Question

The probabliities of blood types O,A,AB and AB in western populations are 0.46 , 0.39 , 0.12 , and 0.03 respectively . Suppose that a clinc is seeking either type O or type B blood. From six randomly selected individuals , what is the probability that at least two people have the desired blood types ?

Answer 1

Solution:

Probabilities of four blood groups in western populations are given as below:

P(O) = 0.46

P(A) = 0.39

P(B) = 0.12

P(AB) = 0.03

All blood groups are independent from each other.

P( O or B) = P(O) + P(B)

P(O or B) = 0.46 + 0.12

P(O or B) = 0.58

Now, we have to find the probability that at least two people have the either type O or type B blood.

We are given

n = 6, p = 0.58

We have to find P(X?2)

P(X?2) = 1 - P(X<2) = 1 – P(X?1)

P(X?1) = P(X=0) + P(X=1)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=0) = 6C0*0.58^0*(1 – 0.58)^(6 – 0)

P(X=0) = 1*1*0.42^6

P(X=0) = 0.005489032

P(X=1) = 6C1*0.58^1*(1 – 0.58)^(6 – 1)

P(X=1) = 6*0.58*0.42^5

P(X=1) = 0.045480549

P(X?1) = P(X=0) + P(X=1)

P(X?1) = 0.005489032 + 0.045480549

P(X?1) = 0.050969581

P(X?2) = 1 – P(X?1)

P(X?2) = 1 – 0.050969581

P(X?2) = 0.949030419

Required Probability = 0.949030419