Question

question 19
if ( p(e)=0.88 ), then calculate and find that (enter one digit per box after simplifying your numbers to the extent possible):
a) the odds for ( e ) are: to
b) the odds against ( e ) are: to
question 20
if ( p(e)=0.45 ), then calculate and find that (enter one digit per box after simplifying your numbers to the extent possible):
a) the odds for ( e ) are to
b) the odds against ( e ) are: to

Explanation:

Step1: Recall the formula for odds

The formula for odds for an event \(E\) is \(\text{Odds for }E=\frac{P(E)}{1 - P(E)}\), and the formula for odds against an event \(E\) is \(\text{Odds against }E=\frac{1 - P(E)}{P(E)}\)

Step2: Solve for Question 19

• Given \(P(E)=0.88\), then \(1 - P(E)=1 - 0.88 = 0.12\)
• Odds for \(E=\frac{0.88}{0.12}=\frac{88}{12}=\frac{22}{3}\approx7.33\) (but as per the requirement of one - digit per box after simplifying, \(\frac{0.88}{0.12}=\frac{88\div4}{12\div4}=\frac{22}{3}\), if we consider the ratio of whole numbers \(22:3\))
• Odds against \(E=\frac{0.12}{0.88}=\frac{12}{88}=\frac{3}{22}\), ratio of whole numbers \(3:22\)

Step3: Solve for Question 20

• Given \(P(E) = 0.45\), then \(1 - P(E)=1 - 0.45=0.55\)
• Odds for \(E=\frac{0.45}{0.55}=\frac{45}{55}=\frac{9}{11}\), ratio of whole numbers \(9:11\)
• Odds against \(E=\frac{0.55}{0.45}=\frac{55}{45}=\frac{11}{9}\), ratio of whole numbers \(11:9\)

Answer:

Question 19:
a) \(22\) to \(3\)
b) \(3\) to \(22\)

Question 20:
a) \(9\) to \(11\)
b) \(11\) to \(9\)