Question

Blood types: The blood type O negative is called the "universal donor" type, because it is the only blood type that may safely be transfused into any person.
Therefore, when someone needs a transfusion in an emergency and their blood type cannot be determined, they are given type O negative blood. For this
reason, donors with this blood type are crucial to blood banks. Unfortunately, this blood type is fairly rare; according to the Red Cross, only 7% of U.S. residents
have type O negative blood. Assume that a blood bank has recruited 19 donors. Round the answers to four decimal places.
a) what is the probability that three or more of them have type O negative blood?
b) what is the probability that fewer than four of them have type O negative blood?
c) It (would/wouldn’t) be unusual if none of the donors had type O negative blood since the probability is __?

Answer 1

a) It is 0.1275 times more probability that fewer than five of them have type o negative blood.

b) The probability that less than five of them are type o negative blood types is 0.9933.

c) 0.2708 probability that there are no donors who have type O negative blood. It would not be odd to have no type o negative donors because this likelihood is higher than 0.05.

There are only two outcomes that can occur for any individual. Type o negative blood is either present or absent. Any other person has no effect on a person's likelihood of having type O negative blood. To answer this question, we thus employ the binomial probability distribution.

Binomial probability distribution

The probability of precisely x successes on n repeated trials is known as a binomial probability, and X can only have two possible outcomes.

P(X=x)= $C_{n,x.P^{x}.(1-P)^{n-x} }$

Additionally, p is the probability that X will occur.

A whopping 7% of Americans have type o negative blood.

This means that P=0.07

18 donors.

This mean that n = 18

the probability that three or more of them have type o negative blood

Either less than three have, or at least three do. The sum of the probabilities of these events is 1. So

P(X<3) + P(X≥3)=1

We want P(X≥3)

So,

P(X≥3) = 1-P(X<3)

In which

P(X< 3)= P(X = 0)+P(X = 1)+P(X = 2) = 0.2708 + 0.3669 +0.2348 = 0.8725

P(X≥3) = 1-P(X<3) = 1- 0.8725=0.1275

It's 0.1275 times more likely that fewer than five of them have type O negative blood.

The probability that fewer than five of them have type o negative blood

P ( x < 5 )= P(X=0)+P(X=1)+P(X=3)+P(X=3)+P(X=4)

P(X=x)= $C_{n,x.P^{x}.(1-P)^{n-x} }$

P(X=0)= $C_{18,0(0.07)^{0}.(0.93)^{18} }= 0.2708$

P(X=1)=   $C_{18,1(0.07)^{1}.(0.93)^{17} }= 0.3669$

P(X=2)= $C_{18,2(0.07)^{2}.(0.93)^{16} }= 0.2348$

P(X=3)= $C_{18,3(0.07)^{3}.(0.93)^{15} }= 0.0942$

P(X=4))= $C_{18,4(0.07)^{4}.(0.93)^{14} }= 0.0266$

P(X < 5)=P(X=0)+P(X=1)+P(X=3)+P(X=3)+P(X=4)

0.2708+0.3669+0.2348+0.0942+0.0266=0.9933

Would it be unusual f none of the donors had type o negative blood

P(X=0)= $C_{18,0(0.07)^{0}.(0.93)^{18} }= 0.2708$

Probability of having no type O negative blood donors is 0.2708. It would not be odd to have no type o negative donors because this likelihood is higher than 0.05.

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