Question

I am stuck on part b of a problem the question is people with O-negative blood are referred to as universal donors. Only 7.2% of the american population have O-negative blood. If 10 people appear at random, what is the probability that at least one of them is a universal donor? I figured this one to be 1-P(not a univ donor) so 1-.928^10 and ^for the 10 random and come up with 0.5263 the problem I am having trouble figuring out is Given that the first 3 were not O-negative, what is the probability that the 4th person is? can anyone help

Answer 1

Let X : Person selected has O negative blood group

Now X ~ Binomial (10 , 0.072)

P(X=x) =

X	P(X=x)
0	0.473674234780715
1	0.367505871812623
2	0.128310239727683
3	0.026546946150555
4	0.003604434498890
5	0.000335585280931
6	0.000021697324198
7	0.000000961950334
8	0.000000027987779
9	0.000000000482548
10	0.000000000003744

Now as you correctly calculated P(X>=1) = P1 - P(X=0) = 1 - 0.928¹⁰ = 0.5263

We have to find the probability:

P(4th person is O negative / first three persons were not O negative)

= P(4th person is O negative AND first 3 person were not o negative) / P(first three persons were not O negative)

Now:

P(4th person is O negative AND first 3 person were not o negative) = P(4th person is O negative) * P(first 3 persons were not O negative)

= 0.072 * (0.928)³

Now

P(first 3 persons were not O negative) = (0.928)³

Therefore

P(4th person is O negative / first three persons were not O negative) = P(4th person is O negative) = 0.072

This follows since a person being O negative is independent of any person being O negative so

P(A/B) = P(A)