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3.01 You have a bag filled with 20 red marbles and 20 blue marbles. You shake the bag so that the marbles are thoroughly mixed.
3.01 You have a bag filled with 20 red marbles and 20 blue marbles. You shake the bag so that the marbles are thoroughly mixed.
(a) You reach in the bag without looking, and draw out one marble. What is the probability that it is red?
(a) You reach in the bag without looking, and draw out one marble. What is the probability that it is red?
Answer: P = 20/40 = 1/2
Answer: P = 20/40 = 1/2
(b) What is the probability of drawing out two blue marbles in a row, assuming that you replace the first marble and remix the bag before drawing the second?
(b) What is the probability of drawing out two blue marbles in a row, assuming that you replace the first marble and remix the bag before drawing the second?
Answer: Use the product rule, P = 1/2 x 1/2 = 1/4
Answer: Use the product rule, P = 1/2 x 1/2 = 1/4
(c) The answer to part (b) would be different if you did not replace the first marble. Why is this true? Consider how the the probability for the second marble changes depending on the color of the first marble drawn from the bag.
(c) The answer to part (b) would be different if you did not replace the first marble. Why is this true? Consider how the the probability for the second marble changes depending on the color of the first marble drawn from the bag.
Answer: Replacing the first marble resets the system so that it is in the same state as when the first marble was drawn. This makes the probability of drawing a blue marble on the second draw that same as for the first draw. If the first marble is not replaced, there are no longer an equal number of blue and red marbles in the bag. The probably of drawing a blue marble has changed. And how it has changed depends on what color the first marble was.
Answer: Replacing the first marble resets the system so that it is in the same state as when the first marble was drawn. This makes the probability of drawing a blue marble on the second draw that same as for the first draw. If the first marble is not replaced, there are no longer an equal number of blue and red marbles in the bag. The probably of drawing a blue marble has changed. And how it has changed depends on what color the first marble was.
(d) Given your answer to part (c), why can't you use the product rule to calculate the probability of drawing two blue marbles in a row, if the first one is not replaced?
(d) Given your answer to part (c), why can't you use the product rule to calculate the probability of drawing two blue marbles in a row, if the first one is not replaced?
Answer: The two events would no longer be independent because taking out the first marble changes the probability of drawing a blue the second time. The product rule is only valid when the events begin considered are independent events.
Answer: The two events would no longer be independent because taking out the first marble changes the probability of drawing a blue the second time. The product rule is only valid when the events begin considered are independent events.
3.02 You are rolling a standard die. Calculate the probability of each of the following events.
3.02 You are rolling a standard die. Calculate the probability of each of the following events.
(a) Rolling a 4.
(a) Rolling a 4.
Answer: P = 1/6
Answer: P = 1/6
(b) Rolling a 5 or a 6.
(b) Rolling a 5 or a 6.
Answer: P = 1/6 + 1/6 = 2/6 = 1/3
Answer: P = 1/6 + 1/6 = 2/6 = 1/3
(c) Rolling a 3 or an odd number.
(c) Rolling a 3 or an odd number.
Answer: P = 3/6 = 1/2 (3 is also an odd number, so there are only 3 outcomes out of the six possible that fit the desired event)
Answer: P = 3/6 = 1/2 (3 is also an odd number, so there are only 3 outcomes out of the six possible that fit the desired event)
(d) Which probability rule can you use to calculate the answer to part (b), and why can you not use the same rule to calculate the answer to part (c)?
(d) Which probability rule can you use to calculate the answer to part (b), and why can you not use the same rule to calculate the answer to part (c)?
Answer: You can use the sum rule for part (b). You add the probabilities of the two individual events to get the answer. Rolling a 5 and rolling a 6 are mutually exclusive events. This rule will not work for part (c) even though there are two events connected by "or" because the events are not mutually exclusive. The occurrence of one even does not exclude the occurrence of the other. Rolling a three and rolling an odd number can happen at the same time.
Answer: You can use the sum rule for part (b). You add the probabilities of the two individual events to get the answer. Rolling a 5 and rolling a 6 are mutually exclusive events. This rule will not work for part (c) even though there are two events connected by "or" because the events are not mutually exclusive. The occurrence of one even does not exclude the occurrence of the other. Rolling a three and rolling an odd number can happen at the same time.
3.03 On the average, about one child in every 10,000 births in the U.S. has a genetic disease called phenylketonuria (PKU).
3.03 On the average, about one child in every 10,000 births in the U.S. has a genetic disease called phenylketonuria (PKU).
(a) What is the probability that the next child born in a particular Mississippi hospital will have PKU? You can assume that the probability of the disease in Mississippi is the same as in the US.
(a) What is the probability that the next child born in a particular Mississippi hospital will have PKU? You can assume that the probability of the disease in Mississippi is the same as in the US.
Answer: P = 1/10,000
Answer: P = 1/10,000
(b) Suppose a PKU child has just been born in this hospital. What is the probability that the next child born in the same hospital will have PKU?
(b) Suppose a PKU child has just been born in this hospital. What is the probability that the next child born in the same hospital will have PKU?
Answer: P = 1/10,000 (the fact the one child has already been born with PKU does not change the probability for the next one)
Answer: P = 1/10,000 (the fact the one child has already been born with PKU does not change the probability for the next one)
(c) What is the probability that two children born in a row will have PKU?
(c) What is the probability that two children born in a row will have PKU?
Answer: P = (1/10,000)(1/10,000) = 1 x 10-8
Answer: P = (1/10,000)(1/10,000) = 1 x 10-8
(d) Question (b) and (c) seem the same, but they are not. Why is the answer to part (c) so different from the answer to part (b)?
(d) Question (b) and (c) seem the same, but they are not. Why is the answer to part (c) so different from the answer to part (b)?
Answer: In part (c) we are asked for the probability that the first child and the next child will both have PKU. We use the product rule. The chance of two rare events happening at the same time is exceedingly rare (assuming that they are independent events). Note that in part (b) the first child has already been born; in part (c), neither of the two events have happened yet.
Answer: In part (c) we are asked for the probability that the first child and the next child will both have PKU. We use the product rule. The chance of two rare events happening at the same time is exceedingly rare (assuming that they are independent events). Note that in part (b) the first child has already been born; in part (c), neither of the two events have happened yet.
3.04 In peas, axillary flowers (A) is dominant to terminal flowers (a), and colored flowers (R) is dominant to white (r). A true-breeding plant with white, terminal flowers is crossed with a true-breeding plant having colored, axillary flowers. The F1 plants are then allowed to self-pollinate.
3.04 In peas, axillary flowers (A) is dominant to terminal flowers (a), and colored flowers (R) is dominant to white (r). A true-breeding plant with white, terminal flowers is crossed with a true-breeding plant having colored, axillary flowers. The F1 plants are then allowed to self-pollinate.
(a) Use the forked-line method to calculate the probability of each possible F2 genotype.
(a) Use the forked-line method to calculate the probability of each possible F2 genotype.
Answer: P generation: aa rr x RR AA : F1 generation: RrAa (self these to get F2)
Answer: P generation: aa rr x RR AA : F1 generation: RrAa (self these to get F2)
(b) Use the forked-line method to calculate the probability of each possible F2 phenotype.
(b) Use the forked-line method to calculate the probability of each possible F2 phenotype.
Answer:
Answer:
3.05 In peas, Green pods (G) is dominant over yellow pods (g), and inflated pods (I) is dominant over constricted pods (i). The following cross is made: Gg Ii x Gg ii. Without constructing a Punnett square, calculate the probability of each of the following.
3.05 In peas, Green pods (G) is dominant over yellow pods (g), and inflated pods (I) is dominant over constricted pods (i). The following cross is made: Gg Ii x Gg ii. Without constructing a Punnett square, calculate the probability of each of the following.
(a) What is the probability that an offspring will be homozygous dominant for the pod color gene
(a) What is the probability that an offspring will be homozygous dominant for the pod color gene
Answer: P(GG) = 1/4
Answer: P(GG) = 1/4
(b) What is the probability that an offspring will be homozygous dominant for the pod shape gene?
(b) What is the probability that an offspring will be homozygous dominant for the pod shape gene?
Answer: P(II) = 0. Since one parent is ii, there is no way for offspring to have two I alleles.
Answer: P(II) = 0. Since one parent is ii, there is no way for offspring to have two I alleles.
(c) What is the probability that an offspring will have green inflated pods or yellow constricted pods?
(c) What is the probability that an offspring will have green inflated pods or yellow constricted pods?
Answer: P(G- I- or gg ii) = (3/4)(1/2) + (1/4)(1/2) = 3/8 + 1/8 = 1/2 (product and sum rules)
Answer: P(G- I- or gg ii) = (3/4)(1/2) + (1/4)(1/2) = 3/8 + 1/8 = 1/2 (product and sum rules)
(d) What proportion of the progeny would you expect to be heterozygous for both genes?
(d) What proportion of the progeny would you expect to be heterozygous for both genes?
Answer: P(Gg Ii) = P(Gg) x P (Ii) = 1/2 x 1/2 = 1/4 (product rule)
Answer: P(Gg Ii) = P(Gg) x P (Ii) = 1/2 x 1/2 = 1/4 (product rule)
3.06 In a certain trihybrid cross, both parents have genotype Aa Bb Cc. Calculate the following without constructing a Punnett square or drawing a forked-line diagram:
3.06 In a certain trihybrid cross, both parents have genotype Aa Bb Cc. Calculate the following without constructing a Punnett square or drawing a forked-line diagram:
(a) What is the probability that this cross will produce an offspring that is shows the recessive phenotype for all three traits?
(a) What is the probability that this cross will produce an offspring that is shows the recessive phenotype for all three traits?
Answer: P(aa bb cc) = P (aa) x P(bb) x P(cc) = (1/4)(1/4)(1/4) = 1/64
Answer: P(aa bb cc) = P (aa) x P(bb) x P(cc) = (1/4)(1/4)(1/4) = 1/64
(b) What is the probability of producing an offspring with the phenotype A- bb C-? (The notation "A-" means that the offspring has the dominant phenotype but might be either homozygous or heterozygous. That is, it has at least one dominant allele.)
(b) What is the probability of producing an offspring with the phenotype A- bb C-? (The notation "A-" means that the offspring has the dominant phenotype but might be either homozygous or heterozygous. That is, it has at least one dominant allele.)
Answer: P(A- bb C-) = P (A-) x P(bb) x P(C-) = (3/4)(1/4)(3/4) = 9/64
Answer: P(A- bb C-) = P (A-) x P(bb) x P(C-) = (3/4)(1/4)(3/4) = 9/64
(c) If you had worked these problems using a Punnett square, how many boxes would it have to contain?
(c) If you had worked these problems using a Punnett square, how many boxes would it have to contain?
Answer: Because both parents are heterozygous for three genes, each one would be produce 8 different gametes (23 = 8). Therefore the Punnett square would have to have 8 columns for the female gametes and 8 rows for the male gametes - a total of 64 (8 x 8) boxes. If you used a Punnett square to determine the answers to (a) and (b) above, you would have had to construct that entire 64-box table first, then count boxes. Using the product rule for these types of questions saves a tremendous amount of time and work.
Answer: Because both parents are heterozygous for three genes, each one would be produce 8 different gametes (23 = 8). Therefore the Punnett square would have to have 8 columns for the female gametes and 8 rows for the male gametes - a total of 64 (8 x 8) boxes. If you used a Punnett square to determine the answers to (a) and (b) above, you would have had to construct that entire 64-box table first, then count boxes. Using the product rule for these types of questions saves a tremendous amount of time and work.
3.07 Three traits in cats are being studied. Long hair (L) is dominant over short (l). White-spotted (S) is dominant over solid colored(s), and dilute color (D) is dominant over non-dilute (d). Two cats with genotypes ll Ss dd and Ll Ss Dd are crossed. Calculate each the following without constructing a Punnett square or drawing a forked-line diagram:
3.07 Three traits in cats are being studied. Long hair (L) is dominant over short (l). White-spotted (S) is dominant over solid colored(s), and dilute color (D) is dominant over non-dilute (d). Two cats with genotypes ll Ss dd and Ll Ss Dd are crossed. Calculate each the following without constructing a Punnett square or drawing a forked-line diagram:
(a) What is the probability of this cross producing an offspring with genotype ll ss dd?
(a) What is the probability of this cross producing an offspring with genotype ll ss dd?
Answer: P(ll ss dd) = P (ll) x P(ss) x P(dd) = (1/2)(1/4)(1/2) = 1/16
Answer: P(ll ss dd) = P (ll) x P(ss) x P(dd) = (1/2)(1/4)(1/2) = 1/16
(b) What is the probability of this cross producing an offspring with the phenotype of short-hair, white-spotted, non-dilute?
(b) What is the probability of this cross producing an offspring with the phenotype of short-hair, white-spotted, non-dilute?
Answer: P(ll S- dd) = P (ll) x P(S-) x P(dd) = (1/2)(3/4)(1/2) = 3/16
Answer: P(ll S- dd) = P (ll) x P(S-) x P(dd) = (1/2)(3/4)(1/2) = 3/16
3.08 Two pure-breeding rats with contrasting phenotypes for two different traits were crossed, AA bb x aa BB, then the F1 rats were crossed among themselves. In the F2 progeny, 50 rats displayed both dominant phenotypes (A- B-). Of these 50 rats, how many you predict to have the double heterozygous genotype (Aa Ba)?
3.08 Two pure-breeding rats with contrasting phenotypes for two different traits were crossed, AA bb x aa BB, then the F1 rats were crossed among themselves. In the F2 progeny, 50 rats displayed both dominant phenotypes (A- B-). Of these 50 rats, how many you predict to have the double heterozygous genotype (Aa Ba)?
Answer: The F1 rats have genotype Aa Ba, so we are crossing two of these. From the Punnett square for this cross, it can be seen that out of the 16 possible outcomes, nine (shaded boxes) would be expected to show both dominant traits. Of those nine, four (bolb type) are double heterozygous. Therefore 4/9 of the double dominant offspring would be expected to be double heterozygous. 50 x 4/9 = 22.22. Approximately 22 of the 50 rats would be heterozygous for both genes.
Answer: The F1 rats have genotype Aa Ba, so we are crossing two of these. From the Punnett square for this cross, it can be seen that out of the 16 possible outcomes, nine (shaded boxes) would be expected to show both dominant traits. Of those nine, four (bolb type) are double heterozygous. Therefore 4/9 of the double dominant offspring would be expected to be double heterozygous. 50 x 4/9 = 22.22. Approximately 22 of the 50 rats would be heterozygous for both genes.
3.09 Assuming that the probability of having a girl on any given birth is 0.5, consider calculating the probability that a family of four will consist of three girls and one boy.
3.09 Assuming that the probability of having a girl on any given birth is 0.5, consider calculating the probability that a family of four will consist of three girls and one boy.
(a) First, calculate the answer to this question by writing out all possible birth orders and determining what fraction of the families fit the required family makeup.
(a) First, calculate the answer to this question by writing out all possible birth orders and determining what fraction of the families fit the required family makeup.
Answer: The possible birth orders: GGGG, GGGB, GGBG, GGBB, GBGG, GBGB, GBBG, GBBB, BGGG, BGGB, BGBG, BGBB, BBGG, BBGB, BBBG, BBBB. Of the 16 possible orders, 4 contain 3 girls and 1 boy, so the probability is 4/16 = 1/4 = 0.25.
Answer: The possible birth orders: GGGG, GGGB, GGBG, GGBB, GBGG, GBGB, GBBG, GBBB, BGGG, BGGB, BGBG, BGBB, BBGG, BBGB, BBBG, BBBB. Of the 16 possible orders, 4 contain 3 girls and 1 boy, so the probability is 4/16 = 1/4 = 0.25.
(b) Now calculate the answer again using the binomial formula. Answer:
(b) Now calculate the answer again using the binomial formula. Answer:
(c) It turns out that the true probability of having a boy or girl on any one birth at not quite equal. The actual proportion of girls at birth is about 0.49, rather than 0.50. Given this more accurate value, recalculate the probability of having a family of three girls and one boy? Answer:
(c) It turns out that the true probability of having a boy or girl on any one birth at not quite equal. The actual proportion of girls at birth is about 0.49, rather than 0.50. Given this more accurate value, recalculate the probability of having a family of three girls and one boy? Answer:
(d) Could you calculate the answer to (c) by counting birth orders as you did in part (a)? Why or why not?
(d) Could you calculate the answer to (c) by counting birth orders as you did in part (a)? Why or why not?
Answer: No, because each birth order would no longer have the same probability.
Answer: No, because each birth order would no longer have the same probability.
3.10 Assuming the probability of having a boy is equal to the probability of having a girl, what is the probability that a family with 5 children will consist of:
3.10 Assuming the probability of having a boy is equal to the probability of having a girl, what is the probability that a family with 5 children will consist of:
For several of the families, it is convenient to just count how many of the possible birth orders fit the situation. Total number of orders = 25 = 32.
For several of the families, it is convenient to just count how many of the possible birth orders fit the situation. Total number of orders = 25 = 32.
(a) 3 daughters and 2 sons?
(a) 3 daughters and 2 sons?
Answer: binomial formula 
Answer: binomial formula 
(b) alternating sexes?
(b) alternating sexes?
Answer: There are 2 ways to get alternate sexes, either starting with a boy or starting with a girl. The probability of any one birth order is 1/32 because their are 32 possible birth orders. Therefore: P = 1/32 + 1/32 = 1/16 = 0.0625
Answer: There are 2 ways to get alternate sexes, either starting with a boy or starting with a girl. The probability of any one birth order is 1/32 because their are 32 possible birth orders. Therefore: P = 1/32 + 1/32 = 1/16 = 0.0625
(c) alternating sexes, starting with a son?
(c) alternating sexes, starting with a son?
Answer: There is only one birth order for alternate sexes starting with a boy, so P = 1/32.
Answer: There is only one birth order for alternate sexes starting with a boy, so P = 1/32.
(d) all daughters?
(d) all daughters?
Answer: There is only one birth order that gives all girls, so P = 1/32.
Answer: There is only one birth order that gives all girls, so P = 1/32.
(e) all the same sex?
(e) all the same sex?
Answer: There are two ways to get all the same sex, all girls or all boys, so P = 1/32 + 1/32= 1/16.
Answer: There are two ways to get all the same sex, all girls or all boys, so P = 1/32 + 1/32= 1/16.
(f) at least 4 daughters?
(f) at least 4 daughters?
Answer: There are two ways to get at least 4 daughters, either 4 girls and 1 boy or all 5 girls. This calls for a combination of the binomial formula and the sum rule:
Answer: There are two ways to get at least 4 daughters, either 4 girls and 1 boy or all 5 girls. This calls for a combination of the binomial formula and the sum rule:
P = P(4 girls, 1 boy) + P(all girls) 
P = P(4 girls, 1 boy) + P(all girls) 
(g) a daughter as the eldest child and a son as the youngest?
(g) a daughter as the eldest child and a son as the youngest?
Answer: Use the product rule, P = (1/2)(1/2) = 1/4
Answer: Use the product rule, P = (1/2)(1/2) = 1/4
3.11 In Drosophila, vestigial (crinkled) wings is recessive to normal wings ("wild type"). A true-breeding wild type female is crossed with a vestigial male. Two F1 flies are then crossed to obtain the F2 generation. If the F2 generation has 8 flies, what is the probability of that 5 will be wild type and 3 will be vestigial?
3.11 In Drosophila, vestigial (crinkled) wings is recessive to normal wings ("wild type"). A true-breeding wild type female is crossed with a vestigial male. Two F1 flies are then crossed to obtain the F2 generation. If the F2 generation has 8 flies, what is the probability of that 5 will be wild type and 3 will be vestigial?
Answer: Using the binomial formula: 
Answer: Using the binomial formula: 
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