Statistics for laboratory scientists

Solutions for the homework problems for lecture 5

1. 1. Pr(F₂ seed is smooth) = 3/4 = 75%

   2. Pr(F₂ seed has genotype Aa) = 1/2 = 50%

   3. Pr(F₂ seed has genotype Aa | it is smooth) = (1/2)/(3/4) = 2/3 = (approx) 67%

   4. Pr(F₃ seed is smooth | F₂ has genotype AA) = 100%

   5. Pr(F₃ seed is smooth | F₂ has genotype Aa) = 3/4 = 75%

   6. Pr(F₃ seed is smooth)
      = Pr(F₃ is smooth | F₂ is AA) x Pr(F₂ is AA) + Pr(F₃ is smooth | F₂ is Aa) x Pr(F₂ is Aa) + Pr(F₃ is smooth | F₂ is aa) x Pr(F₂ is aa)
      = [1 x (1/4)] + [(3/4) x (1/2)] + [0 x (1/4)] = 5/8 = 62.5%

   7. Pr(F₃ seed is smooth | F₂ is smooth)
      = Pr(F₃ is smooth | F₂ is AA) x Pr(F₂ is AA | F₂ is smooth) + Pr(F₃ is smooth | F₂ is Aa) x Pr(F₂ is Aa | F₂ is smooth)
      = [1 x (1/3)] + [(3/4) x (2/3)] = 5/6 = (approx) 83.3%

2. 1. (5/10) x (5/10) = 1/4 = 25%

   2. 100%

   3. If 1st ball was blue, then we must be drawing from urn A, and so Pr(2nd is red | 1st is blue) = 50%.

   4. Pr(2nd is red and 1st is red) = Pr(both are red and coin was heads) + Pr(both are red and coin was tails) = Pr(coin was heads) x Pr(both red | coin heads) + Pr(coin tails) x Pr(both red | coin tails) = (1/2) x (1/4) + (1/2) x 1 = 1/8 + 1/2 = 5/8 = 63%.

      Similarly, Pr(1st is red) = (1/2) x (1/2) + (1/2) x 1 = 1/4 + 1/2 = 3/4 = 75%.

      Thus, Pr(2nd is red | 1st is red) = (5/8) / (3/4) = 5/6 = 83%.

3. 1. 4/52 = 1/13 = 7.7%

   2. 1/13 = 7.7%

   3. 2/50 = 1/25 = 4%

   4. Pr(all are Jacks) = Pr(1st is Jack) x Pr(2nd is Jack | 1st is Jack) x Pr(3rd is Jack | 1st two are Jacks) = (4/52) x (3/51) x (2/50) = 3/16575.

   5. (48/52)x(47/51)x(46/50) = 78%

   6. Pr(at least one is a Jack) = 1 - Pr(none are Jacks) = 22%.