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.

its an acute right triangle

step-by-step explanation:

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