Question

Suppose that the proportion of blood phenotypes in a particular population are as follows: type Proportion A 0.42 B 0.10 AB 0.04 O 0.44 Assuming that the phenotypes of two randomly selected individuals are independent of each other, what is the probability that both phenotypes of two randomly selected individuals match?

Answer 1

`P(AA \cup BB \cup ABAB\cup OO)= 0.42*0.42 + 0.1*0.1+ 0.04*0.04+ 0.44*0.44 = 0.3816`

So then we can conclude that the probability that both with the same phenotypes of two randomly selected individuals match is 0.3816

Explanation:

For this case we have the following probabilities givenL

`P(A) = 0.42, P(B)=0.1, P(AB)= 0.04, P(O)=0.44`

And for this case we want to find the probability that two phenotypes will mach, so then can be both A, B, AB or O, so we want to find this probability:

`P(AA \cup BB \cup ABAB\cup OO)`

Using the condition of independence we can express this probability like this:

`P(AA \cup BB \cup ABAB\cup OO)= P(A)*P(A) + P(B)*P(B)+ P(AB)*P(AB)+ P(O)*P(O)`

And replacing the valus given we have:

`P(AA \cup BB \cup ABAB\cup OO)= 0.42*0.42 + 0.1*0.1+ 0.04*0.04+ 0.44*0.44 = 0.3816`

So then we can conclude that the probability that both with the same phenotypes of two randomly selected individuals match is 0.3816