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.