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Let's break this down step by step to solve each part of the problem.

### Given information:
- He has two hockey sticks: Brand A and Brand B.
- The probability of breaking **Brand A** in a game is 0.06.
- The probability of breaking **Brand B** in a game is 0.15.
- He is 3 times more likely to use **Brand A** at the start of a game than **Brand B**.

Let's define:
- \( P(\text{A}) \) as the probability that he starts with Brand A.
- \( P(\text{B}) \) as the probability that he starts with Brand B.
- \( P(\text{A breaks}) = 0.06 \)
- \( P(\text{B breaks}) = 0.15 \)

Since he is 3 times more likely to use Brand A, we have:
\[
P(\text{A}) = 3 \times P(\text{B})
\]
We also know that \( P(\text{A}) + P(\text{B}) = 1 \).

### Part (a): Probability that he will use his Brand A hockey stick at the start of a hockey game

From the equation \( P(\text{A}) = 3 \times P(\text{B}) \) and \( P(\text{A}) + P(\text{B}) = 1 \), we can solve this system of equations.

Substitute \( P(\text{A}) = 3P(\text{B}) \) into \( P(\text{A}) + P(\text{B}) = 1 \):
\[
3P(\text{B}) + P(\text{B}) = 1
\]
\[
4P(\text{B}) = 1
\]
\[
P(\text{B}) = 0.25
\]
Thus, \( P(\text{A}) = 3 \times 0.25 = 0.75 \).

So, the probability that he starts with **Brand A** is **0.7500**.

### Part (b): Probability that the stick used at the start of the game breaks

This is a conditional probability based on which stick he chooses. The total probability of the stick breaking is given by the law of total probability:
\[
P(\text{break}) = P(\text{A}) \times P(\text{A breaks}) + P(\text{B}) \times P(\text{B breaks})
\]
Substituting the values:
\[
P(\text{break}) = (0.75 \times 0.06) + (0.25 \times 0.15)
\]
\[
P(\text{break}) = 0.045 + 0.0375 = 0.0825
\]

So, the probability that the stick breaks is **0.0825**.

### Part (c): If he breaks the stick, what is the probability he started with Brand B?

This is a conditional probability, \( P(\text{B} \mid \text{break}) \), which can be calculated using Bayes' theorem:
\[
P(\text{B} \mid \text{break}) = \frac{P(\text{break} \mid \text{B}) \times P(\text{B})}{P(\text{break})}
\]
We already know \( P(\text{break} \mid \text{B}) = 0.15 \), \( P(\text{B}) = 0.25 \), and \( P(\text{break}) = 0.0825 \). Substituting these values:
\[
P(\text{B} \mid \text{break}) = \frac{0.15 \times 0.25}{0.0825} = \frac{0.0375}{0.0825} \approx 0.4545
\]

So, the probability that he started with Brand B given that the stick broke is **0.4545**.

---

Do you want more details on any of the steps or explanations? Here are 5 related questions you could explore:

1. What is the probability that the stick **doesn't break** during the game?
2. If he started the game with Brand A, what is the probability that it didn't break?
3. What would the probabilities be if he were only 2 times more likely to choose Brand A?
4. How would the probability calculations change if the breaking probability for Brand B was 0.10 instead of 0.15?
5. How would we calculate the expected number of games until he breaks a stick?

**Tip:** When solving problems using Bayes' theorem, always ensure you carefully identify the prior probabilities and likelihoods before applying the formula.

A Statistics professor takes two hockey sticks - Brand A and Brand B - to each game he plays. Since Brand A is a newer stick than Brand B, he believes the chance he will break his Brand A stick in a game is 6%; he also believes the probability he will break Brand B in a game is 0.15. Due to his various superstitions, he is 3 times more likely to use his Brand A hockey stick to start a game than the Brand B stick. Part (a) Find the probability that he will use his Brand A hockey stick at the start of a hockey game. Part (b) Keeping in mind that he could use either Brand A or Brand B to start the game, what is the probability that the stick he uses at the start of the game breaks? Part (c) If he breaks the stick he started the game with, what is the probability that he started the game with Brand B?

A detailed probability problem involving a statistics professor who uses two hockey sticks, Brand A and Brand B, with different breaking probabilities. Learn how to solve this problem using concepts like conditional probability and Bayes' theorem. Suitable for high school and introductory college-level students.

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