Ken plays basketball. He makes free throw shots 50% of the time. Ken must now attempt two free throws. The probability that Ken makes the second free throw given that he made the first is 0.65.
What is the probability that Ken makes his first free throw or his second free throw?
Keep in mind Im In an intro to stats class

To find the probability that Ken makes **either his first free throw or his second free throw**, we need to break the problem down using the formula for the probability of the union of two events.

The formula for finding the probability of either event \( A \) or event \( B \) happening is:

\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]

Where:
- \( P(A) \) is the probability that Ken makes his first free throw.
- \( P(B) \) is the probability that Ken makes his second free throw.
- \( P(A \cap B) \) is the probability that Ken makes both the first and the second free throws.

### Step 1: Identify the probabilities
- Ken has a 50% chance of making the first free throw:  
  \[
  P(A) = 0.50
  \]
  
- The problem states that the probability Ken makes the second free throw given that he made the first is 0.65. This means we know \( P(B|A) = 0.65 \).

- The probability of making both the first and second free throws is:
  \[
  P(A \cap B) = P(A) \times P(B|A) = 0.50 \times 0.65 = 0.325
  \]

### Step 2: Calculate the probability of Ken making the second free throw
To calculate \( P(B) \), we use the law of total probability:
\[
P(B) = P(B|A) \times P(A) + P(B|\text{not } A) \times P(\text{not } A)
\]
Where \( P(B|\text{not } A) \) is the probability Ken makes the second free throw given he missed the first, which is just the original 50% probability because nothing changes if he misses the first. Therefore:
\[
P(B) = (0.65 \times 0.50) + (0.50 \times 0.50) = 0.325 + 0.25 = 0.575
\]

### Step 3: Apply the formula
Now, we can use the union formula:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.50 + 0.575 - 0.325 = 0.75
\]

Thus, the probability that Ken makes **either the first or the second free throw** is:

\[
\boxed{0.75}
\]

This means there is a 75% chance that Ken will make at least one of his free throws.

Would you like further details or have any questions?

### Related Questions:
1. What is the probability that Ken makes both free throws?
2. How would the probability change if Ken had an 80% chance of making his first shot?
3. What is the probability that Ken misses both free throws?
4. If Ken attempted three free throws, how would the calculations change?
5. How would you calculate the expected number of free throws Ken would make in two attempts?

### Tip: 
When dealing with conditional probabilities, always keep track of how one event affects another and use the law of total probability to account for all possibilities.

Ken plays basketball. He makes free throw shots 50% of the time. Ken must now attempt two free throws. The probability that Ken makes the second free throw given that he made the first is 0.65. What is the probability that Ken makes his first free throw or his second free throw?

Find the probability that Ken makes either his first or second free throw, using conditional probability and the law of total probability. This problem explores the relationship between independent events and conditional events in probability, suitable for intro to stats students.

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