Questions: Helen plays basketball. For free throws, she makes the shot 66% of the time. Helen must now attempt two free throws. The probability that Helen makes the second free throw given that she made the first is 0.76. C= the event that Helen makes the first shot. D= the event Helen makes the second shot. a. What notation would you use when finding the probability that Helen makes both free throws? P(C) and P(D) P(C D) P(C or D) P(C and D) b. What is the probability that Helen makes both free throws? Use a decimal rounded to 4 places where possible.

Solution Steps

To find the probability that Helen makes both free throws, we denote:

• \( C \): the event that Helen makes the first shot.
• \( D \): the event that Helen makes the second shot.

The appropriate notation for the probability that Helen makes both free throws is given by: \[ P(C \text{ and } D) \]

We know the following probabilities:

• \( P(C) = 0.66 \): the probability that Helen makes the first free throw.
• \( P(D \mid C) = 0.76 \): the probability that Helen makes the second free throw given that she made the first.

Using the multiplication rule for conditional probabilities, we can calculate \( P(C \text{ and } D) \): \[ P(C \text{ and } D) = P(C) \times P(D \mid C) \] Substituting the known values: \[ P(C \text{ and } D) = 0.66 \times 0.76 \]

Calculating this gives: \[ P(C \text{ and } D) = 0.5016 \]

Final Answer