Question

question 9
1 pts
events ( a_1 ) and ( a_2 ) are mutually exclusive and form a complete partition of a sample space ( s ) with ( p(a_2) = 0.65 ). if ( e ) is an event in ( s ) with ( p(e | a_1) = 0.08 ) and ( p(e | a_2) = 0.14 ), compute ( p(a_2 | e) = ? ) (hint: because ( a_1 ) and ( a_2 ) are mutually exclusive and form a complete partition of the sample space, ( p(a_1) = 1 - p(a_2) )).
note: if your final answer has up to four decimal places, please enter your full answer without rounding it. if your answer contains more than four decimal places, please round it to four decimal places before entering it in the box below.

question 10
1 pts
suppose events ( e ) and ( f ) are independent, with ( p(f) = 0.12 ) and ( p(e cap f) = 0.072 ). calculate the following and enter your answer one digit per box (please simplify your numbers to the extent possible).
a) the odds for ( e ) are:

to

b) the odds against ( e ) are:

to

Explanation:

Question 9

Step1: Find \( P(A_1) \)

Since \( A_1 \) and \( A_2 \) form a complete partition, \( P(A_1)=1 - P(A_2) \). Given \( P(A_2) = 0.65 \), so \( P(A_1)=1 - 0.65 = 0.35 \).

Step2: Apply Bayes' Theorem

Bayes' Theorem for two - partition case: \( P(A_2|E)=\frac{P(E|A_2)P(A_2)}{P(E|A_1)P(A_1)+P(E|A_2)P(A_2)} \)
Substitute the values: \( P(E|A_1) = 0.08 \), \( P(E|A_2)=0.14 \), \( P(A_1)=0.35 \), \( P(A_2)=0.65 \)
First, calculate the numerator: \( P(E|A_2)P(A_2)=0.14\times0.65 = 0.091 \)
Then, calculate the denominator: \( P(E|A_1)P(A_1)+P(E|A_2)P(A_2)=0.08\times0.35 + 0.14\times0.65=0.028+0.091 = 0.119 \)

Step3: Calculate \( P(A_2|E) \)

\( P(A_2|E)=\frac{0.091}{0.119}\approx0.7647 \)

Step1: Recall the property of independent events

If \( E \) and \( F \) are independent, then \( P(E\cap F)=P(E)\times P(F) \)
Given \( P(F) = 0.12 \) and \( P(E\cap F)=0.072 \), we can solve for \( P(E) \):
\( P(E)=\frac{P(E\cap F)}{P(F)}=\frac{0.072}{0.12}=0.6 \)
Then \( P(\overline{E})=1 - P(E)=1 - 0.6 = 0.4 \)

Step2: Calculate odds for \( E \)

Odds for an event \( E \) is defined as \( \frac{P(E)}{P(\overline{E})} \)
So odds for \( E \) is \( \frac{0.6}{0.4}=\frac{3}{2} \), which means 3 to 2.

Step3: Calculate odds against \( E \)

Odds against an event \( E \) is defined as \( \frac{P(\overline{E})}{P(E)} \)
So odds against \( E \) is \( \frac{0.4}{0.6}=\frac{2}{3} \), which means 2 to 3.

Answer:

\( 0.7647 \)

Question 10