Problem 1.4.3 Solution
The first generation consists of two plants each with genotype yg or gy. They are crossed to
produce the following second generation genotypes, S =
{
yy, yg, gy, gg
}
. Each genotype is just
as likely as any other so the probability of each genotype is consequently 1/4. A pea plant has
yellow seeds if it possesses at least one dominant y gene. The set of pea plants with yellow
seeds is
Y =
{
yy, yg, gy
}
So the probability of a pea plant with yellow seeds is
P [Y ] = P [yy] + P [yg] + P [gy] = 3/4.
Problem 1.5.4 Solution
Define D as the event that a pea plant has two dominant y genes. To find the conditional
probability of D given the event Y , corresponding to a plant having yellow seeds, we look to
evaluate
P [D
|
Y ] = P [DY ] / P [Y ]
Note that P[DY ] is just the probability of the genotype yy. From Problem 1.4.3, we found that
with respect to the color of the peas, the genotypes yy, yg, gy, and gg were all equally likely. This
implies
P [DY ] = P [yy] = 1/4 P [Y ] = P [yy, gy, yg] = ¾
Thus, the conditional probability can be expressed as
P [D
|
Y ] = P [DY ] / P [Y ] = (¼)/ (3/4) = 1/3
Problem 1.6.4 Solution
(a) Since A
∩
B =
∅
, P[A
∩
B] = 0. To find P[B], we can write
P [A
∪
B] = P [A] + P [B]
−
P [A
∩
B]
5/8 = 3/8 + P [B]
−
0
Thus, P[B] = 1/4. Since A is a subset of B
c
, P[A
∩
B
c
] = P[A] = 3/8. Furthermore, since
A is a subset of B
c
, P[A
∪
B
c
] = P[B
c
] = 3/4.
(b) The events A and B are dependent because
