Free Throw Freddie is an 80% free throw shooter. At the end of practice he shoots until he makes 10 shots.
a) What is the probability that Freddie makes 10 in a row?
b) What is the probability that Freddie needs 15 or more attempts?
I understand how to do part a), but I'm lost with part b).
For part a) I did:
0.8^10=0.107.
Could anyone help me out with part b? Thank you.
To find the probability that Freddie needs 15 or more attempts, you can consider the complementary event. The complementary event in this case is the probability that Freddie makes all 10 shots within the first 14 attempts.
Since Freddie has an 80% chance of making a shot, the probability of making a shot on a given attempt is 0.8. Hence, the probability of missing a shot on a given attempt is the complementary event, which is 1 - 0.8 = 0.2.
To calculate the probability of Freddie making all 10 shots within the first 14 attempts, you can use the binomial probability formula. The formula is:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
where:
- P(X = k) is the probability of k successes (Freddie making k shots in a row),
- n is the number of trials (attempts),
- p is the probability of success (making a shot), and
- (n C k) is the binomial coefficient, which represents the number of ways to choose k successes from n trials.
In this case, you want to find the probability of Freddie making all 10 shots in the first 14 attempts, so n = 14 and k = 10. Plugging these values into the formula:
P(X = 10) = (14 C 10) * 0.8^10 * 0.2^(14 - 10)
Calculating this:
P(X = 10) = (14 C 10) * 0.8^10 * 0.2^4
= (14! / (10! * 4!)) * 0.8^10 * 0.2^4
Using a calculator or software, you can compute (14 C 10) = 1001, then substitute the value back into the formula:
P(X = 10) = 1001 * 0.8^10 * 0.2^4
≈ 0.0574 (rounded to four decimal places)
This is the probability that Freddie makes all 10 shots within the first 14 attempts. To find the probability that Freddie needs 15 or more attempts, you subtract this value from 1, since it represents the complementary event:
P(15 or more attempts) = 1 - P(X = 10)
= 1 - 0.0574
≈ 0.9426 (rounded to four decimal places)
So, the probability that Freddie needs 15 or more attempts is approximately 0.9426.