Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**genericname****Member**- Registered: 2012-05-16
- Posts: 52

How do you go about solving problems like this? Is there a trick to it?

*A class consists of 25 students, of whom 10 are women and the rest are men. We take a random sample of 5 students from this class. (Without replacement)*

-What is the probability that the sample will include at least one woman?

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

You use the hypergeometric distribution.

You will have to sum 5 different terms all involving the hypergeometric distribution. In summation notation it looks like this:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**genericname****Member**- Registered: 2012-05-16
- Posts: 52

Is it safe to assume that we use the formula only if it has "at least" in the question?

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Is it safe to assume that we use the formula only if it has "at least" in the question?

An "at least" in the problem implies a range of values in this case1 to 5. So, you will have to sum 5 hypergeometric terms that is why there is a sum there.

This is the formula:

You could us for say exactly 5 women by saying low = 5 and high = 5. Then you would be summing only one term.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**genericname****Member**- Registered: 2012-05-16
- Posts: 52

Thank you, Bobby.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Your welcome. May the computing be with you.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**genericname****Member**- Registered: 2012-05-16
- Posts: 52

Hi, another question:

An instructor gives an exam with 14 questions. Students are allowed to choose any 10 of them to answer. Suppose 6 questions require proof and 8 do not:

-How many groups of 10 questions contain at least one that require a proof?

I got 1736 as an answer. Is that correct?

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi genericname;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline