A tennis team has a probability of 76% of winning a match when they are at home and 45% of winning a match when they are away. If the team plays 58% of their matches away, find the probability that the team wins their home match given that they are away

To solve this problem, we need to use the concept of conditional probability and Bayes' theorem. However, it's important to clarify the question to make sure we understand what is being asked. The question asks for the probability that the team wins a home match given that they are away. If we interpret this strictly based on the phrasing, it seems contradictory to ask for the probability of winning a home match when we know the match is being played away. This probability would technically be zero because a match cannot be both a home and away match simultaneously.

However, if we reinterpret the problem as needing to find the probability that the team is playing at home, given that they have won a match, we can proceed with the following approach:

### Given Data

- Probability of winning a home match: \( P(\text{Win} \mid \text{Home}) = 0.76 \)
- Probability of winning an away match: \( P(\text{Win} \mid \text{Away}) = 0.45 \)
- Probability of playing an away match: \( P(\text{Away}) = 0.58 \)
- Probability of playing a home match: \( P(\text{Home}) = 1 - P(\text{Away}) = 1 - 0.58 = 0.42 \)

### Finding \( P(\text{Home} \mid \text{Win}) \)

This is the probability that the match was played at home, given that the team won the match.

Using Bayes' theorem, we have:

\[
P(\text{Home} \mid \text{Win}) = \frac{P(\text{Win} \mid \text{Home}) \cdot P(\text{Home})}{P(\text{Win})}
\]

To find \( P(\text{Win}) \), the total probability of winning a match, we use the law of total probability:

\[
P(\text{Win}) = P(\text{Win} \mid \text{Home}) \cdot P(\text{Home}) + P(\text{Win} \mid \text{Away}) \cdot P(\text{Away})
\]

Substituting the given values:

\[
P(\text{Win}) = (0.76 \times 0.42) + (0.45 \times 0.58)
\]

\[
P(\text{Win}) = 0.3192 + 0.261
\]

\[
P(\text{Win}) = 0.5802
\]

### Now Calculate \( P(\text{Home} \mid \text{Win}) \)

\[
P(\text{Home} \mid \text{Win}) = \frac{0.76 \times 0.42}{0.5802}
\]

\[
P(\text{Home} \mid \text{Win}) = \frac{0.3192}{0.5802}
\]

\[
P(\text{Home} \mid \text{Win}) \approx 0.55
\]

### Conclusion

The probability that the team played their home match, given that they won the match, is approximately 0.55 or 55%.

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:

1. What is the probability that the team won a match given that it was played away?
2. How would the probabilities change if the team played 70% of their matches away?
3. What is the probability that the team wins a match given they play at home?
4. If the team plays only home matches, what is the probability of winning 3 consecutive matches?
5. How would these probabilities change if the win probability for away matches increased to 60%?

**Tip:** When dealing with conditional probabilities, carefully define your events and use Bayes' theorem to calculate the probability of interest.

Explore the probability that a tennis team wins their home match given that they are playing away. Use Bayes' theorem and the law of total probability to calculate the conditional probability. Includes detailed steps and calculations.

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