Chapter 11.2, Problem 36PPSE

Solution

a.

To state: The probability of the event if the odds in favor of the event are $a\text{ to }b,\text{ or }\frac{a}{b}$ and the odds in favor of the event are the ratio of the number of favorable outcomes to the number of unfavorable outcomes.

The resultant probability is $P\left(E\right)=\frac{a}{a+b}$ .

Given information:

The odds in favor of the event are the ratio of the number of favorable outcomes to the number of unfavorable outcomes and odds in favor of the event are $a\text{ to }b,\text{ or }\frac{a}{b}$ .

Explanation:

Odds in favor of the event: $a\text{ to }b,\text{ or }\frac{a}{b}$

Then the probability of the event is:

  $\begin{array}{c}P\left(E\right)=\frac{\text{odds in favor}}{\text{Odds in favor}+\text{odds in against}}\\ =\frac{a}{a+b}\end{array}$

Therefore, the probability is $P\left(E\right)=\frac{a}{a+b}$ .

b.

To state: The odds in favor of the event if the probability of the event is $\frac{a}{b}$ .

The resultant answer is $b-a$ .

Given information:

The probability of the event is $\frac{a}{b}$ .

Explanation:

The probability of the event is: $P\left(E\right)=\frac{a}{b}$

The probability of the event is:

  $\begin{array}{c}P\left(E\right)=\frac{\text{odds in favor}}{\text{Odds in favor}+\text{odds in against}}\\ =\frac{a}{b}\end{array}$

The odds in favor of the event will become: $b-a$

Therefore, the odds in favor of the event is $b-a$ .

c.

To state: Whether one will play a game in which his odds of winning are $\frac{1}{2}$ , or a game in which his probability of winning is $\frac{1}{2}$ .

The probability of winning the game should be preferred.

Given information:

The probability of the event is $\frac{a}{b}$ .

Explanation:

The probability of winning a game is: $\frac{1}{2}$

  $\begin{array}{c}P\left(W_1\right)=\frac{1}{2}\\ =50\text{ percent}\end{array}$

The probability of the event is having an odd of winning is: $\frac{1}{2}$

  $\begin{array}{c}P\left(W_2\right)=\frac{\text{odds in favor}}{\text{Odds in favor}+\text{odds in against}}\\ =\frac{1}{1+2}\\ =\frac{1}{3}\\ =33.33\text{ percent}\end{array}$

Therefore, the probability of winning the game is more than the probability of odds in favor then probability of winning the game is preferred.