20. Royal Flush: 4
3.1 CONDITIONAL PROBABILITY
Chapter 3
C onditional P robability
and I ndependence
9.
40 2
65 6
105 8
1− 2
i=0
40 8− i
65 i
105 8
= 0.239.
10. P (α= i | β = 0) =
⎧⎪
⎪⎨
⎪⎪
⎩
1/19 if i= 0
2/19 if i= 1, 2, 3, . . . , 9 0 if i= 10, 11, 12, . . . , 18.
11. Let b∗gbmean that the oldest child of the family is a boy, the second oldest is a girl, the youngest is a boy, and the boy found in the family is the oldest child, with similar representations for other cases. The reduced sample space is
S=
ggb∗, gb∗g, b∗gg, b∗bg, bb∗g, gb∗b, gbb∗, bgb∗, b∗gb, b∗bb, bb∗b, bbb∗ . Note that the outcomes of the sample space are not equiprobable. We have that
P
{ggb∗}
= P
{gb∗g}
= P
{b∗gg}
= 1/7 P
{b∗bg}
= P
{bb∗g}
= 1/14 P
{gb∗b}
= P
{gbb∗}
= 1/14 P
{bgb∗}
= P
{b∗gb}
= 1/14 P
{b∗bb}
= P {bb∗b}
= P {bbb∗}
= 1/21.
The solutions to (a), (b), (c) are as follows.
(a) P
{bb∗g}
= 1/14;
(b) P
{bb∗g, gbb∗, bgb∗, bb∗b, bbb∗}
= 13/42;
(c) P
{b∗bg, bb∗g, gb∗b, gbb∗, bgb∗, b∗gb}
= 3/7.
12. P (A)= 1 implies that P (A ∪ B) = 1. Hence, by
P (A∪ B) = P (A) + P (B) − P (AB), we have that P (B)= P (AB). Therefore,
P (B | A) = P (AB)
P (A) =P (B)
1 = P (B).
Section 3.1 Conditional Probability 37 13. P (A| B) = P (AB)
b ,where
P (AB)= P (A) + P (B) − P (A ∪ B) ≥ P (A) + P (B) − 1 = a + b − 1.
14. (a) P (AB)≥ 0, P (B) > 0. Therefore, P (A | B) = P (AB) P (B) ≥ 0.
(b) P (S| B) = P (SB)
P (B) = P (B) P (B)= 1.
(c) P
∞
i=1
Ai B
=
P ∞
i=1Ai
B
P (B) = P ∞ i=1AiB
P (B)
= ∞
i=1
P (AiB)
P (B) =
∞ i=1
P (AiB) P (B) =
∞ i=1
P (Ai | B).
Note that P (∪∞i=1AiB) = ∞
i=1P (AiB), since mutual exclusiveness of Ai’s imply that of AiB’s; i.e., AiAj = ∅, i = j, implies that (AiB)(AjB)= ∅, i = j.
15. The given inequalities imply that P (EF )≥ P (GF ) and P (EFc)≥ P (GFc).Thus P (E)= P (EF ) + P (EFc)≥ P (GF ) + P (GFc)= P (G).
16. Reduce the sample space: Marlon chooses from six dramas and seven comedies two at random.
What is the probability that they are both comedies? The answer is
7 2
13 2
= 0.269.
17. Reduce the sample space: There are 21 crayons of which three are red. Seven of these crayons are selected at random and given to Marty. What is the probability that three of them are red?
The answer is
18 4
21 7
= 0.0263.
18. (a) The reduced sample space is S= {1, 3, 5, 7, 9, . . . , 9999}. There are 5000 elements in S. Since the set{5, 7, 9, 11, 13, 15, . . . , 9999} includes exactly 4998/3 = 1666 odd numbers that are divisible by three, the reduced sample space has 1667 odd numbers that are divisible by 3. So the answer is 1667/5000= 0.3334.
(b) Let O be the event that the number selected at random is odd. Let F be the event that it is divisible by 5 and T be the event that it is divisible by 3. The desired probability is calculated as follows.
P (FcTc| O) = 1 − P (F ∪ T | O) = 1 − P (F | O) − P (T | O) + P (F T | O)
= 1 −1000
5000−1667
5000 + 333
5000 = 0.5332.
19. Let A be the event that during this period he has hiked in Oregon Ridge Park at least once. Let Bbe the event that during this period he has hiked in this park at least twice. We have
P (B | A) = P (B) P (A), where
P (A)= 1 −510
610 = 0.838 and
P (B)= 1 − 510
610 −10× 59
610 = 0.515.
So the answer is 0.515/0.838= 0.615.
20. The numbers of 333 red and 583 blue chips are divisible by 3. Thus the reduced sample space has 333+ 583 = 916 points. Of these numbers, [1000/15] = 66 belong to red balls and are divisible by 5 and[1750/15] = 116 belong to blue balls and are divisible by 5. Thus the desired probability is 182/916= 0.199.
21. Reduce the sample space: There are two types of animals in a laboratory, 15 type I and 13 type II. Six animals are selected at random; what is the probability that at least two of them are Type II? The answer is
1−
15 6
+
13 1
15 5
28 6
= 0.883.
22. Reduce the sample space: 30 students of which 12 are French and nine are Korean are divided randomly into two classes of 15 each. What is the probability that one of them has exactly four French and exactly three Korean students? The solution to this problem is
12 4
9 3
9 8
30 15
15 15
= 0.00241.
23. This sounds puzzling because apparently the only deduction from the name “Mary” is that one of the children is a girl. But the crucial difference between this and Example 3.2 is reflected in the implicit assumption that both girls cannot be Mary. That is, the same name cannot be used for two children in the same family. In fact, any other identifying feature that cannot be shared by both girls would do the trick.
Section 3.2 Law of Multiplication 39 3.2 LAW OF MULTIPLICATION
1. Let G be the event that Susan is guilty. Let L be the event that Robert will lie. The probability that Robert will commit perjury is
P (GL)= P (G)P (L | G) = (0.65)(0.25) = 0.1625.
2. The answer is
11 14 ×10
13× 9 12× 8
11 × 7 10 ×6
9 = 0.15.
3. By the law of multiplication, the answer is 52
52 ×50 51 ×48
50×46 49×44
48 ×42
47 = 0.72.
4. (a) 8 20× 7
19× 6 18 × 5
17 = 0.0144;
(b) 8 20× 7
19×12 18 + 8
20 ×12 19× 7
18+12 20 × 8
19× 7 18+ 8
20 × 7 19 × 6
18 = 0.344.
5. (a) 6 11× 5
10×5 9 ×4
8×4 7×3
6×3 5 ×2
4×2 3×1
2 ×1
1 = 0.00216.
(b) 5 11× 4
10×3 9 ×2
8 ×1
7 = 0.00216.
6. 3 8× 5
10 × 5 13× 8
15+5 8× 3
11 × 8 13 × 5
16 = 0.0712.
7. Let Ai be the event that the ith person draws the “you lose” paper. Clearly, P (A1)= 1
200,
P (A2)= P (Ac1A2)= P (Ac1)P (A2| Ac1)= 199 200 · 1
199 = 1 200, P (A3)= P (Ac1Ac2A3)= P (Ac1)P (Ac2| Ac1)P (A3| Ac1Ac2)=199
200 ·198 199 · 1
198 = 1 200, and so on. Therefore, P (Ai)= 1/200 for 1 ≤ i ≤ 200. This means that it makes no difference if you draw first, last or anywhere in the middle. Here is Marilyn Vos Savant’s intuitive solution to this problem:
It makes no difference if you draw first, last, or anywhere in the middle. Look at it this way: Say the robbers make everyone draw at once. You’d agree that everyone has the same change of losing (one in 200), right? Taking turns just makes that same event happen in a slow and orderly fashion. Envision a raffle at a church with 200 people in attendance, each person buys a ticket. Some buy a ticket when they arrive, some during the event, and some just before the winner is drawn. It doesn’t matter. At the party the end result is this: all 200 guests draw a slip of paper, and, regardless of when they look at the slips, the result will be identical: one will lose.
You can’t alter your chances by looking at your slip before anyone else does, or waiting until everyone else has looked at theirs.
8. Let B be the event that a randomly selected person from the population at large has poor credit report. Let I be the event that the person selected at random will improve his or her credit rating within the next three years. We have
P (B | I) = P (BI )
P (I ) = P (I | B)P (B)
P (I ) = (0.30)(0.18)
0.75 = 0.072.
The desired probability is 1−0.072 = 0.928. Therefore, 92.8% of the people who will improve their credit records within the next three years are the ones with good credit ratings.
9. For 1 ≤ n ≤ 39, let Enbe the event that none of the first n− 1 cards is a heart or the ace of spades. Let Fn be the event that the nth card drawn is the ace of spades. Then the event of “no heart before the ace of spades” is39
n=1EnFn. Clearly,{EnFn, 1≤ n ≤ 39} forms a sequence of mutually exclusive events. Hence
P
39
n=1
EnFn =
39 n=1
P (EnFn)= 39
n=1
P (En)P (Fn| En)
= 39 n=1
38 n− 1
52 n− 1
× 1
53− n = 1 14,
a result which is not unexpected.
10. P (F )P (E| F ) =
13 3
39 6
52 9
×10
43 = 0.059.
11. By the law of multiplication, P (An)= 2
3×3 4×4
5 × · · · ×n+ 1 n+ 2 = 2
n+ 2.
Section 3.3 Law of Total Probability 41 Now since A1⊇ A2⊇ A3⊇ · · · ⊇ An⊇ An+1 ⊇ · · · , by Theorem 1.8,
P
∞
i=1
Ai
= lim
n→∞P (An)= 0.