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Explanation & Answer
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Week 6 Assignment - Answers
1.
a)
P(Both Are O) → P(O) × P(O)
The probability is 0.44 × 0.44=0.1936. So, there is a 19.36% chance that both phenotypes
are O.
b)
The probability that the phenotypes of two randomly selected individuals will match
is as follows:
The probability that both the phenotypes match is the sum of square of probabilities
of each phenotype.
P(Both Match) = P(Both Are A) + P(Both Are B) + P(Both Are AB) + P(Both
AreO)
P(Both Match) = (0.42×0.42) + (0.10×0.10) + (0.04×0.04) + (0.44×0.44)
P(Both Match) = 0.1764 + 0.01 + 0.0016 + 0.1936 = 0.3816
Thus, there is a 38.16% chance that that the phenotypes of two randomly selected
individuals will match.
c)
The probability that the phenotype is A given that it is already known that the individual
is of type A, B, or AB but not O is based on the following calculation:
P(A∣A,B,AB) = P(A,BorAB) / P(A∩(A,BorAB)) = P(A) / P(A)+P(B)+P(AB) =
0.42 / 0.42+0.10+0.04 → 0.75
Thus, there is a 75% chance that the phenotype is A given that you already know that this
individual is of type A, B, or AB but not O.
2.
a)
P(D) = 6/156 = 0.038
P(D|A) = 2/52 = 0.038
b)
P(D and A) = P(D) × P(A)
Checking if D and A are independent events:
P(D and A) = 2/156 = 0.012 (1)
P(A) = 52/156 = 0....
