Coram is at basketball practice preparing to shoot free throws. The probability of Coram making a free throw is 77%. What is the probability that he makes his first free throw on the second shot?
A. 79%
B. 95%
C. 88%
D. 83%
I think the answer is C
not sure but leaning to C to
the answer is 95%
Well, I have to admit, predicting the future is not my strongest suit. But I'm always ready to clown around and give it a shot!
If the probability of making a free throw is 77%, then the probability of missing it would be 23%. Since Coram is shooting for the first time, there are two possibilities: he makes it on the first shot (77%) or he misses it (23%).
Now, if he misses the first shot, he will have to make it on the second shot. So, the probability of him making it on the second shot would be 77%, because that's the only opportunity left for him to make it.
So, the probability that he makes his first free throw on the second shot would be 23% * 77%, which equals 0.23 * 0.77 = 0.1771.
Well, that's not an option given in the answer choices. So, I'm afraid I can't clown around anymore with this one!
To find the probability that Coram makes his first free throw on the second shot, we need to consider two scenarios: he misses the first shot and makes the second shot, or he makes the first shot and makes the second shot as well.
Let's calculate the probability for each scenario separately:
1. Probability of missing the first shot and making the second shot:
Since the probability of making a shot is 77%, the probability of missing a shot is 100% - 77% = 23%.
So, the probability of missing the first shot and making the second shot is 23% * 77% = 0.23 * 0.77 = 0.1771 or approximately 17.71%.
2. Probability of making the first shot and making the second shot:
The probability of making a shot is 77% as given.
Therefore, the probability of making the first shot and making the second shot is 77% * 77% = 0.77 * 0.77 = 0.5929 or approximately 59.29%.
To find the total probability of Coram making his first free throw on the second shot, we sum up the probabilities from both scenarios:
Total probability = Probability of missing the first shot and making the second shot + Probability of making the first shot and making the second shot
Total probability = 17.71% + 59.29%
Total probability = 77%
Therefore, the correct answer is D. 83%.
prob(hit) = .77
so prob(miss) = .23
My interpretation of the event: he misses the first shot, but makes the second
so you want prob(miss, then hit) = (.23)(.77) = .1771 or app 18%
none of the choices are correct