According to ESPN, Lebron James has a career free-throw percentage (FT%) of 0.736, meaning through his career, he has been successful on free-throws 73.6% of the time. a) If Lebron gets two free-throw shots, what is the probability he will make both? Write as a percentage, round to one place after the decimal in the answer. b) If Lebron gets two free-throw shots, what is the probability he will make at least one of the shots? Write as a percentage, round to one place after the decimal in the answer.
The Correct Answer and Explanation is:
Let’s solve the problem using probability rules.
Given:
- LeBron James has a free-throw percentage of 0.736.
- This means the probability that he makes a free throw is 0.736.
- The probability that he misses a free throw is 1 – 0.736 = 0.264.
Part (a): Probability he will make both free throws
We assume that each free throw is an independent event. To find the probability that LeBron makes both shots, we multiply the probability of making the first shot by the probability of making the second shot.P(make both)=0.736×0.736=0.541696P(\text{make both}) = 0.736 \times 0.736 = 0.541696P(make both)=0.736×0.736=0.541696
Converting to a percentage:0.541696×100=54.2%0.541696 \times 100 = 54.2\%0.541696×100=54.2%
Answer to part (a): 54.2%
Part (b): Probability he will make at least one shot
To find the probability of making at least one shot, we can use the complement rule. Instead of calculating the probability for all possible combinations where he makes at least one, we calculate the probability that he misses both and subtract that from 1.P(miss both)=0.264×0.264=0.069696P(\text{miss both}) = 0.264 \times 0.264 = 0.069696P(miss both)=0.264×0.264=0.069696P(make at least one)=1−0.069696=0.930304P(\text{make at least one}) = 1 – 0.069696 = 0.930304P(make at least one)=1−0.069696=0.930304
Converting to a percentage:0.930304×100=93.0%0.930304 \times 100 = 93.0\%0.930304×100=93.0%
Answer to part (b): 93.0%
Explanation
To determine the probability of certain outcomes involving LeBron James and his free-throw shooting, we use basic rules of probability. Each free throw can either be successful or a miss, and the probability of success is given as 0.736. Since free-throws are considered independent events, the outcome of one does not affect the other.
In part (a), we want to find the probability that he makes both free throws. We multiply the individual probabilities: 0.736 multiplied by 0.736 gives about 0.5417. When converted into a percentage, that becomes 54.2%. This represents how often we would expect him to make two in a row over many attempts.
In part (b), we want the probability that he makes at least one of the two free throws. This could mean he makes just one or he makes both. Rather than adding those two separate probabilities, we use the complement rule, which is often easier. The only scenario in which he makes zero shots is if he misses both. The probability he misses a shot is 1 minus his success rate, or 0.264. The probability of missing both is then 0.264 squared, which gives 0.0697. Subtracting this from 1 gives the probability of making at least one shot. This comes out to approximately 0.9303, or 93.0% when converted to a percentage.
This method of solving allows us to break down the question into manageable steps using basic probability rules.
