A committe of 5 members will be chosen from a group of 10 teachers and 5 students. What is the probability that the committee

Question

will have
A) all teachers?
B) 3 teachers and 2 students?
C) 3 or 4 teachers?

Answer 1

You can treat this as a set of Bernoulli Trials and use the binomial formula to find your probability. If we define a 'success' as selecting a teacher and a 'failure' as selecting a student,

n = number of selections (5)
r = number of successes
p = probability of success per trial = 2/3
q = probability of failure per trial = 1/3

For these symbols, the probably of 'r' successes is given by:

P(r) = nCr * p^r * q^(n-r)

Answer 2

a) There are five successes, r = 5

P = 5C5 * (2/3)^5 * (1/3)^(5-5)
= (2/3)^5
= 32/243

b) There are three successes, r = 3

P = 5C3 * (2/3)^3 * (1/3)^(5-3)
= 10 * (2/3)^3 * (1/3)^2
= 80/243

Give the third one a go!

Answer 3

Binomial distribution requires replacement, constant probability.
See:
https://en.wikipedia.org/wiki/Hypergeometric_distribution

Answer 4

I answered this question yesterday when it was posted under "Angel"
https://www.jiskha.com/display.cgi?id=1515201345
Why did you repost it ??

I stand by those answers for a) and b)
c) is open for debate since it is ambiguous.

here is the other way to do a)
Prob(all 5 are teachers)
= (10/15)(9/14)(8/13)(7/12)(6/11) = 12/143

which agrees with my
C(10,5)/C(15,5) = 252/3003 = 12/143

b) alternate way:
a particular case would be TSTTS
that particular probability of that is
(10/15)(5/14)(9/13)(8/12)(4/11) = 40/1001

but those 3 teachers and 2 students can be arranged in
5!/(3!2!) or 10 ways, so the prob(3 teachers with 2 students) = 10(40/1001) = 400/1001

which agrees with my
C(10,3)*C(5,2)/C(15,5) = 400/1001