In a class of 100 students, 25 students have hardcover and 75 students have paperback textbooks for the course. If you randomly choose 10 students in this class, fnd the probability that 2 of them have hardcover texts in the following ways: a. the exact probability b. approximate probability using a binomial distribution

h = prob of hardcover =.25

p = prob of paperback =.75

a.

Prob (2 hardcover of 10) = (10 choose 2) (.25) ^2 (.75) ^8

= (10 (9) / 2) (1/4^10) (1^2 3^8)

= 295245/1048576

≈ 28.15%

b.

I think I just did the binomial distribution. We can approximate it as a normal distribution. The mean and variance in this case are computed thus:

N=10, n=2

μ = Np = 2.5

σ² = Npq = 1.875

σ = 1.3693

P (n) = 1 / (σ√2π) e^ - (n-μ) / (2σ²)

P (2) = 1 / (1.3693√2π) e^ - (2 - 2.5) / (2 (1.875))

P (2) = 33.3%

Pretty good, not great.