Homework Set — February 20, 2015
Solutions
Graded problems are in red
Chapter 3
Problem 56
P (new) =
m
X
pi P (new|the n-th coupon is of type i) =
i=1
m
X
pi (1−pi )n−1 .
i=1
2
1
p(1 − p) = 2p(1 − p)
(b) P (increase by one after three days) = 32 p2 (1 − p) = 3p2 (1 − p)
Problem 57 (a) P (original price after two days) =
(c) P (increase on first day|increase by one after three days) =
p·2p(1−p)
3p2 (1−p)
=
2
3
Problem 59 (a) P (HHHH) = p4
(b) P (T HHH) = p3 (1 − p)
(c) The pattern HHHH can only occur before T HHH if the first four
coin flips come up heads. Hence, P (T HHH occurs before HHHH) =
1 − p4 .
Problem 64 Let E be the event that the wife answers correctly, and let F be the
event that the husband answers correctly.
(a) If only one of them answers, then the probability of a correct
answer is P (E) = P (F ) = p.
(b) P (correct answer) = P (EF ) + 21 · 2 · p(1 − p) = p2 + p − p2 = p
Problem 66 Let Ei be the event that the i-th switch is on.
(a)
P (current flows from A to B)
= (P (E1 E2 ) + P (E3 E4 ) − P (E1 E2 E3 E4 )) P (E5 )
= (p1 p2 + p3 p4 − p1 p2 p3 p4 )p5
1
(b)
P (current flows from A to B)
= P (E1 E4 ∪ E1 E3 E5 ∪ E2 E5 ∪ E2 E3 E4 )
= p 1 p4 + p1 p3 p5 + p2 p5 + p2 p3 p 4
−p1 p3 p4 p5 − p1 p2 p4 p5 − p1 p2 p3 p4
−p1 p2 p3 p5 − p1 p2 p3 p4 p5 − p2 p3 p4 p5
+4p1 p2 p3 p4 p5 − p1 p2 p3 p4 p5
= p1 p4 + p 1 p3 p5 + p2 p5 + p2 p3 p4
−p1 p3 p4 p5 − p1 p2 p4 p5 − p1 p2 p3 p4
−p1 p2 p3 p5 − p2 p3 p4 p5 + 2p1 p2 p3 p4 p5
Chapter 4
Problem 1 Possible values of X: 0,2,4,-1,-2,1 Probabilities:
2
1
2
P {X = 0} = 14 =
91
2
4 2
8
P {X = 2} = 1 141 =
91
2
4
6
2
=
P {X = 4} = 14
91
2
8 2
16
P {X = −1} = 1 141 =
91
2
8
28
2
=
P {X = −2} = 14
91
2
8 4
32
P {X = 1} = 1 141 =
91
2
2
Problem 4
P {X
P {X
P {X
P {X
P {X
P {X
P {X
5
1
9!
1
= 1} =
=
10!
2
5 5
8!
5
=
= 2} = 1 1
18
10! 5
5
2! 1 7!
5
= 3} = 2
=
36
10!5
5
3!
6!
5
1
=
= 4} = 3
84
10!5
5
4!
5!
5
1
= 5} = 4
=
252
10!5
5
5!
4!
1
1
= 6} = 5
=
10!
252
= 7} = P {X = 8} = P {X = 9} = P {X = 10} = 0
Problem 5 The possible values are n, n − 2, n − 4, ..., −n + 4, −n + 2, −n.
Problem 13 Let X be the total dollar value of all sales. Then X can take the values
0, 500, 1000, 1500, 2000, and we have
P {X = 0} = 0.7 · 0.4 = 0.28
1
P {X = 500} = (0.3 · 0.4 + 0.7 · 0.6) = 0.27
2
1
1
P {X = 1000} = (0.3 · 0.4 + 0.7 · 0.6) + 0.3 · 0.6 = 0.315
2
4
1
P {X = 1500} = 2 0.3 · 0.6 = 0.09
4
1
P {X = 2000} = 0.3 · 0.6 = 0.045
4
3
Problem 14
P {X = 0} =
P {X = 1} =
P {X = 2} =
P {X = 3} =
P {X = 4} =
0!
2!
1!
3!
2!
4!
3!
5!
4!
5!
=
=
=
=
=
1
2
1
6
1
12
1
20
1
5
Problem 17 (a)
P {X = 1} = P {X ≤ 1} − P {X < 1} =
11 3
1
− =
12 4
6
11
1
P {X = 3} = 1 −
=
12
12
(b) P 21 < X < 32 = 58 − 81 = 12 .
P {X = 2} =
Problem 19
1
2
3 1
1
P {X = 1} = − =
5 2
10
4 3
1
P {X = 2} = − =
5 5
5
9
4
1
P {X = 3} =
− =
10 5
10
9
1
P {X = 3.5} = 1 −
=
10
10
P {X = 0} =
4
1 1
1
− =
2 4
4