The probability that a tennis set will go to a tiebreaker is 14%. What is the probability that two of three sets will go to tie-breakers? Round the answer to the nearest thousandth. O A. 0.311 B. 0.020 O C. 0.140 O D. 0.051
The probability that a tennis set will go to a tiebreaker is 14%. What is the probability that two of three sets will go to tie-breakers? Round the answer to the nearest thousandth. O A. 0.311 B. 0.020 O C. 0.140 O D. 0.051
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Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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![**Probability in Tennis: Tie-Breaker Sets**
The probability that a tennis set will go to a tiebreaker is 14%. What is the probability that two out of three sets will go to tiebreakers? Round the answer to the nearest thousandth.
**Options:**
- **A.** 0.311
- **B.** 0.020
- **C.** 0.140
- **D.** 0.051
**Detailed Explanation:**
Given:
- Probability of a single set going to a tiebreaker (\( p \)) = 14% = 0.14
- Probability of a single set not going to a tiebreaker (\( q \)) = 1 - 0.14 = 0.86
We are looking for the probability that exactly two out of three sets will go to tiebreakers. This situation can be modeled using the binomial distribution formula:
\[ P(X = k) = \binom{n}{k} p^k q^{(n-k)} \]
where:
- \( n \) = number of sets = 3
- \( k \) = number of sets that go to tiebreakers = 2
- \( p \) = probability of a set going to a tiebreaker = 0.14
- \( q \) = probability of a set not going to a tiebreaker = 0.86
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \)
Calculate the binomial coefficient for \( \binom{3}{2} \):
\[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3 \]
Now, use the binomial distribution formula:
\[ P(X = 2) = \binom{3}{2} (0.14)^2 (0.86)^1 \]
\[ P(X = 2) = 3 \times (0.14)^2 \times (0.86) \]
\[ P(X = 2) = 3 \times 0.0196 \times 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F84228c5d-16cd-4b45-8700-712bad5c302e%2F14585547-b86a-4b7f-bc48-8dd904915ebe%2Fs9gw8v2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Probability in Tennis: Tie-Breaker Sets**
The probability that a tennis set will go to a tiebreaker is 14%. What is the probability that two out of three sets will go to tiebreakers? Round the answer to the nearest thousandth.
**Options:**
- **A.** 0.311
- **B.** 0.020
- **C.** 0.140
- **D.** 0.051
**Detailed Explanation:**
Given:
- Probability of a single set going to a tiebreaker (\( p \)) = 14% = 0.14
- Probability of a single set not going to a tiebreaker (\( q \)) = 1 - 0.14 = 0.86
We are looking for the probability that exactly two out of three sets will go to tiebreakers. This situation can be modeled using the binomial distribution formula:
\[ P(X = k) = \binom{n}{k} p^k q^{(n-k)} \]
where:
- \( n \) = number of sets = 3
- \( k \) = number of sets that go to tiebreakers = 2
- \( p \) = probability of a set going to a tiebreaker = 0.14
- \( q \) = probability of a set not going to a tiebreaker = 0.86
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \)
Calculate the binomial coefficient for \( \binom{3}{2} \):
\[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3 \]
Now, use the binomial distribution formula:
\[ P(X = 2) = \binom{3}{2} (0.14)^2 (0.86)^1 \]
\[ P(X = 2) = 3 \times (0.14)^2 \times (0.86) \]
\[ P(X = 2) = 3 \times 0.0196 \times 0
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