A committe of 5 members will be chosen from a group of 10 teachers and 5 students. What is the probability that the committee
will have
A) all teachers?
B) 3 teachers and 2 students?
C) 3 or 4 teachers?
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)
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!
Binomial distribution requires replacement, constant probability.
See:
https://en.wikipedia.org/wiki/Hypergeometric_distribution
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
To find the probability for each scenario, we need to determine the total number of possible committees and the total number of favorable outcomes for each case.
A) Probability of all teachers in the committee:
Total number of possible committees = C(15, 5) = 3003, because we need to choose 5 members from a pool of 15 (10 teachers + 5 students).
Total number of favorable outcomes = C(10, 5) = 252, because we need to choose all 5 members from the group of 10 teachers.
Probability = favorable outcomes / possible outcomes = 252 / 3003 = 0.084 or 8.4%.
B) Probability of 3 teachers and 2 students in the committee:
Total number of possible committees = C(15, 5) = 3003.
Total number of favorable outcomes = C(10, 3) * C(5, 2) = 120 * 10 = 1200.
Here, we choose 3 teachers from 10 and 2 students from 5.
Probability = favorable outcomes / possible outcomes = 1200 / 3003 = 0.399 or 39.9%.
C) Probability of 3 or 4 teachers in the committee:
Total number of possible committees = C(15, 5) = 3003.
Total number of favorable outcomes for 3 teachers = C(10, 3) * C(5, 2) = 120 * 10 = 1200, same as in the previous case.
Total number of favorable outcomes for 4 teachers = C(10, 4) * C(5, 1) = 210 * 5 = 1050.
Total number of favorable outcomes = 1200 + 1050 = 2250.
Probability = favorable outcomes / possible outcomes = 2250 / 3003 = 0.749 or 74.9%.
To summarize:
A) The probability of all teachers is 0.084 (8.4%).
B) The probability of 3 teachers and 2 students is 0.399 (39.9%).
C) The probability of having 3 or 4 teachers is 0.749 (74.9%).