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.

 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)

Probability neither the A nor the B is given by

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

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step-by-step explanation: