Question

Blood types A, B, O, and AB have percentages 40%, 20%, 30% and 10%. If two individuals are chosen at random, assuming independence, how do you find probability that one is blood type A and the other is type O?

Answer 1

Let's first translate percentages into fractions, meaning `100%harr1.00`

Explanation:

Then `p(A)=0.40,p(B)=0.20, p(O)=0.30,p(AB)=0.10)`
Since they're independent , we may multiply :

The probability of taking an `A` and then an `O` is:
`p(A,O)=p(A)xxp(O)=0.4xx0.3=0.12`

The same goes for the other way aound:
`p(O,A)=p(O)xxp(A)=0.3xx0.4=0.12`

So the total probability (since it's either O-A or A-O we may add )

`p(O^^A)=p(A,O)+p(O,A)=0.12+0.12=0.24`

Which translates to `p(O^^A)=24%`