# A couple of probability homework questions

## Homework Statement

The first one seems too easy and I'm not sure I did it right (it looks like they are nearly asking what they gave in different words.) I'm relatively comfortable about what I did in the second, note that we did not cover probabilities of intersections of sets yet so I removed those and used other theorems.

1.) Consider, once again, the four bloodtypes A, B, AB, and O, together with the two antigens anti-a and anti-b. Suppose that, for a given person, the probability of type O blood is 0.5, the probability of type A blood is 0.34, and the probility of type B blood is 0.12.

a.) Find the probability that each of the antigens will react with this person's blood.

b.) Find the probability that both antigens will react with this person's blood.

2.) If the probability that student A will fail a certain statistic examination is 0.5, the probability that student B will fail the examination is 0.2, and the probability that both student A and student B will fail the examinatyion is 0.1, what is the probability that:

a.) at least one of these two students will fail the examination?
b.) neither student fails the examination?
c.) exactly one student fails the examination?

## The Attempt at a Solution

1.a)

Antigen A's probability of reacting is the probability that the person has type A blood OR type AB blood.

Type AB's probability is easily found by considering that our sample space must have probability 1, therefore AB has a probability of .04.

Since all of these events are disjoint, the probability of AuAB is just .34 + .04 = .38.

Antigen B is found the same way, .12 + .04 = .16.

1.b)

This is just the probability of having type AB blood, .12.

2.) a and b I am relatively confident in, but c was a bit more difficult for me.

a.) This is Pr(AuB), which is Pr(A) + Pr(b) - Pr(AnB) = 0.6

b.) This is Pr((AuB)'), which is 1 - Pr(AuB) = 0.4

c.) I called this set (AuB)n(AnB)', to denote in the union and not in the intersection.

Since I want Pr((AuB)n(AnB)'), I first took the complement as that set of a whole in order to get a union:

((AuB)n(AnB)')' = (AuB)'u(AnB)

Then
Pr((AuB)n(AnB)') = 1 - Pr[(AuB)'u(AnB)]

Knowing that this is the union of two disjoint sets, the sum of the union is just the sum of their probabilities. Since I found Pr((AuB)') above, and am given Pr(AnB), I found the result to be

Pr((AuB)n(AnB)') = 1 - (.4 + .1)
= .5

Thanks for any help.

Last edited:

vela
Staff Emeritus
Homework Helper

## Homework Statement

The first one seems too easy and I'm not sure I did it right (it looks like they are nearly asking what they gave in different words.) I'm relatively comfortable about what I did in the second, note that we did not cover probabilities of intersections of sets yet so I removed those and used other theorems.

1.) Consider, once again, the four bloodtypes A, B, AB, and O, together with the two antigens anti-a and anti-b. Suppose that, for a given person, the probability of type O blood is 0.5, the probability of type A blood is 0.34, and the probility of type B blood is 0.12.

a.) Find the probability that each of the antigens will react with this person's blood.

b.) Find the probability that both antigens will react with this person's blood.

2.) If the probability that student A will fail a certain statistic examination is 0.5, the probability that student B will fail the examination is 0.2, and the probability that both student A and student B will fail the examinatyion is 0.1, what is the probability that:

a.) at least one of these two students will fail the examination?
b.) neither student fails the examination?
c.) exactly one student fails the examination?

## The Attempt at a Solution

1.a)

Antigen A's probability of reacting is the probability that the person has type A blood OR type AB blood.

Type AB's probability is easily found by considering that our sample space must have probability 1, therefore AB has a probability of .04.

Since all of these events are disjoint, the probability of AuAB is just .34 + .04 = .38.

Antigen B is found the same way, .12 + .04 = .16.

1.b)

This is just the probability of having type AB blood, .12.
You mean 0.04, which you said above was the probability of AB.

2.) a and b I am relatively confident in, but c was a bit more difficult for me.

a.) This is Pr(AuB), which is Pr(A) + Pr(b) - Pr(AnB) = 0.6

b.) This is Pr((AuB)'), which is 1 - Pr(AuB) = 0.4

c.) I called this set (AuB)n(AnB)', to denote in the union and not in the intersection.

Since I want Pr((AuB)n(AnB)'), I first took the complement as that set of a whole in order to get a union:

((AuB)n(AnB)')' = (AuB)'u(AnB)

Then
Pr((AuB)n(AnB)') = 1 - (AuB)'u(AnB)

Knowing that this is the union of two disjoint sets, the sum of the union is just the sum of their probabilities. Since I found Pr((AuB)') above, and am given Pr(AnB), I found the result to be

Pr((AuB)n(AnB)') = 1 - (.4 + .1)
= .5