Technique 2: We have that
3.6 APPLICATIONS OF PROBABILITY TO GENETICS
1. Clearly, Kim and Dan both have genotype OO. With a genotype other than AO for John, it is impossible for Dan to have blood type O. Therefore, the probability is 1 that John’s genotype is AO.
2. The answer is
k 2
+ k = k(k+ 1)
2 .
3. The genotype of the parent with wrinkled shape is necessarily rr. The genotype of the other parent is either Rr or RR. But, RR will never produce wrinkled offspring. So it must be Rr.
Therefore, the parents are rr and Rr.
4. Let A represent the dominant allele for free earlobes and a represent the recessive allele for attached earlobes. Let B represent the dominant allele for freckles and b represent the recessive allele for no freckles. Since Dan has attached earlobes and no freckles, Kim and John both must be AaBb. This implies that Kim and John’s next child is AA with probability 1/4, Aa
Section 3.6 Applications of Probability to Genetics 57 with probability 1/2, and aa with probability 1/4. Therefore, the next child has free earlobes with probability 3/4. Similarly, the next child is BB with probability 1/4, Bb with probability 1/2, and bb with probability 1/4. Hence he or she will have no freckles with probability 1/4.
By independence, the desired probability is (3/4)(1/4)= 3/16.
5. If the genes are not linked, 25% of the offspring are expected to be BbV v, 25% are expected to be bbvv, 25% are expected to be Bbvv, and 25% are expected to be bbV v. The observed data shows that the genes are linked.
6. Clearly, John’s genotype is either Dd or dd. Let E be the event that it is dd. Then Ec is the event that John’s genotype is Dd. Let F be the event that Dan is deaf. That is, his genotype is dd. We use Bayes’ theorem to calculate the desired probability.
P (E| F ) = P (F | E)P (E)
P (F | E)P (E) + P (F | Ec)P (Ec)
= 1· (0.01)
1· (0.01) + (1/2)(0.99) = 0.0198.
Therefore, the probability is 0.0198 that John is also deaf.
7. A person who has cystic fibrosis carries two mutant alleles. Applying the Hardy-Weinberg law, we have that q2 = 0.0529, or q = 0.23. Therefore, p = 0.77. Since q2+ 2pq = 1− p2 = 0.4071, the percentage of the people who carry at least one mutant allele of the disease is 40.71%.
8. Dan inherits all of his sex-linked genes from his mother. Therefore, John being normal has no effect on whether or not Dan has hemophilia or not. Let E be the event that Kim is H h. Then Ecis the event that Kim is H H . Let F be the event that Dan has hemophilia. By the law of total probability,
P (F )= P (F | E)P (E) + P (F | Ec)P (Ec)
= (1/2)
2(0.98)(0.02)
+ 0 · (0.98)(0.98) = 0.0196.
9. Dan has inherited all of his sex-linked genes from his mother. Let E1be the event that Kim is CC, E2be the event that she is Cc, and E3be the event that she is cc. Let F be the event that Dan is color-blind. By Bayes’ formula, the desired probability is
P (E3| F ) = P (F | E3)P (E3)
P (F | E1)P (E1)+ P (F | E2)P (E2)+ P (F | E3)P (E3)
= 1· (0.17)(0.17)
0· (0.83)(0.83) + (1/2)
2(0.83)(0.17)
+ 1 · (0.17)(0.17) = 0.17.
10. Since Ann is hh and John is hemophiliac, Kim is either H h or hh. Let E be the event that she is H h. Then Ecis the event that she is hh. Let F be the event that Ann has hemophilia. By
Bayes’ formula, the desired probability is
P (E| F ) = P (F | E)P (E)
P (F | E)P (E) + P (F | Ec)P (Ec)
= (1/2)
2(0.98)(0.02) (1/2)
2(0.98)(0.02)
+ 1 · (0.02)(0.02) = 0.98.
11. Clearly, both parents of Mr. J must be Cc. Since Mr. J has survived to adulthood, he is not cc.
Therefore, he is either CC or Cc. We have
P (he is CC| he is CC or Cc) = P (he is CC)
P (he is CC or Cc) =1/4 3/4 = 1
3. P (he is Cc| he is CC or Cc) = 2
3.
Mr. J’s wife is either CC with probability 1− p or Cc with probability p. Let E be the event that Mr. J is Cc, F be the event that his wife is Cc, and H be the event that their next child is cc. The desired probability is
P (H )= P (HEF ) = P (H | EF )P (EF )
= P (H | EF )P (E)P (F ) = 1 4· 2
3· p = p 6.
12. Let E1be the event that both parents are of genotype AA, let E2be the event that one parent is of genotype Aa and the other of genotype AA, and let E3be the event that both parents are of genotype Aa. Let F be the event that the man is of genotype AA. By Bayes’ formula,
P (E1| F ) = P (F | E1)P (E1)
P (F | E1)P (E1)+ P (F | E2)P (E2)+ P (F | E3)P (E3)
= 1· p4
1· p4+ (1/2) · 4p3q+ (1/4) · 4p2q2 = p2
(p+ q)2 = p2.
Similarly, P (E2| F ) = 2pq and P (E3| F ) = q2. Let B be the event that the brother is AA.
We have
P (B | F ) = P (B | FE1)P (E1| F ) + P (B | FE2)P (E2| F ) + P (B | FE3)P (E3| F )
= P (B | E1)P (E1| F ) + P (B | E2)P (E2| F ) + P (B | E3)P (E3| F )
= 1 · p2+1
2 · 2pq +1
4· q2= (2p+ q)2
4 =(1+ p)2
4 .
Chapter 3 Review Problems 59 REVIEW PROBLEMS FOR CHAPTER 3
1. 12 30· 13
30 +13 30 · 12
30 = 26
75 = 0.347.
2. 1− (0.97)6= 0.167.
3. (0.48)(0.30)+ (0.67)(0.53) + (0.89)(0.17) = 0.65.
4. (0.5)(0.05)+ (0.7)(0.02) + (0.8)(0.035) = 0.067.
5. (a) (0.95)(0.97)(0.85)= 0.783; (b) 1 − (0.05)(0.03)(0.05) = 0.999775;
(c) 1− (0.95)(0.97)(0.85) = 0.217; (d) (0.05)(0.03)(0.15) = 0.000225.
6. 103/132= 0.780.
7. (0.08)(0.20)
(0.2)(0.3)+ (0.25)(0.5) + (0.08)(0.20) = 0.0796.
8. 1−
26 6
39 6
= 0.929.
9. 1/6.
10.
1−5 6
10
− 105 6
91 6
1−5 6
10 = 0.615.
11.
2 7· 4
7 2 7· 4
7+5 7· 3
7
= 8
23 = 0.35.
12. Let A be the event of “head on the coin.” Let B be the event of “tail on the coin and 1 or 2 on the die.” Then A and B are mutually exclusive, and by the result of Exercise 36 of Section 3.5, the answer is 1/2
(1/2)+ (1/6) =3 4.
13. The probability that the number of 1’s minus the number of 2’s will be 3 is P (four 1’s and one 2)+ P (three 1’s and no 2’s)
=
6 4
1 6
4 2 1
1 6
4 6
+
6 3
1 6
34 6
3
= 0.03.
14. The probability that the first urn was selected in the first place is 20
45 · 1 2 20
45· 1 2 +10
25· 1 2
= 10 19. The desired probability is
20 45· 10
19+10 25 · 9
19 ≈ 0.42.
15. Let B be the event that the ball removed from the third urn is blue. Let BR be the event that the ball drawn from the first urn is blue and the ball drawn from the second urn is red. Define BB, RB, and RR similarly. We have that
P (B)= P (B | BB)P (BB) + P (B | RB)P (RB) + P (B | RR)P (RR) + P (B | BR)P (BR)
= 4 14· 1
10 5 6+ 5
14· 9 10
5 6+ 6
14· 9 10
1 6+ 5
14· 1 10
1 6 = 38
105 = 0.36.
16. Let E be the event that Lorna guesses correctly. Let R be the event that a red hat is placed on Lorna’s head, and B be the event that a blue hat is placed on her head. By the law of total probability,
P (E)= P (E | R)P (R) + P (E | B)P (B)
= α ·1
2 + (1 − α) ·1 2 = 1
2
This shows that Lorna’s chances are 50% to guess correctly no matter what the value of α is.
This should be intuitively clear.
17. Let F be the event that the child is found; E be the event that he is lost in the east wing, and W be the event that he is lost in the west wing. We have
P (F )= P (F | E)P (E) + P (F | W)P (W)
=
1− (0.6)3
(0.75)+
1− (0.6)2
(0.25)= 0.748.
18. The answer is that it is the same either way. Let W be the event that they win one of the nights to themselves. Let F be the event that they win Friday night to themselves. Then
P (W )= P (W | F )P (F ) + P (W | Fc)P (Fc)= 1 ·1 3+1
2· 2 3 =2
3. 19. Let A be the event that Kevin is prepared. We have that
P (R| BcSc)= P (RBcSc)
P (BcSc) =P (RBcSc | A)P (A) + P (RBcSc| Ac)P (Ac) P (BcSc | A)P (A) + P (BcSc | Ac)P (Ac)
= (0.85)(0.15)2(0.85)+ (0.20)(0.80)2(0.15)
(0.15)2(0.85)+ (0.80)2(0.15) = 0.308.
Chapter 3 Review Problems 61 Note that
P (R)= P (R | A)P (A) + P (R | Ac)P (Ac)= (0.85)(0.85) + (0.20)(0.15) = 0.7525.
Since P (R| BcSc)= P (R), the events R, B, and S are not independent. However, it must be clear that R, B, and S are conditionally independent given that Kevin is prepared and they are conditionally independent given that Kevin is unprepared. To explain this, suppose that we are given that, for example, Smith and Brown both failed a student. This information will increase the probability that the student was unprepared. Therefore, it increases the probability that Rose will also fails the student. However, if we know that the student was unprepared, the knowledge that Smith and Brown failed the student does not affect the probability that Rose will also fail the student.
20. (a) Let A be the event that Adam has at least one king; B be the event that he has at least two kings. We have
P (B | A) = P (AB)
P (A) = P (Adam has at least two kings) P (Adam has at least one king)
= 1−
48 13
52 13
−
48 12
4 1
52 13
1−
48 13
52 13
= 0.3696.
(b) Let A be the event that Adam has the king of diamonds. Let B be the event that he has the king of diamonds and at least one other king. Then
P (B | A) = P (BA) P (A) =
48 11
3 1
+
48 10
3 2
+
48 9
3 3
52 13
51 12
52 13
= 0.5612.
Knowing that Adam has the king of diamonds reduces the sample space to a size considerably smaller than the case in which we are given that he has a king. This is why the answer to
part (b) is larger than the answer to part (a). If one is not convinced of this, he or she should solve the problem in a simpler case. For example, a case in which there are four cards, say, king of diamonds, king of hearts, jack of clubs, and eight of spade. If two cards are drawn, the reduced sample space in the case Adam announces that he has a king is
{KdKh, KdJc, Kd8s, KhJc, Kh8s},
while the reduced sample space in the case Adam announces that he has the king of diamonds is
{KdKh, KdJc, Kd8s}.
In the first case, the probability of more kings is 1/5; in the second case the probability of more kings is 1/3.