大學基礎概率問題

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