Probability
Math 421
Selected Homework Solutions 3 – 6
September 24, 2015
Dr. Pendergrass
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Section 1.4, problem 4. Let Hi be the event that a heart is drawn on the ith draw, i = 1, 2, . . . . Also let
Ci and Ai be the events that a club and an ace are drawn on the ith draw respectively.
(a) P (H1 H2 ) = P (H1 ) · P (H2 |H1 ) =
13 12
1
·
= .
52 51
17
(b) P (H1 C2 ) = P (H1 ) · P (C2 |H1 ) =
13
13 13
·
=
.
52 51
204
(c) Note that H1 A2 = H1 A1 A2 ∪ H1 A01 A2 , and the union is disjoint. So
P (H1 A2 ) = P (H1 A1 A2 ) + P H1 A01 A2
= P (H1 A1 ) · P (A2 |H1 A1 ) + P H1 A01 · P A2 |H1 A01
1 3
12 4
=
·
+
·
52 52 52 52
437
=
18564
≈ 0.02354
Section 1.4, problem 6.
Let X be the number of kings in a thirteen-card hand. Then we have
P (X ≥ 3|X ≥ 2) =
P (X ≥ 3 and X ≥ 2)
P (X ≥ 3)
=
P (X ≥ 2)
P (X ≥ 2)
Now
P (X ≥ 2) = P (X = 2) + P (X = 3) + P (X = 4)
48
48
48
4
4
4
3 · 10
4 · 9
2 · 11
+
+
=
52
52
52
13
13
13
Similarly
P (X ≥ 3) = P (X = 3) + P (X = 4)
48
48
4
4
3 · 10
4 · 9
+
=
52
52
13
So
4
3
48
11
P (X ≥ 3|X ≥ 2) =
4
2
·
·
+
48
10 +
4
3 ·
13
4
4
48
10
·
+
48
9
4
4
·
=
48
9
913
≈ 0.1704
5359
Section 1.4, problem 10. Assume that Player 1 draws first, followed by Player 2. Consider the event Ei
that a winner is declared on the ith draw, where i ∈ {3, 4, 5, · · · 20}. (Clearly if i is odd Player 1 is the winner,
while if i is even Player 2 is the winner.) Let Ai be the event that exactly 2 WINS appear in draws 1 through
i − 1, and let Bi be the event that a WIN appears on draw i. Then Ei = Ai Bi , and so
P (Ei ) = P (Ai ) · P (Bi |Ai )
Page 1 of 4
Math 421
Selected Homework Solutions 3 – 6
Dr. Pendergrass
Clearly
1
1
=
20 − (i − 1)
21 − i
P (Bi |Ai ) =
In addition
17
2 2! i−3
20
i−1 (i −
3
i−1
2
P (Ai ) =
(i − 3)!
1)!
Simplifying, we get
3
P (Ei ) =
=
i−1
2
17
2 2! i−3
20
i−1 (i −
(i − 3)!
1)!
3(i − 1)(i − 2)
20 · 19 · 18
·
1
21 − i
for i = 3, 4, · · · 20.
(a) The probability that Player 1 wins on his second draw is
P (E3 ) =
3·2·1
≈ 0.0008772
20 · 19 · 18
(b) The probability that Player 2 wins on his second draw is
P (E4 ) =
3·3·2
≈ 0.002632
20 · 19 · 18
(c) Player 1 wins if and only if a winner is declared on trial 3, 5, 7, · · · , or 19. Since these events are mutually
exclusive,
P (Player 1 wins) = P (E3 ∪ E5 ∪ E7 ∪ · · · ∪ E19 )
= P (E3 ) + P (E5 ) + P (E7 ) + · · · P (E19 )
X
3(i − 1)(i − 2)
=
20 · 19 · 18
i∈{3,5,7,···19}
35
76
≈ 0.4605
=
(d) Since Player 2 wins with probability 1 − 35/76 ≈ 0.5395, I would prefer to draw second.
Section 1.4, problem 12. The sample space can be thought of as all the 10! arrangements of the 10
(distinguishable) milk bottles, two of which are sour. All arrangements are equally likely, so
2
2
1 1! 9!
P (sour in position 6) =
= .
10!
10
Section 1.4, problem 14. Note that this is equivalent to randomly drawing 6 cards sequentially from a
single well-shuffled deck with replacement. The sample-space is all 526 equally likely possible outcomes.
(a)
P (all 6 cards are different) =
52 · 51 · 50 · 49 · 48 · 47
≈ 0.7414
526
(b)
P (at least two cards match) = 1 − P (all 6 are different)
52 · 51 · 50 · 49 · 48 · 47
=1−
526
≈ 0.2586
Page 2 of 4
Math 421
Selected Homework Solutions 3 – 6
Dr. Pendergrass
Section 1.4, problem 16.
(a) The sample space can be thought of as all 18! orderings of 18 distinct chips, one of which is blue, the rest
of which are red. All orderings are equally likely. Let Ai = {student i wins} for i = 1, 2, · · · 18. Then we
have
1 · 17!
1
P (Ai ) = P (blue chip appears in position i) =
= .
18!
18
Therefore all students have the same chance of winning.
(b) The sample space can now be thought of as all 18! orderings of 18 distinct chips, two of which are blue,
the rest of which are red. All orderings are equally likely.
2
2
1 · 1! · 17!
P (Ai ) = P (blue chip appears in position i) =
= .
18!
18
Again, all students have an equal chance of winning.
Section 1.4, problem 18.
By the law of total probability
P (RII ) = P (RI ) P (RII |RI ) + P (WI ) P (RII |WI )
3 5 2 4
= · + ·
5 8 5 8
23
=
40
= 0.575
Section 1.4, problem 20.
(a) P (A1 ) =
30
100
(b) P (A3 B2 ) =
9
100
(c) P (A2 ∪ B3 ) =
60
100
(d) P (A1 |B2 ) =
11
41
(e) P (B1 |A3 ) =
13
29
Section 1.5, problem 8. Let A = {die A shows orange}, B = {die B shows orange}, and C = {die C shows orange}
Then the events A, B, and C are independent, and
P (exactly two dice show orange) = P ABC 0 + P AB 0 C + P A0 BC
= P (A) P (B) P C 0 + P (A) P B 0 P (C) + P A0 P (B) P (C)
1 2 3 1 4 3 5 2 3
= · · + · · + · ·
6 6 6 6 6 6 6 6 6
48
=
216
≈ 0.2222
Page 3 of 4
Math 421
Selected Homework Solutions 3 – 6
Dr. Pendergrass
Section 1.6, problem 12. Let D be the event that the selected person has the disease, and let T be the
event that the selected person tests positive for the disease. We are given
P (T |D) = 0.90
P T |D0 = 0.02
P (D) = 0.005
By Bayes’ Theorem
P (D) P (T |D)
P (D) P (T |D) + P (D0 ) P (T |D0 )
0.005 · 0.90
=
0.005 · 0.90 + 0.995 · 0.02
45
=
244
≈ 0.1844
P (D|T ) =
Section 2.1, problem 10
The probability mass function for X is
47 3
P (X = x) =
x
10−x
50
10
,
x = 0, 1, 2, 3.
(a)
3
1
P (X = 1) =
47
10−1
50
10
≈ 0.3980
(b)
3
0
P (X = 0 or X = 1) =
47
10−0
50
10
Page 4 of 4
3
1
+
47
10−1
50
10
≈ 0.9020