To solve this problem, we can use conditional probability.

Problem Summary:

• Let $A$ be the event that you get grounded.
• Let $B$ be the event that you have extra chores.
• We are given the following probabilities:
  • $P(A) = 0.85$ (probability of being grounded)
  • $P(A \cap B) = 0.50$ (probability of being grounded and having extra chores)

We need to find the probability of having extra chores given that you are already grounded. This is represented by:

$P(B \mid A)$

Conditional Probability Formula:

The formula for conditional probability is:

$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$

Plugging in the given values:

$P(B \mid A) = \frac{0.50}{0.85} \approx 0.588$

Thus, the probability that you will have extra chores given that you are already grounded is approximately 0.588 or 58.8%.

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Would you like more details or have any questions about the solution? Here are 5 follow-up questions you can explore:

1. What are the properties of conditional probability?
2. How does conditional probability differ from independent probability?
3. Can you solve this using a Venn diagram?
4. How would the problem change if the probability of being grounded and having extra chores was different?
5. How do you calculate the probability of events that are mutually exclusive?

Tip: Conditional probability helps refine the likelihood of an event when you're given new information, such as knowing that one event has already occurred.