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發問:
人的血型大致可分為4種 : A, B, AB, O 每一種血型的分佈為下 A : 42% B : 10% AB : 4% O : 44% if 4 people are picked randomly, let P(k) be the probability that there are exactly k different blood types among them. Find P(k) for k = 1,2,3,4? I guess P(1) is just that all 4 people are of the same blood type
最佳解答:
P(1) is straight forward = 0.424 + 0.14 + 0.044 + 0.444 = 0.0687 P(2) are P(A,B) + P(A, AB) + P(A, O) + P(B, AB) + P(B,O) + P(AB,O) Consider P(A,B), let's look at the binomial expansion, (0.42 + 0.1)4 = 4C0(0.424) + 4C1(0.423)(0.1) + 4C2(0.422)(0.12) + 4C3(0.42)(0.13) + 4C4(0.1)4 The middle three terms are exactly the probability of P(A,B) So P(A,B) = (0.42 + 0.1)4 – 0.424 – 0.14 Similarly P(A, AB) = (0.42 + 0.04)4 – 0.424 – 0.044 P(A, O) = (0.42 + 0.44)4 – 0.424 – 0.444 P(B, AB) = (0.1 + 0.04)4 – 0.14 – 0.044 P(B,O) = (0.1 + 0.44)4 – 0.14 – 0.444 P(AB,O) = (0.04 + 0.44)4 – 0.044 – 0.44 P(2) = 0.5973 Let's now look at the probable ways to arrange 3 blood groups for 4 persons (a + b + c)4 = [(a + b) + c]4 = 4C0(a + b)4 + 4C1(a + b)3(c) + 4C2(a + b)2(c2) + 4C3(a + b)(c3) + 4C4(c)4 = (a + b)4 + 4(a + b)3c + 6(a + b)2c2 + 4(a + b)c3 + c4 Our task is to find our the terms that contain all a, b and c, they are: 4(3a2b + 2ab2)c + 6(2ab)c2 = 12(a2bc + ab2c + abc2) = 12abc(a + b + c) So P(3) = P(A, B, AB) + P(A, B, O) + P(A, AB, O) + P(B, AB, O) = 12(0.42)(0.1)(0.04)(0.42+0.1+0.04) + 12(0.42)(0.1)(0.44)(0.42+0.1+0.44) + 12(0.42)(0.04)(0.44)(0.42+0.04+0.44) + 12(0.1)(0.04)(0.44)(0.1+0.04+).44) = 0.3163 P(4) is straight forward = (4!)(0.42)(0.1)(0.04)(0.44) = 0.0177
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大學基礎概率問題發問:
人的血型大致可分為4種 : A, B, AB, O 每一種血型的分佈為下 A : 42% B : 10% AB : 4% O : 44% if 4 people are picked randomly, let P(k) be the probability that there are exactly k different blood types among them. Find P(k) for k = 1,2,3,4? I guess P(1) is just that all 4 people are of the same blood type
最佳解答:
P(1) is straight forward = 0.424 + 0.14 + 0.044 + 0.444 = 0.0687 P(2) are P(A,B) + P(A, AB) + P(A, O) + P(B, AB) + P(B,O) + P(AB,O) Consider P(A,B), let's look at the binomial expansion, (0.42 + 0.1)4 = 4C0(0.424) + 4C1(0.423)(0.1) + 4C2(0.422)(0.12) + 4C3(0.42)(0.13) + 4C4(0.1)4 The middle three terms are exactly the probability of P(A,B) So P(A,B) = (0.42 + 0.1)4 – 0.424 – 0.14 Similarly P(A, AB) = (0.42 + 0.04)4 – 0.424 – 0.044 P(A, O) = (0.42 + 0.44)4 – 0.424 – 0.444 P(B, AB) = (0.1 + 0.04)4 – 0.14 – 0.044 P(B,O) = (0.1 + 0.44)4 – 0.14 – 0.444 P(AB,O) = (0.04 + 0.44)4 – 0.044 – 0.44 P(2) = 0.5973 Let's now look at the probable ways to arrange 3 blood groups for 4 persons (a + b + c)4 = [(a + b) + c]4 = 4C0(a + b)4 + 4C1(a + b)3(c) + 4C2(a + b)2(c2) + 4C3(a + b)(c3) + 4C4(c)4 = (a + b)4 + 4(a + b)3c + 6(a + b)2c2 + 4(a + b)c3 + c4 Our task is to find our the terms that contain all a, b and c, they are: 4(3a2b + 2ab2)c + 6(2ab)c2 = 12(a2bc + ab2c + abc2) = 12abc(a + b + c) So P(3) = P(A, B, AB) + P(A, B, O) + P(A, AB, O) + P(B, AB, O) = 12(0.42)(0.1)(0.04)(0.42+0.1+0.04) + 12(0.42)(0.1)(0.44)(0.42+0.1+0.44) + 12(0.42)(0.04)(0.44)(0.42+0.04+0.44) + 12(0.1)(0.04)(0.44)(0.1+0.04+).44) = 0.3163 P(4) is straight forward = (4!)(0.42)(0.1)(0.04)(0.44) = 0.0177
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