Math Is Fun Forum

Tangram

Member

Registered: 2014-07-25

Posts: 12

Probability problem

What is the probability that a random sample of 4 from the set {A, B, C, D, E} will contain at least two A's? (obviously, repetition of elements is permitted).

I tried to solve this using the on-site combinations calculator but I'm getting an error. My method is to work out the number of outcomes giving at least two A's and divide by the total outcomes.

The total outcomes is the number of ways without any restrictions, which is C(n-1+r, r) where n = 5 and r = 4, so this gives C(8,4) = 70. No problem with the site calculator on this step.

The tricky bit is finding the number of outcomes giving at least two A's. If at least two A's are to be present then each sample of 4 will have the form AAXY, where XY is a sample of size 2 (repetition allowed) from {A, B, C, D, E}. So using the same formula, n = 5 and r = 2, and C(6, 2) = 15. So the probability is 15/70. I think this is correct but I wanted to check using the calculator with the "has" rule. This is what I put in the box:

a,b,c,d,e
has 2 a

But I'm getting the error message "Warning: rule 'has 2 a' NOT understood".

Any suggestions? Thanks.

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What gets us into trouble isn't what we don't know; it's what we know for sure that just ain't so! - Mark Twain

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability problem

Hi;

I am getting 113 / 625 for the answer so that you can check.

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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.

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Probability problem

Tangram, you forgot that the two As can be in different places.

I would count it this way:

625 ways

4^4 = 256 ways with 0 A

4 * 4^3 = 256 ways with 1 A

So I get 1 - (256 + 256)/625 = 113/625

Last edited by phanthanhtom (2015-10-27 04:23:47)

Tangram

Member

Registered: 2014-07-25

Posts: 12

Re: Probability problem

Thanks guys for the responses. Sorry about the late reply.

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What gets us into trouble isn't what we don't know; it's what we know for sure that just ain't so! - Mark Twain

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability problem

Hi;

You are welcome.

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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.