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Assume you have applied for two scholarships, a Merit scholarship (M) and an Athletic scholarship (A). The probability that you receive an Athletic scholarship is 0.18. The probability of receiving both scholarships is 0.11. The probability of getting at least one of the scholarships is 0.3.

a. What is the probability that you will receive a Merit scholarship?
b. Are events A and M mutually exclusive? Why or why not? Explain.
c. Are the two events A, and M, independent? Explain, using probabilities.
d. What is the probability of receiving the Athletic scholarship given that you have been awarded the Merit scholarship?
e. What is the probability of receiving the Merit scholarship given that you have been awarded the Athletic scholarship?

Respuesta :

Answer:

a)[tex] P(A \cup M)= P(A) + P(M) -P(A \cap M)[/tex]

And if we solve for P(M) we got:

[tex] P(M) = P(A\cup B) +P(A \cap B) -P(A)[/tex]

And replacing we got:

[tex] P(M) = 0.3 +0.11-0.18= 0.23[/tex]

b) In order to A and M be mutually exclusive we need to satisfy:

[tex] P(A \cap M ) =0[/tex]

And for this case since [tex] P(A \cap M) = 0.11 \neq 0[/tex] the events A and M are NOT mutually exclusive

c) In order to satisfy independence we need to have the following relation:

[tex] P(A \cap B) = P(A) *P(B)[/tex]

And for this case we have that:

[tex] 0.11 \neq 0.23*0.18[/tex]

So then A and M are NOT independent

d) [tex] P(A|M) [/tex]

And we can use the Bayes theorem and we got:

[tex] P(A|M) = \frac{P(A \cap M)}{P(M)}[/tex]

And replacing we got:

[tex] P(A|M)= \frac{0.11}{0.23}= 0.478[/tex]

e) [tex] P(M|A) [/tex]

And we can use the Bayes theorem and we got:

[tex] P(M|A) = \frac{P(M \cap A)}{P(A)}[/tex]

And replacing we got:

[tex] P(M|A)= \frac{0.11}{0.18}= 0.611[/tex]

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:

[tex] P(A) =0.18, P(A \cap M) =0.11, P(A \cup M) =0.3[/tex]

Part a

For this case we can use the total rule of probability and we have this:

[tex] P(A \cup M)= P(A) + P(M) -P(A \cap M)[/tex]

And if we solve for P(M) we got:

[tex] P(M) = P(A\cup B) +P(A \cap B) -P(A)[/tex]

And replacing we got:

[tex] P(M) = 0.3 +0.11-0.18= 0.23[/tex]

Part b

In order to A and M be mutually exclusive we need to satisfy:

[tex] P(A \cap M ) =0[/tex]

And for this case since [tex] P(A \cap M) = 0.11 \neq 0[/tex] the events A and M are NOT mutually exclusive

Part c

In order to satisfy independence we need to have the following relation:

[tex] P(A \cap B) = P(A) *P(B)[/tex]

And for this case we have that:

[tex] 0.11 \neq 0.23*0.18[/tex]

So then A and M are NOT independent

Part d

For this case we want this probability:

[tex] P(A|M) [/tex]

And we can use the Bayes theorem and we got:

[tex] P(A|M) = \frac{P(A \cap M)}{P(M)}[/tex]

And replacing we got:

[tex] P(A|M)= \frac{0.11}{0.23}= 0.478[/tex]

Part e

For this case we want this probability:

[tex] P(M|A) [/tex]

And we can use the Bayes theorem and we got:

[tex] P(M|A) = \frac{P(M \cap A)}{P(A)}[/tex]

And replacing we got:

[tex] P(M|A)= \frac{0.11}{0.18}= 0.611[/tex]

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