Question

in each case, find the probability of an event e having the given odds. (a) the odds in favor of e are 8 to 5. (b) the odds against e are 14 to 9. (c) the odds in favor of e are the same as the odds against e. (a) the probability of an event e occurring is \\(\frac{8}{13}\\) (type an integer or a simplified fraction.) (b) the probability of an event e occurring is \\(\frac{9}{23}\\) (type an integer or a simplified fraction.) (c) the probability of an event e occurring is \\(\square\\) (type an integer or a simplified fraction.)

Explanation:

Step1: Recall Odds-Probability Relationship

Odds against \( E \) are \( a:b \) means \( P(\text{not } E)=\frac{a}{a + b} \), so \( P(E)=1 - \frac{a}{a + b}=\frac{b}{a + b} \). For part (b), odds against \( E \) are \( 14:9 \), so \( a = 14 \), \( b = 9 \).

Step2: Calculate Probability for (b)

Using the formula \( P(E)=\frac{b}{a + b} \), substitute \( a = 14 \), \( b = 9 \). Then \( a + b=14 + 9 = 23 \), so \( P(E)=\frac{9}{23} \)? Wait, no, wait: Odds against \( E \) is \( 14:9 \), so favorable to not \( E \) is 14, favorable to \( E \) is 9. So total outcomes \( 14 + 9 = 23 \), so \( P(E)=\frac{9}{23} \). Wait, but the user's (b) in the problem says "The odds against E are 14 to 9" (wait, maybe typo, 14 to 9? Wait, the user's (b) in the written part: "The odds against E are 14 to 9" (maybe 14 to 9? Wait, the user's (b) in the problem: "The odds against E are 14 to 9" (maybe 14 to 9, but in the image, (b) is written as "The odds against E are 14 to 9" (wait, the user's (b) in the problem: "The odds against E are 14 to 9" (maybe 14 to 9, but in the image, (b) is "The odds against E are 14 to 9" (wait, the user's (b) in the problem: "The odds against E are 14 to 9" (maybe 14 to 9, but let's check (c): "The odds in favor of E are the same as the odds against E". So for (c), odds in favor \( a:a \), so \( P(E)=\frac{a}{a + a}=\frac{1}{2} \).

Wait, let's re-express:

Part (c):

Odds in favor of \( E \) = odds against \( E \). Let odds in favor be \( k:k \). So number of favorable outcomes to \( E \) is \( k \), unfavorable is \( k \). Total outcomes \( k + k = 2k \). Thus, \( P(E)=\frac{k}{2k}=\frac{1}{2} \).

Step3: Calculate Probability for (c)

For part (c), odds in favor of \( E \) equal odds against \( E \). Let odds in favor be \( m:m \). Then number of favorable outcomes (for \( E \)) is \( m \), unfavorable is \( m \). Total possible outcomes \( m + m = 2m \). So probability \( P(E)=\frac{\text{Number of favorable to } E}{\text{Total outcomes}}=\frac{m}{m + m}=\frac{m}{2m}=\frac{1}{2} \).

Wait, the user's (c) is: "The odds in favor of E are the same as the odds against E". So we need to find \( P(E) \) for (c).

So let's solve (c):

Let odds in favor of \( E \) be \( a:a \) (since they are same as odds against, which would be \( a:a \) as well? Wait, odds against \( E \) is (unfavorable):(favorable). So if odds in favor of \( E \) is \( a:b \), odds against is \( b:a \). So if odds in favor = odds against, then \( a:b = b:a \), which implies \( a = b \). So odds in favor is \( a:a \), odds against is \( a:a \). So number of favorable outcomes (to \( E \)) is \( a \), unfavorable is \( a \). Total outcomes \( a + a = 2a \). Thus, \( P(E)=\frac{a}{2a}=\frac{1}{2} \).

So for (c), the probability is \( \frac{1}{2} \).

Wait, the user's (c) in the problem: "The odds in favor of E are the same as the odds against E". So we need to compute \( P(E) \) for (c).

So let's clarify:

• Odds in favor of \( E \): \( \frac{\text{Number of favorable to } E}{\text{Number of unfavorable to } E} = \frac{m}{n} \)
• Odds against \( E \): \( \frac{\text{Number of unfavorable to } E}{\text{Number of favorable to } E} = \frac{n}{m} \)

Given that odds in favor = odds against, so \( \frac{m}{n}=\frac{n}{m} \) → \( m^2 = n^2 \) → \( m = n \) (since \( m,n \) are positive integers).

Thus, \( m = n \), so total outcomes \( m + n = 2m \), and \( P(E)=\frac{m}{2m}=\frac{1}{2} \).

So for part (c), the probability is \( \frac{1}{2} \).

Answer:

For part (c), the probability is \(\boxed{\dfrac{1}{2}}\)