Ch 2 - 9, 11, 13

9

Question

In Ch 1 - 15, 27#15, calculate the following probabilities:
a. $P (A_{2} | A_{1})$
b. $P (A_{2} \cap A_{3} | A_{1})$
c. $P (A_{2} \cup A_{3} | A_{1})$
d. $P (A_{1} \cap A_{2} \cap A_{3} | A_{1} \cup A_{2} \cup A_{3})$

Proof
a.

P (A_{2} | A_{1}) = \frac{P (A_{2} \cap A_{1})}{P (A_{1})} = \frac{.11}{.22} = .5

P (A_{2} \cap A_{3} | A_{1}) = \frac{P (A_{1} \cap A_{2} \cap A_{3})}{P (A_{1})} = \frac{.01}{.22} = .045

\begin{aligned} P (A_{2} \cup A_{3} | A_{1}) & = \frac{P ((A_{2} \cup A_{3}) \cap A_{1})}{P (A_{1})} \\ = \frac{P (A_{1} \cap A_{2}) + P (A_{1} \cap A_{3}) - P (A_{1} \cap A_{2} \cap A_{3})}{.22} \\ = \frac{.11 + .05 - .01}{.22} \\ = .682 \end{aligned}

\begin{aligned} P (A_{1} \cap A_{2} \cap A_{3} | A_{1} \cup A_{2} \cup A_{3}) & = \frac{P ((A_{1} \cap A_{2} \cap A_{3}) \cap (A_{1} \cup A_{2} \cup A_{3})}{P (A_{1} \cup A_{2} \cup A_{3})} \\ = \frac{P (A_{1} \cap A_{2} \cap A_{3})}{.53} \\ = \frac{.01}{.53} \\ = .019 \end{aligned}

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11

Question

A certain university accepts 16% of students who apply and
offers academic scholarships to 5% of students they accept.
Of those receiving academic scholarships, 20% receive a “full
ride” (i.e. their scholarships cover tuition, room and board, and
all textbooks and other course supplies).
a. What is the chance that a randomly selected applicant to
this university will be accepted and offered an academic scholarship?
b. What is the probability that a randomly selected applicant
will be accepted and offered a “full ride” scholarship?
c. Given that someone has been accepted to the university,
what is the probability s/he receives a “full ride” scholarship?

Proof
Say $A$ is the event of being accepted, and $S$ the event that someone got a scholarship, where $F$ is the event they got a full ride. Obviously you need $A$ in order to have $S$ and $S$ for $F$ so then:

P (A) = .16, P (S | A) = .05, P (F | S) = .20

a. $P (A \cap S) = P (A) \cdot P (S | A) = .16 \cdot .05 = .008$
b. $P (A \cap F) = P (A) \cdot P (S | A) \cdot P (F | S \underset{can ignore}{\underset{⏟}{\cap A}}) = .16 \cdot .05 \cdot .20 = .0016$
c. $P (F | A) = \frac{P (F \cap A}{P (A)} = \frac{.0016}{.16} = .01$
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13

Question

Pasted image 20240927131956.png
Pasted image 20240927132004.png

Proof
For these, say that $R, D, I$ represent political parties, and $M, F$ represent genders.
a. $P (F) = \frac{N (F)}{100} = \frac{17 + 8}{100} = .25$
$P (D) = \frac{17 + 28}{100} = .45$
b. $P (D | F) = \frac{N (D \cap F)}{N (F)} = \frac{17}{25} = .68$
c. $P (M | R) = \frac{N (M \cap R)}{N (R)} = \frac{45}{53} = .849$
d. $P (M | R^{'}) = \frac{N (M \cap R^{'})}{N (R^{'})} = \frac{28 + 2}{47} = .638$

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