Number six please and thanks sm

Assume you have applied for two scholarships, a Merit scholarship (M) and an Athletic scholarship (A). The probability that you

Brut [27]

Answer:

a) $P(A \cup M)= P(A) + P(M) -P(A \cap M)$

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

$P(M) = P(A\cup B) +P(A \cap B) -P(A)$

And replacing we got:

$P(M) = 0.3 +0.11-0.18= 0.23$

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

$P(A \cap M ) =0$

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

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

$P(A \cap B) = P(A) *P(B)$

And for this case we have that:

$0.11 \neq 0.23*0.18$

So then A and M are NOT independent

d) $P(A|M)$

And we can use the Bayes theorem and we got:

$P(A|M) = \frac{P(A \cap M)}{P(M)}$

And replacing we got:

$P(A|M)= \frac{0.11}{0.23}= 0.478$

e) $P(M|A)$

And we can use the Bayes theorem and we got:

$P(M|A) = \frac{P(M \cap A)}{P(A)}$

And replacing we got:

$P(M|A)= \frac{0.11}{0.18}= 0.611$

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:

$P(A) =0.18, P(A \cap M) =0.11, P(A \cup M) =0.3$

Part a

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

$P(A \cup M)= P(A) + P(M) -P(A \cap M)$

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

$P(M) = P(A\cup B) +P(A \cap B) -P(A)$

And replacing we got:

$P(M) = 0.3 +0.11-0.18= 0.23$

Part b

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

$P(A \cap M ) =0$

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

Part c

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

$P(A \cap B) = P(A) *P(B)$

And for this case we have that:

$0.11 \neq 0.23*0.18$

So then A and M are NOT independent

Part d

For this case we want this probability:

$P(A|M)$

And we can use the Bayes theorem and we got:

$P(A|M) = \frac{P(A \cap M)}{P(M)}$

And replacing we got:

$P(A|M)= \frac{0.11}{0.23}= 0.478$

Part e

For this case we want this probability:

$P(M|A)$

And we can use the Bayes theorem and we got:

$P(M|A) = \frac{P(M \cap A)}{P(A)}$

And replacing we got:

$P(M|A)= \frac{0.11}{0.18}= 0.611$

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