Asked by idk
Based on the 2017 season, the Houston Astros have a winning percentage of .623. Use the binomial model to find the probability that the Astros will win 4 of their next 6 games. P(x)=[n!x!(n−x)!]pxqn−x (1 point) Responses 12.4% 12.4% 24.7% 24.7% 32.1% 32.1% 62.3%
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Answered by
GPT-4o mini
To compute the probability that the Houston Astros will win exactly 4 out of their next 6 games using the binomial model, we can use the formula for the binomial probability:
\[
P(x) = \binom{n}{x} p^x (1-p)^{n-x}
\]
Where:
- \( n \) = total number of trials (games) = 6
- \( x \) = number of successful outcomes (wins) = 4
- \( p \) = probability of success (win rate) = 0.623
- \( 1-p \) = probability of failure (loss rate) = 1 - 0.623 = 0.377
1. Calculate \( \binom{n}{x} \) (the number of combinations of winning \( x \) games out of \( n \)):
\[
\binom{n}{x} = \frac{n!}{x!(n-x)!} = \frac{6!}{4! \cdot 2!} = \frac{720}{24 \cdot 2} = \frac{720}{48} = 15
\]
2. Calculate \( p^x \) (the probability of winning \( x \) games):
\[
p^4 = (0.623)^4
\]
Calculate \( 0.623^4 \):
\[
0.623^4 \approx 0.0966
\]
3. Calculate \( (1-p)^{n-x} \) (the probability of losing \( n-x \) games):
\[
(1-p)^{2} = (0.377)^2
\]
Calculate \( 0.377^2 \):
\[
0.377^2 \approx 0.1425
\]
4. Now, combine all parts to find \( P(4) \):
\[
P(4) = \binom{6}{4} p^4 (1-p)^{2} = 15 \cdot 0.0966 \cdot 0.1425
\]
Calculate:
\[
15 \cdot 0.0966 \cdot 0.1425 \approx 15 \cdot 0.01376 \approx 0.2064 \text{ (approximately)}
\]
5. Finally, convert to percentage:
\[
P(4) \approx 0.2064 \times 100 \approx 20.64\%
\]
From the response options provided (12.4%, 24.7%, 32.1%, 62.3%), it seems none of them closely match the computed value of approximately 20.64%.
Given common issues in such calculations, you should carefully check for rounding and calculation steps. However, based on the computed values and approximations, the probability of the Astros winning 4 out of their next 6 games using the binomial model is approximately **20.64%**, which does not align with the provided options directly.
\[
P(x) = \binom{n}{x} p^x (1-p)^{n-x}
\]
Where:
- \( n \) = total number of trials (games) = 6
- \( x \) = number of successful outcomes (wins) = 4
- \( p \) = probability of success (win rate) = 0.623
- \( 1-p \) = probability of failure (loss rate) = 1 - 0.623 = 0.377
1. Calculate \( \binom{n}{x} \) (the number of combinations of winning \( x \) games out of \( n \)):
\[
\binom{n}{x} = \frac{n!}{x!(n-x)!} = \frac{6!}{4! \cdot 2!} = \frac{720}{24 \cdot 2} = \frac{720}{48} = 15
\]
2. Calculate \( p^x \) (the probability of winning \( x \) games):
\[
p^4 = (0.623)^4
\]
Calculate \( 0.623^4 \):
\[
0.623^4 \approx 0.0966
\]
3. Calculate \( (1-p)^{n-x} \) (the probability of losing \( n-x \) games):
\[
(1-p)^{2} = (0.377)^2
\]
Calculate \( 0.377^2 \):
\[
0.377^2 \approx 0.1425
\]
4. Now, combine all parts to find \( P(4) \):
\[
P(4) = \binom{6}{4} p^4 (1-p)^{2} = 15 \cdot 0.0966 \cdot 0.1425
\]
Calculate:
\[
15 \cdot 0.0966 \cdot 0.1425 \approx 15 \cdot 0.01376 \approx 0.2064 \text{ (approximately)}
\]
5. Finally, convert to percentage:
\[
P(4) \approx 0.2064 \times 100 \approx 20.64\%
\]
From the response options provided (12.4%, 24.7%, 32.1%, 62.3%), it seems none of them closely match the computed value of approximately 20.64%.
Given common issues in such calculations, you should carefully check for rounding and calculation steps. However, based on the computed values and approximations, the probability of the Astros winning 4 out of their next 6 games using the binomial model is approximately **20.64%**, which does not align with the provided options directly.
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