Question

Assume the blood type distribution to be A: 41 %, B:9 % , AB: 4%, O: 46 % . What is the probability that the blood of randomly selected individual will contain either the A antigen or the B antigen? That it will contain neither the A nor the B antigen? 2.

Answer 1

Answer:  a) 50% , and b) 50%.

Step-by-step explanation:

Since we have given that

P(A) = 41%

P(B) = 9%

P(AB) = 4%

P(O) = 46%

Since antigen A or antigen B are independent events.

So, P(A ∪ B) = P(A) + P(B)

$P(A\cup B) = 41\%+9\%=50\%$

Probability neither the A nor the B is given by

$P(A'\cap B')=P(A\cup B)'=1-0.50=0.50=50\%$

Hence, a) 50% , and b) 50%.