Algebra 2 probability!

Question

A large college class has 54 women and 42 men as students. 10 of the students are chosen at random.
a) What is the probability that the group of 10 will include 5 women and 5 men. Answer to 4 decimal places.
b) List all possible makeups of the group that have a probability of less than 1%. Answer to 4 decimal places.
c) Is the probability distribution symmetric or skewed? Justify the answer.

Answer 1

There are 96 students in a college class (54 women and 42 men).

10 students are chosen at random.

a.) We want to use the probability with combinations where order doesn't matter. 10 students who are chosen at random will include 5 women and 5 men.

C(54, 5) represents the number of ways of choosing 5 women to be part of the college classes out of 54.

C(42, 5) represents the number of ways of choosing 5 men to be part of the college classes out of 42.

C(96, 10) represents the number of ways of choosing 10 students to be part of the college classes out of 96.

P(the group of 10 will have 5 women and 5 men) = [C(54, 5) × C(42, 5)] / C(96, 10)

= (3,162,510)*(850,668) / 11,279,926,456,656

≈ 0.2385

The probability that the group of 10 will have 5 women and 5 men in a college class is about 0.2385.

b.) x represents a selection from a group of 10 students. p is the success probability = 0.01, and q is the failure probability = 0.99.

X ∼ Bin(10, 0.01)

P(X = x) = C(10, x) * (0.01)^x * (0.99)^{10 - x}

P(X = 2) = C(10, 2) * (0.01)² * (0.99)^{10 - 2} = 45*(0.01)²*(0.99)⁸ ≈ 0.004152

P(X = 3) = C(10, 3) * (0.01)³ * (0.99)^{10 - 3} = 120*(0.01)³*(0.99)⁷ ≈ 0.0001118

P(X = 4) = C(10, 4) * (0.01)⁴ * (0.99)^{10 - 4} = 210*(0.01)⁴*(0.99)⁶ ≈ 0.000001977

P(X = 5) = C(10, 5) * (0.01)⁵ * (0.99)^{10 - 5} = 252*(0.01)⁵*(0.99)⁵ ≈ 0.00000002396

P(X = 6) = C(10, 6) * (0.01)⁶ * (0.99)^{10 - 6} = 210*(0.01)⁶*(0.99)⁴ ≈ 0.0000000002017

P(X = 7) = C(10, 7) * (0.01)⁷ * (0.99)^{10 - 7} = 120*(0.01)⁷*(0.99)³ ≈ 0.000000000001164

P(X = 8) = C(10, 8) * (0.01)⁸ * (0.99)^{10 - 8} = 45*(0.01)⁸*(0.99)² ≈ 0.00000000000000441

P(X = 9) = C(10, 9) * (0.01)⁹ * (0.99)^{10 - 9} = 10*(0.01)⁹*(0.99)¹ ≈ 0.0000000000000000099

P(X = 10) = C(10, 10) * (0.01)¹⁰ * (0.99)^{10 - 10} = 1*(0.01)¹⁰*(0.99)⁰ ≈ 0.00000000000000000001

These are the possible makeups of the group with the probability of less than 1%.

c.) This is a binomial distribution. We need to find the mean and standard deviation for a group of 10.

E(X) = np = (10)(0.01) = 0.1 < 5

σ(X) = √npq = √(10)(0.01)(0.99) = 0.3146 < 5

For small p and small n, the binomial distribution is skewed right. That is, the bulk of the probability falls in the smaller numbers, x = 0 and 1, and the distribution tails off to the right.