It is certainly possible to �t infi�nitely
many slots into a �nite space. Do you have any insights if the slot sizes were
1/2 ft, 1/4 ft, 1/8 ft, 1/16 ft,....? (1=2+1=4+1=8+::: = 1) We need to assume
that the tokens are infi�nitely small. Clearly, there are in�finitely may possible
combinations. However, one is interested in determining the more probable
outcomes. Tip: Divide one bin into two and do the calculation. Then take the
right bin and divide it into two, calculate...etc.

It's our final review questions, and I really don't know how to solve these. Can anyone help me?
THX so much!!!

Sure, I'd be happy to help you solve these questions! Let's go through each question step by step:

1. To calculate the number of ways you can drop n tokens into k slots, we can use the concept of combinations. For each token, we have k choices of slots to drop it into. So, for n tokens, we have k choices for the first token, k choices for the second token, and so on. Therefore, the total number of ways to drop n tokens into k slots is k^n.

For marked tokens, the number of ways is the same as the total number of ways, which is k^n.

2. The probability of getting all the tokens in the unique "winning" slot would depend on whether you are using the marked or unmarked count from the previous question. If you are considering the marked count, i.e., the total number of ways (k^n), then the probability would be 1/(k^n), as there is only one "winning" slot out of k^n total possibilities.

3. To find the probability of getting at least two tokens in the same slot, we can use the concept of complementary probability. The probability of getting at least two tokens in the same slot is equivalent to 1 minus the probability of getting all tokens in different slots.

The probability of all tokens in different slots can be calculated by considering the first token, having k choices of slots, the second token, having (k-1) choices (excluding the slot already occupied by the first token), and so on. Therefore, the probability would be (k/k)(k-1/k)(k-2/k)...(k-(n-1)/k), which simplifies to (k/k)(k-1/k)...(k-n+1/k).

So, the probability of getting at least two tokens in the same slot would be 1 - (k/k)(k-1/k)...(k-n+1/k).

4. The probability of getting at most one token in each slot is simply the complementary probability of getting at least two tokens in the same slot. So, using the same approach as in question 3, the probability would be 1 - (k/k)(k-1/k)...(k-n+1/k).

The probability of getting at least one token in each slot, when n≥k, is 1 because we have more tokens than slots, guaranteeing at least one token in each slot.

5. As we let k approach infinity, the probability of getting at most one token in each slot (question 4) would approach 1, as the odds of having two or more tokens in the same slot decrease as the number of available slots increases.

In the case of infinitely small slot sizes, such as 1/2 ft, 1/4 ft, 1/8 ft, 1/16 ft, and so on, we can use the concept of limits to analyze the probability. As the slot sizes become infinitely small, the probability of getting at most one token in each slot would approach 1, similar to the case with an infinite number of slots.

I hope this helps you understand how to solve these questions! If you have any further doubts or need more clarification, feel free to ask.