Question

Color blindness is the decreased ability to see colors and clearly distinguish different colors of the visible spectrum. It is known that color blindness affects 11% of the European population. A sample of 12 subjects is taken for research purposes by a team of optometrists.

What is the probability that one subject is color blind?

What is the probability that more than one subject is color blind?

What is the probability that at most 3 subjects are color blind?

What is the probability that the number of color blind subjects is between 2 and 4 (both included)?

What is the probability that at least 10 subjects are not color blind?

Answer 1

n = 12

p = 0.11

It is a binomial distribution

P(X = x) = nCx * p^x * (1 - p)^{n - x}

a) P(X = 1) = 12C1 * (0.11)^1 * (0.89)^11 = 0.3663

b) P(X > 1) = 1 - P(X < 1)

                 = 1 - (P(X = 0) + P(X = 1))

                 = 1 - (12C0 * (0.11)^0 * (0.89)^12 + 12C1 * (0.11)^1 * (0.89)^11)

                 = 1 - 0.6133 = 0.3867

c) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

                 = 12C0 * (0.11)^0 * (0.89)^12 + 12C1 * (0.11)^1 * (0.89)^11 + 12C2 * (0.11)^2 * (0.89)^10 + 12C3 * (0.11)^3 * (0.89)^9 = 0.9649

d) P(2 < X < 4) = P(X = 2) + P(X = 3) + P(X = 4)

                      = 12C2 * (0.11)^2 * (0.89)^10 + 12C3 * (0.11)^3 * (0.89)^9 + 12C4 * (0.11)^4 * (0.89)^8 = 0.3801

e) P(X > 10) = P(X = 10) + P(X = 11) + P(X = 12)

                  = 12C10 * (0.89)^10 * (0.11)^2 + 12C11 * (0.89)^11 * (0.11)^1 + 12C12 * (0.89)^12 * (0.11)^0 = 0.8623