Statistics for laboratory scientists

Solutions for the homework problems for lecture 6

1. 1. Suppose that A and B are mutually exclusive.

      1. Pr(A or B) = Pr(A) + Pr(B)

      2. Pr(A and B) = 0, since they can't both happen.

   2. Suppose that A and B are independent.

      1. Pr(A and B) = Pr(A) × Pr(B)

      2. Pr(A or B) = Pr(A) + Pr(B) - Pr(A) × Pr(B)

   3. Suppose that A and B are both mutually exclusive and independent.

      Since A and B are independent, Pr(A and B) = Pr(A) × Pr(B). But since A and B are mutually exclusive, Pr(A and B) = 0. Thus Pr(A) × Pr(B) = 0. And so either Pr(A) = 0 or Pr(B) = 0 or both. In other words, either A or B (or both) cannot happen!

      The point: generally when we talk about independent events, they are not mutually exclusive, and vice versa.

2. Recall that we pick a woman at random from the population. Let U = {She is unaffected}, C = {She is a carrier}, A = {her brother is affected and her parents are both unaffected}, B = {her first child is affected}, D = {at least one of her five children is affected}. Note that by "carrier," I mean "has genotype +d."

   Calculate the following.

   1. Pr(C | U) = 2p(1-p) / [2p(1-p) + (1-p)²] = ... = 2p / (1 + p) = approx 2.0%.

   2. Pr(C | U and A) = 2/3 = approx 67%, since her parents must both be carriers.

   3. Pr(B | U and A) = Pr(C | U and A) × Pr(B | C and U and A) = (2/3) × [(1/2) × p] = p/3 = approx 3/1000.

      Note that Pr(B | C and U and A) = Pr(B | C). We're assuming here that the woman is mating at random (with respect to genotype at this gene). Thus the allele contributed by the father has probability p of being the disease allele.

   4. Let E = {none of the five children is affected}, F₀ = {father has genotype ++}, F₁ = {father has genotype +d}, and F₂ = {father has genotype ++}.

      Note that Pr(E | C) = Pr(F₀) × Pr(E | C and F₀) + Pr(F₁) × Pr(E | C and F₁) + Pr(F₂) × Pr(E | C and F₂)
      = (1-p)² × 1 + 2 p (1-p) × (3/4)⁵ + p² × (1/2)⁵ = approx 98.5%.

      Thus, Pr(D | U and A) = Pr(C | U and A) × Pr(D | C and U and A)
      = Pr(C | U and A) × [1 - Pr(E | C)]
      = approx 1%.

3. 1. Pr(G₁) = Pr(A) × Pr(G₁ | A) + Pr(B) × Pr(G₁ | B) = (1/3) × (1/2) + (2/3) × (1/5) = 3/10 = 30%

   2. Pr(G₂ | A and G₁) = 4/9

   3. Pr(G₂ | A) = 1/2

   4. Pr(G₂ | B and G₁) = 1/9

   5. Pr(G₂ | G₁) = Pr(G₂ and G₁) / Pr(G₁)

      Note that Pr(G₂ and G₁) = Pr(A) × Pr(G₁ and G₂ | A) + Pr(B) × Pr(G₁ and G₂ | B) = (1/3) × (5/10) × (4/9) + (2/3) × (2/10) × (1/9) = 4/45 = approx 8.9%

      We found that Pr(G₁) = 3/10 (part a).

      Thus, Pr(G₂ | G₁) = (4/45) / (3/10) = 8/27 = approx 30%

   6. Pr(exactly one green ball) = Pr(G₁ and not G₂) + Pr(G₂ and not G₁) = 2 × Pr(G₁ and not G₂) = 2 × [Pr(A) × Pr(G₁ and not G₂) + Pr(B) × Pr(G₁ and not G₂)] = 2 × [(1/3) × (5/10) × (5/9) + (2/3) × (2/10) × (8/9)] = 114/270 = approx 42%.

   7. Note that Pr(at least one green ball | A) = 1 - Pr(no green balls | A) = 1 - (1/2) × (4/9) = 7/9 = approx 78%. Pr(at least one green ball | B) = 1 - (4/5) × (7/9) = 17/45 = approx 38%.

      Thus Pr(at least one green ball) = (1/3) × (7/9) + (2/3) × (17/45) = approx 51%.

      And so Pr(A | at least one green ball) = Pr (A) × Pr(at least one green ball | A) / Pr(at least one green ball) = (1/3) × (7/9) / [(1/3) × (7/9) + (2/3) × (17/45)] = approx 51%.

4. Let F = {the champion is the colt's father} and C = {the colt carries the Y marker}.

   1. Pr(F) = 1/2.

   2. Pr(F | C) = Pr(F) Pr(C|F) / [Pr(F) Pr(C|F) + Pr(not F) Pr(C | not F)] = (1/2) × (1) / [(1/2) × 1 + (1/2) × 0.02] = approx 98%.

   3. Given that the mare was exposed to a total of 1000 stallions, we have Pr(F) = 1/1000. Thus Pr(F | C) = Pr(F) Pr(C|F) / [Pr(F) Pr(C|F) + Pr(not F) Pr(C | not F)] = (1/1000) × (1) / [(1/1000) × 1 + (999/1000) × 0.02] = approx 4.8%.