Question

In the United States, 44% of the people have type O, 42% have type A, 10% have type B, and 4% have type AB blood. Find the probability that there are 4 people with type O blood, 4 people with type A blood, 1 person with type B blood, and 1 person with type AB blood in a randomly chosen 10 people in the US (round off to third decimal place).

Answer 1

the probability is 0.7053 (70.53%)

Explanation:

assuming that the probability of having each type of blood are independent from the others , then the probability that there are 4 people with type O blood, 4 people with type A blood, 1 person with type B blood, and 1 person with type AB blood in a randomly chosen 10 people can be calculated with multinomial probability :

P(A,B,C,D)= possible combinations of A,B,C and D * probability of getting A,B,C and D

P(A,B,C,D)= n!/(A!*B!*C!*D!)*pa^A*pb^B*pc^C*pd^D

where

A= people with A blood = 4

B= people with B blood = 1

C= people with AB blood = 1

D= people with O blood = 4

n = sample size = 10

pa,pb,pc,pd = probability of having each A,B,AB and O type of blood respectively = 0.42 , 0.10 , 0.04 , 0.44

replacing values

P(A,B,C,D) = 10!/(4!*4!*1!*1!)* 0.42^4* 0.10^1* 0.04^1 *0.44^4 = 0.7053

then the probability is 0.7053 (70.53%)