Answer:
a)![P(A \cup M)= P(A) + P(M) -P(A \cap M)](https://tex.z-dn.net/?f=%20P%28A%20%5Ccup%20M%29%3D%20P%28A%29%20%2B%20P%28M%29%20-P%28A%20%5Ccap%20M%29)
And if we solve for P(M) we got:
![P(M) = P(A\cup B) +P(A \cap B) -P(A)](https://tex.z-dn.net/?f=%20P%28M%29%20%3D%20P%28A%5Ccup%20B%29%20%2BP%28A%20%5Ccap%20B%29%20-P%28A%29)
And replacing we got:
![P(M) = 0.3 +0.11-0.18= 0.23](https://tex.z-dn.net/?f=%20P%28M%29%20%3D%200.3%20%2B0.11-0.18%3D%200.23)
b) In order to A and M be mutually exclusive we need to satisfy:
![P(A \cap M ) =0](https://tex.z-dn.net/?f=%20P%28A%20%5Ccap%20M%20%29%20%3D0)
And for this case since
the events A and M are NOT mutually exclusive
c) In order to satisfy independence we need to have the following relation:
![P(A \cap B) = P(A) *P(B)](https://tex.z-dn.net/?f=%20P%28A%20%5Ccap%20B%29%20%3D%20P%28A%29%20%2AP%28B%29)
And for this case we have that:
![0.11 \neq 0.23*0.18](https://tex.z-dn.net/?f=%200.11%20%5Cneq%200.23%2A0.18)
So then A and M are NOT independent
d) ![P(A|M)](https://tex.z-dn.net/?f=%20P%28A%7CM%29%20)
And we can use the Bayes theorem and we got:
![P(A|M) = \frac{P(A \cap M)}{P(M)}](https://tex.z-dn.net/?f=%20P%28A%7CM%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20M%29%7D%7BP%28M%29%7D)
And replacing we got:
![P(A|M)= \frac{0.11}{0.23}= 0.478](https://tex.z-dn.net/?f=%20P%28A%7CM%29%3D%20%5Cfrac%7B0.11%7D%7B0.23%7D%3D%200.478)
e) ![P(M|A)](https://tex.z-dn.net/?f=%20P%28M%7CA%29%20)
And we can use the Bayes theorem and we got:
![P(M|A) = \frac{P(M \cap A)}{P(A)}](https://tex.z-dn.net/?f=%20P%28M%7CA%29%20%3D%20%5Cfrac%7BP%28M%20%5Ccap%20A%29%7D%7BP%28A%29%7D)
And replacing we got:
![P(M|A)= \frac{0.11}{0.18}= 0.611](https://tex.z-dn.net/?f=%20P%28M%7CA%29%3D%20%5Cfrac%7B0.11%7D%7B0.18%7D%3D%200.611)
Step-by-step explanation:
For this case we define the following events:
A denote the event of receiving an Athletic Scholarship.
M denote the event of receiving a Merit scholarship.
For this case we have the following probabilities given:
![P(A) =0.18, P(A \cap M) =0.11, P(A \cup M) =0.3](https://tex.z-dn.net/?f=%20P%28A%29%20%3D0.18%2C%20P%28A%20%5Ccap%20M%29%20%3D0.11%2C%20P%28A%20%5Ccup%20M%29%20%3D0.3)
Part a
For this case we can use the total rule of probability and we have this:
![P(A \cup M)= P(A) + P(M) -P(A \cap M)](https://tex.z-dn.net/?f=%20P%28A%20%5Ccup%20M%29%3D%20P%28A%29%20%2B%20P%28M%29%20-P%28A%20%5Ccap%20M%29)
And if we solve for P(M) we got:
![P(M) = P(A\cup B) +P(A \cap B) -P(A)](https://tex.z-dn.net/?f=%20P%28M%29%20%3D%20P%28A%5Ccup%20B%29%20%2BP%28A%20%5Ccap%20B%29%20-P%28A%29)
And replacing we got:
![P(M) = 0.3 +0.11-0.18= 0.23](https://tex.z-dn.net/?f=%20P%28M%29%20%3D%200.3%20%2B0.11-0.18%3D%200.23)
Part b
In order to A and M be mutually exclusive we need to satisfy:
![P(A \cap M ) =0](https://tex.z-dn.net/?f=%20P%28A%20%5Ccap%20M%20%29%20%3D0)
And for this case since
the events A and M are NOT mutually exclusive
Part c
In order to satisfy independence we need to have the following relation:
![P(A \cap B) = P(A) *P(B)](https://tex.z-dn.net/?f=%20P%28A%20%5Ccap%20B%29%20%3D%20P%28A%29%20%2AP%28B%29)
And for this case we have that:
![0.11 \neq 0.23*0.18](https://tex.z-dn.net/?f=%200.11%20%5Cneq%200.23%2A0.18)
So then A and M are NOT independent
Part d
For this case we want this probability:
![P(A|M)](https://tex.z-dn.net/?f=%20P%28A%7CM%29%20)
And we can use the Bayes theorem and we got:
![P(A|M) = \frac{P(A \cap M)}{P(M)}](https://tex.z-dn.net/?f=%20P%28A%7CM%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20M%29%7D%7BP%28M%29%7D)
And replacing we got:
![P(A|M)= \frac{0.11}{0.23}= 0.478](https://tex.z-dn.net/?f=%20P%28A%7CM%29%3D%20%5Cfrac%7B0.11%7D%7B0.23%7D%3D%200.478)
Part e
For this case we want this probability:
![P(M|A)](https://tex.z-dn.net/?f=%20P%28M%7CA%29%20)
And we can use the Bayes theorem and we got:
![P(M|A) = \frac{P(M \cap A)}{P(A)}](https://tex.z-dn.net/?f=%20P%28M%7CA%29%20%3D%20%5Cfrac%7BP%28M%20%5Ccap%20A%29%7D%7BP%28A%29%7D)
And replacing we got:
![P(M|A)= \frac{0.11}{0.18}= 0.611](https://tex.z-dn.net/?f=%20P%28M%7CA%29%3D%20%5Cfrac%7B0.11%7D%7B0.18%7D%3D%200.611)