Question

Consider a husband and wife with brown eyes who have 0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes.

(a) What is the probability that their first child will have green eyes and the second will not?

(b) What is the probability that exactly one of their two children will have green eyes?

(c) If they have 7 children, what is the probability that exactly 2 will have blue eyes?

(d) If they have 7 children, what is the probability that at least 2 will have green eyes?

(e) What is the probability that the first brown eyed child will be child number 3?

Answer 1

P(brown eyes) = 0.75

P(blue eyes) = 0.125

P(green eyes) = 0.125

a) P(first child have green eyes and second will not) = P(first child have green eyes)*P(second child do not have green eyes)

P(first child have green eyes and second will not) = 0.125 * (1-0.125) = 0.125*0.875

P(first child have green eyes and second will not) = 7/64 = 0.109375

b) This is a binomial distribution question with
n = 2
p = 0.125
q = 1 - p = 0.875
where
c) This is a binomial distribution question with
n = 7
p = 0.125 (probablity of child having blue eyes)
q = 1 - p = 0.875
where
d) This is a binomial distribution question with
n = 7
p = 0.125 (probablity of child having green eyes)
q = 1 - p = 0.875
where

e) P(first brown eyed child will be child number 3) = P(first child do not have brown eyes) * P(second child do not have brown eyes) * P(third child have brown eyes)

P(first brown eyed child will be child number 3) = (1-0.75) * (1-0.75) * (0.75)

P(first brown eyed child will be child number 3) = 0.25*0.25*0.75

P(first brown eyed child will be child number 3) = 3/64 = 0.046875

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