# Statistics problem-exponential approximation

• nuagerose
In summary: The next thing is to write out the factorials in the actual form they'll take on. I don't think I'll do it here, but I'll point out one thing, that for large n, \frac{n}{n+1} = 1 - \frac{1}{n+1} \approx 1 - \frac{1}{n} . You should be able to use that to simplify the formula.For part b), how do I set up an exponential approximation? To get started, I think that it would be e^(-1) + e^(-2) +...+ e^(1-k)... am I on the right track?The formula you wrote is an exponential approximation, but it

#### nuagerose

Statistics problem---exponential approximation

## Homework Statement

A box contains 2n balls of n different colors, with 2 of each color. Balls are picked at random from the box with replacement until two balls of the same color have appeared. Let X be the number of draws made.

a) Find a formula for P(X>k) k=2,3,...

b) Assuming n is large, use an exponential approximation to find a formula for k in terms of n such that P(X>k) is approximately 1/2. Evaluate k for n equal to one million.

## The Attempt at a Solution

For part a), I got that P(X>k) = (2n-2)/2n * (2n-4)/2n *...* (2n-2k+2n)/2n for k terms.

For part b), how do I set up an exponential approximation? To get started, I think that it would be e^(-1) + e^(-2) +...+ e^(1-k)... am I on the right track?

nuagerose said:
For part a), I got that P(X>k) = (2n-2)/2n * (2n-4)/2n *...* (2n-2k+2n)/2n for k terms.

I haven't done the counting, but this formula looks wrong since it appears to have ##n-k## terms. Once you work out the right form, I would clean it up by canceling common factors of 2 and also using the factorial function.

For part b), how do I set up an exponential approximation? To get started, I think that it would be e^(-1) + e^(-2) +...+ e^(1-k)... am I on the right track?

They could mean use http://en.wikipedia.org/wiki/Stirling's_approximation]Stirling's[/PLAIN] [Broken] approximation, which is useful whenever you have an expression for the factorial of a large number.

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nuagerose said:

## Homework Statement

A box contains 2n balls of n different colors, with 2 of each color. Balls are picked at random from the box with replacement until two balls of the same color have appeared. Let X be the number of draws made.

a) Find a formula for P(X>k) k=2,3,...

b) Assuming n is large, use an exponential approximation to find a formula for k in terms of n such that P(X>k) is approximately 1/2. Evaluate k for n equal to one million.

## The Attempt at a Solution

For part a), I got that P(X>k) = (2n-2)/2n * (2n-4)/2n *...* (2n-2k+2n)/2n for k terms.

For part b), how do I set up an exponential approximation? To get started, I think that it would be e^(-1) + e^(-2) +...+ e^(1-k)... am I on the right track?

You really ought to explain the logic: your formula is the probability that the first k colours are all different. Of course, you can cancel out all the 2s to get
$$P(X > k) = \frac{n-1}{n} \frac{n-2}{n} \cdots \frac{n-k +1}{n}$$
(Note: the final factor is NOT what you wrote, but I assume that was just a 'typo', since you otherwise seemed to know what you were doing.)

You can do something similar to what Feller would do in his probability textbook, and re-write the result as
$$P(X > k) = \left(1 - \frac{1}{n}\right)\left( 1 - \frac{2}{n}\right) \cdots \left( 1 - \frac{k-1}{n}\right)$$
That provides a convenient starting point.