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).

the probability is 0.7053 (70.53%)

Step-by-step 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%)