mathematics data

Question

a committee of six people is formed to investigate connections between business and the education system. it consists of people chosen from 5 students, 8 teachers, 5 principles, 10 parents, 4 professors, and 6 business people. if the members are chosen at random, what is the probability that there is one person from each group on the committee?

Answer 1

Hello Hassan - greetings from Holyke MA

 

  I think Richard A may have misunderstood your problem (or perhaps I have):

It sounds like there is a pool of 38 people and 6 will be chosen at random to be on a committee.

Those 38 are from six categories: "5 students, 8 teachers, 5 principals, 10 parents, 4 professors, and 6 business people."

We want to know the chance that the randomly-selected committee will have exactly one from each category.

 

Let's consider selecting from the pool one at a time and start with the probability that we choose, in order: a student, a teacher, a principal, a parent, a prof. and finally a business person.  That probability is:

 

(5/38)*(8/37)*(5/36)*(10/35)*(4/34)*(6/33)

 

Note that with each pick there is one less person in the pool, so the demoninator decreaes.

If we want to pick them in a different order we get a similar expression, for example the chance to pick them in the reverse order is:

 

(6/38)*(4/37)*(10/36)*(5/35)*(8/34)*(5/33)

 

This is equal to the first case: the numerator consists of the product of the numbers of each type, and the denominator is the product of 38 ... 33.

 

Now, as Richard A points out, there are 6! (=720) different ways (orders) we could pick one from each category; from the above cases it makes sense that each ordering has the same probability, so all together the chance is:

 

720 * (5/38)*(8/37)*(5/36)*(10/35)*(4/34)*(6/33)

         =  (5*4*5*8*10*2)/(19*37*7*17*11)  <-- it's not a simple fraction

  ~ 0.0174

 

Let me know if this is not correect, thanks.

 

Answer 2

Hi Hassan,

The individual probabilities for one from each group would need to be multiplied.  (1/5)(1/8)(1/5)(1/10)(1/4)(1/6) = 1/48000. But this could happen in 6 possibilities times 5 possibilities times 4 possibilities times 3 possibilities times 2 possibilities times one possibilities or 720 different ways. Therefore 1/48000 times 720 = .015