Combination + Permutation Question

[avanm]

Hi there,

I'm having trouble with this question below:

A special committee of five professionals is formed from a group of four lawyers, three doctors and three teachers. At least three lawyers are needed in the special committee. Define X as the number of teachers in the special committee. Construct the probability distribution of the random variable X.

I get initially that n=10 and k=5 as there are 10 people in total and 5 spots. But 3 of the spots are occupied by 3 lawyers therefore the n and k change to, n=7 and k=2.

From here I am lost.

Do I use a combination or a permutation? and is it 7C2 = 21 or 7P2 = 42. Also, because I am looking for the random variable X, am I just looking at the combinations of teachers so is that 3P2?

I've attached my work below,

Thanks in advance

 

[Dr.Peterson]

A special committee of five professionals is formed from a group of four lawyers, three doctors and three teachers. At least three lawyers are needed in the special committee. Define X as the number of teachers in the special committee. Construct the probability distribution of the random variable X.

I get initially that n=10 and k=5 as there are 10 people in total and 5 spots. But 3 of the spots are occupied by 3 lawyers therefore the n and k change to, n=7 and k=2.

Do I use a combination or a permutation? and is it 7C2 = 21 or 7P2 = 42. Also, because I am looking for the random variable X, am I just looking at the combinations of teachers so is that 3P2?

First, what probability are you calculating? I expect a calculation for each possible value of X.

In any case, you have the wrong denominator. It is a given that there must be at least 3 lawyers, so you should be finding the probability of a particular number of teachers, given that there are at least three lawyers.

Suppose you're finding P(X=1); that will be the probability that the committee has one teacher, given that it has at least three lawyers.

It doesn't necessarily matter, in general, whether you use permutations or combinations, as long as you're consistent; but for a committee, combinations seems more natural. The important thing is to specify carefully what you are counting.

How many ways are there to choose a committee with at least three lawyers? How many ways are there to choose one with at least three lawyers and exactly one teacher?

 

[pka]

I can not read your attachment. But here is a list of possible distribution.
[imath]\underline{ L,D,T}\\ \{3,2,0\}\\ \{3,1,1\}\\ \{3,0,2\}\\ \{4,1,0\}\\ \{4,0,1\}[/imath]
The committee must contain three or four lawyers
The number of ways to have three lawyers, one doctor, and one teacher is
[imath]\dbinom{4}{3}\cdot\dbinom{3}{1}\cdot\dbinom{3}{1}[/imath]

 

[avanm]

First, what probability are you calculating? I expect a calculation for each possible value of X.

In any case, you have the wrong denominator. It is a given that there must be at least 3 lawyers, so you should be finding the probability of a particular number of teachers, given that there are at least three lawyers.

Suppose you're finding P(X=1); that will be the probability that the committee has one teacher, given that it has at least three lawyers.

It doesn't necessarily matter, in general, whether you use permutations or combinations, as long as you're consistent; but for a committee, combinations seems more natural. The important thing is to specify carefully what you are counting.

How many ways are there to choose a committee with at least three lawyers? How many ways are there to choose one with at least three lawyers and exactly one teacher?

So does this mean that X = 0, 1, 2 teachers that can occupy that spot?

P(X=1) given that there are at least three lawyer: so does that mean if there is 1 teacher that n=7 becomes n=5 and k=2 which is 5C2 = 10 combinations so that the probability is 5C2/ 7C2?

 

[avanm]

I dont understand how you got the last line?
also how did you come up with those combinations?

I can not read your attachment. But here is a list of possible distribution.
[imath]\underline{ L,D,T}\\ \{3,2,0\}\\ \{3,1,1\}\\ \{3,0,2\}\\ \{4,1,0\}\\ \{4,0,1\}[/imath]
The committee must contain three or four lawyers
The number of ways to have three lawyers, one doctor, and one teacher is
[imath]\dbinom{4}{3}\cdot\dbinom{3}{1}\cdot\dbinom{3}{1}[/imath]

 

[Dr.Peterson]

So does this mean that X = 0, 1, 2 teachers that can occupy that spot?

P(X=1) given that there are at least three lawyer: so does that mean if there is 1 teacher that n=7 becomes n=5 and k=2 which is 5C2 = 10 combinations so that the probability is 5C2/ 7C2?

Your saying things like "n=7 becomes n=5" suggests that you are thinking improperly; n isn't some number that changes. You need to think about the numerator and denominator separately.

The denominator, as I suggested, is the total number of ways to choose a committee with either 3 or 4 lawyers. One way to do this is to find the probability of each event in pka's table, and add them up.

I dont understand how you got the last line?
also how did you come up with those combinations?

Please give it some thought, and tell us how you would find the number of ways to choose 3 lawyers, 1 doctor, and 1 teacher from the people available.

 

[avanm]

Your saying things like "n=7 becomes n=5" suggests that you are thinking improperly; n isn't some number that changes. You need to think about the numerator and denominator separately.

The denominator, as I suggested, is the total number of ways to choose a committee with either 3 or 4 lawyers. One way to do this is to find the probability of each event in pka's table, and add them up.


Please give it some thought, and tell us how you would find the number of ways to choose 3 lawyers, 1 doctor, and 1 teacher from the people available.

Hi! I think I got it now -- please review my attached file and let me know if my though process is on the right track. I think I understood what pka did which I applied

 

[Dr.Peterson]

Hi! I think I got it now -- please review my attached file and let me know if my though process is on the right track. I think I understood what pka did which I applied

It's a little hard to follow even when I enlarge the image greatly to be able to read it. But it looks like you'd fill out his table like this:

X { L,D,T}
0 {3,2,0} (your DD) 12
1 {3,1,1} (your DT) 36
2 {3,0,2} (your TT) 12
0 {4,1,0} (your LD)  3
1 {4,0,1} (your LT)  3
total               66

so P(X=0) = 15/66, P(X=1) = 39/66, and P(X=2) = 12/66.

And your work looks appropriate, avoiding the error suggested by your notation like LD, that you would be just picking a lawyer and a doctor for the last two slots, which would be wrong. You did correctly count the one way to choose all four lawyers, and 3 ways to choose a doctor.

As a final check, I'll count the total number of possible committees a different way. To pick 3 lawyers and two others, you have C(4,3)*C(6,2) = 60 ways (your 12+36+12), and to pick all 4 lawyers and one other, you have C(4,4)*C(6,1) = 6 ways (your 3+3). So everything looks good.