Probability

[Maybe_Memorie]

1. The problem statement, all variables and given/known data

1. A group of people contains 10 with blood group O, 5 with blood group A and 5 with blood group B. Give a formula for the probability that a random sample of size 6 will contain two people from each group.

2. An athlete conceals 2 performance enhancing drugs in a bottle containing 8 vitamin tablets. If three tablets are selected at random for testing what is the probability cheating will be detected?

2. Relevant equations



3. The attempt at a solution

To get the six people having two from each group would it be
(10/20)(9/19)(5/18)(4/17)(5/16)(4/15) and there are 6! ways this can be arranged so my answer would be
P = 6!(10/20)(9/19)(5/18)(4/17)(5/16)(4/15)

Is this correct?


For part 2, there is 10 tablets in total. We take a sample of 3 tablets.
I'm sure there's a fancier way of doing this but all I can think of is this
P(drugs detected) = P(1 drug found) + P(2 drugs found)
= 3!(2/10)(8/9)(7/8) + 3!(2/10)(1/9)(8/8)

[vela]

1. The problem statement, all variables and given/known data

1. A group of people contains 10 with blood group O, 5 with blood group A and 5 with blood group B. Give a formula for the probability that a random sample of size 6 will contain two people from each group.

3. The attempt at a solution

To get the six people having two from each group would it be
(10/20)(9/19)(5/18)(4/17)(5/16)(4/15) and there are 6! ways this can be arranged so my answer would be
P = 6!(10/20)(9/19)(5/18)(4/17)(5/16)(4/15)

Is this correct?
Almost. You need to divide by (2!)3 because you're double-counting each pair. That is, you're counting X getting picked first and then Y separately from Y getting picked first and then X, for each pair.

[Maybe_Memorie]

Almost. You need to divide by (2!)3 because you're double-counting each pair. That is, you're counting X getting picked first and then Y separately from Y getting picked first and then X, for each pair.

Ah yes I see! :smile:

Is my part 2 correct? Is there any other way of going about it?

[vela]

1. The problem statement, all variables and given/known data

2. An athlete conceals 2 performance enhancing drugs in a bottle containing 8 vitamin tablets. If three tablets are selected at random for testing what is the probability cheating will be detected?

3. The attempt at a solution

For part 2, there is 10 tablets in total. We take a sample of 3 tablets.
I'm sure there's a fancier way of doing this but all I can think of is this
P(drugs detected) = P(1 drug found) + P(2 drugs found)
= 3!(2/10)(8/9)(7/8) + 3!(2/10)(1/9)(8/8)
Same mistake as before. You're double-counting the pairs.

[Ray Vickson]

1. The problem statement, all variables and given/known data

1. A group of people contains 10 with blood group O, 5 with blood group A and 5 with blood group B. Give a formula for the probability that a random sample of size 6 will contain two people from each group.

2. An athlete conceals 2 performance enhancing drugs in a bottle containing 8 vitamin tablets. If three tablets are selected at random for testing what is the probability cheating will be detected?

2. Relevant equations



3. The attempt at a solution

To get the six people having two from each group would it be
(10/20)(9/19)(5/18)(4/17)(5/16)(4/15) and there are 6! ways this can be arranged so my answer would be
P = 6!(10/20)(9/19)(5/18)(4/17)(5/16)(4/15)

Is this correct?


For part 2, there is 10 tablets in total. We take a sample of 3 tablets.
I'm sure there's a fancier way of doing this but all I can think of is this
P(drugs detected) = P(1 drug found) + P(2 drugs found)
= 3!(2/10)(8/9)(7/8) + 3!(2/10)(1/9)(8/8)

In your part 2: your answer adds up to 16/15 > 1!

In both cases you have a hypergeometric-type problem: in 1, it is a 3-category hypergeometric, with N1 = 10 of type 1, N2 = 5 of type 2 and N3 = 5 of type 3.

In case 2 you have a standard (2-category) hypergeometric, with N1 = 2 of type 1 and N2 = 8 of type 2. You can easily get P{no type 1 in sample of size 3}.

See. eg., http://stattrek.com/lesson2/hypergeometric.aspx or
http://www.stat.ufl.edu/~berg/sta5325/files/sta5325-12.pdf for the hypergeometric.

In part 1 you have a multi-class version of the hypergeometric, which is not usually done in easily-accessible web pages. Think of it like this: P{k1,k2,k3} = P{k1 type 1, k2+k3 type 2 or 3}*P{k2 type 2, k3 type 3|k2 + k3 type 2 or 3}. Each factor is a simple hypergeometric probability as found in the cited web pages.

RGV

[vela]

Is there any other way of going about it?
Do you know about binomial coefficients? Both problems can be solved easily using them.

In problem 2, you could also find the probability of not getting caught, i.e. the probability of choosing 3 vitamin tablets, and then subtract that from 1 to find the probability of getting caught.

[Maybe_Memorie]

Do you know about binomial coefficients? Both problems can be solved easily using them.

In problem 2, you could also find the probability of not getting caught, i.e. the probability of choosing 3 vitamin tablets, and then subtract that from 1 to find the probability of getting caught.

Yes, I was trying originally to apply the binomial method but couldn't see how to...

[Maybe_Memorie]

In your part 2: your answer adds up to 16/15 > 1!

In both cases you have a hypergeometric-type problem: in 1, it is a 3-category hypergeometric, with N1 = 10 of type 1, N2 = 5 of type 2 and N3 = 5 of type 3.

In case 2 you have a standard (2-category) hypergeometric, with N1 = 2 of type 1 and N2 = 8 of type 2. You can easily get P{no type 1 in sample of size 3}.

See. eg., http://stattrek.com/lesson2/hypergeometric.aspx or
http://www.stat.ufl.edu/~berg/sta5325/files/sta5325-12.pdf for the hypergeometric.

In part 1 you have a multi-class version of the hypergeometric, which is not usually done in easily-accessible web pages. Think of it like this: P{k1,k2,k3} = P{k1 type 1, k2+k3 type 2 or 3}*P{k2 type 2, k3 type 3|k2 + k3 type 2 or 3}. Each factor is a simple hypergeometric probability as found in the cited web pages.

RGV

Thanks! :smile:

[vela]

Yes, I was trying originally to apply the binomial method but couldn't see how to...
For the second problem, you could calculate the probability of selecting three vitamin tablets. That would be equal to the number of ways you can select 3 tablets from the 8 vitamin tablets divided by the total number of ways to select 3 tablets from the 10 tablets.

You could also calculate the desired probability directly. For example, if you wanted to calculate the number of ways to select 1 drug and 2 vitamin tablets, for example, the total number of ways to do that is the number of ways to select 1 drug tablet from the 2 possible multiplied by the number of ways to draw 2 tablets from the 8 vitamin tablets.