AP Statistics – Chapter 7 Exercises Solutions:
7.11: Rolling Two Dice
a. The 36 possibilities of “up faces” are:
(1, 1), (1, 2), … , (1, 6)
(2, 1), (2, 2), … , (2, 6)
*
*
*
*
*
*
*
*
*
(6, 1), (6, 2), … , (6,6)
b. Each pair has a probability of 1/36
c. Let X = sum of “up faces” [ex: X = 2 for the pair (1, 1) and X = 12 for the pair (6, 6)]
X
P(X)
2
1/36
3
2/36
4
3/36
5
4/36
6
5/36
7
6/36
8
5/36
9
4/36
10
3/36
11
2/36
12
1/36
d. P( X = 7 or X = 11) = 6/36 + 2/36 = 8/36 = 2/9
e. P(X ≠ 7) = 1 – P(X = 7) = 1 – 6/36 = 30/36 = 5/6
7.14: Car Ownership
a. All probabilities are between 0 and 1 and the sum of all probabilities equals 1.
b. The event {X ≥ 1} means that the household owns at least one car.
P(X ≥ 1) = 1 – P(X = 0) = 0.91
c. P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5) = 0.20
20% of households own more cars than a two-car garage can hold.
7.16: Student Fees
a. Let S = {student supports funding}
Let O = {student opposes funding}
P(SSO) = .6 * .6 * .4 = 0.144
b. 8 possibilities: SSS, SSO, SOS, SOO, OOO, OOS, OSO, OSS
P(SSS) = .216
P(SSO) = P(SOS) = P(OSS) = .144
P(SOO) = P(OSO) = P(OOS) = .096
P(OOO) = .064
c. Probability Distribution for X = number of students opposed
X
0
1
P(X)
.216
.432
d. {X ≥ 2} or {X > 1}
P(X > 1) = .352
2
.288
3
.064