# Probability And Statistical Methods For Engineers Worksheets With Answer Key - Mae 108, 2011

NAME:
MAE 108 – Probability and Statistical Methods for Engineers - Winter 2011
Midterm # 2 - February 25, 2011
50 minutes, open book, open notes, calculator allowed, no cell phones.
(1) (10 points)
The eye color of a person is classiﬁed on the basis of one pair of genes. Suppose d represents a
dominant gene (brown eyes) and r a recessive gene (blue eyes): a person with dd genes is pure
dominance, one with rr is pure recessive, and one with rd or dr is hybrid. The pure dominance
and hybrid have brown eyes, while the recessive have blue eyes. Children receive one random gene
from each parent, independently from the genes received by their siblings.
(a) Two hybrid parents have 4 children. What is the probability that 3 out of the 4 children have
brown eyes?
Each parent is rd so children can be, with equal probability: rr, rd, dr, or dd. Hence a child has
a 75% probability of having brown eyes. The probability of having 3 out of 4 with brown eyes is
4
3
thus a binomial: P = (
)(3/4)
1/4
42%.
3
(b) One hybrid and one recessive parent have 3 children. What is the probability that at least one
of their children have blue eyes?
Parent 1 is rd and parent 2 is rr hence children will be rr or dr with equal probability. The
probability of any given child to have blue eyes is thus p = 1/2. The probability to have at least
one out of three having blue eyes is 1 minus the probability of having 3 children with brown eyes
3
and thus P = 1
(1/2)
= 87.5%.
1