Question

Enrique is shooting two free throws during a basketball game, with an 85% success rate for making each free throw. Assuming that his free throw attempts are independent events, what is the probability that he:

1. Makes both free throws?
2. Makes the first and then misses the second?
3. Misses the first and then makes the second?
4. Misses both of his free throws?

Calculate each probability individually and provide your answers as percentages. Lastly, sum up these four probabilities.

Answer 1

The probability of making both free throws is 72.25%. The probability of making the first and then missing the second is 12.75%. The probability of missing both free throws is 2.25%. The sum of these probabilities is 100%.

Step-by-step explanation:

When a player has an 85% success rate of making a free throw, the probability of making a free throw is 0.85. The probability of missing a free throw is 1 - 0.85 = 0.15. Since the free throw attempts are independent events, we can calculate the probabilities as follows:

1. To find the probability of making both free throws: P(Making both) = P(Making first) * P(Making second) = 0.85 * 0.85 = 0.7225 = 72.25%
2. To find the probability of making the first and then missing the second: P(Making first and missing second) = P(Making first) * P(Missing second) = 0.85 * 0.15 = 0.1275 = 12.75%
3. To find the probability of missing the first and then making the second: P(Missing first and making second) = P(Missing first) * P(Making second) = 0.15 * 0.85 = 0.1275 = 12.75%
4. To find the probability of missing both free throws: P(Missing both) = P(Missing first) * P(Missing second) = 0.15 * 0.15 = 0.0225 = 2.25%

To sum up these four probabilities:

P(Making both) + P(Making first and missing second) + P(Missing first and making second) + P(Missing both) = 72.25% + 12.75% + 12.75% + 2.25% = 100%