Suppose a basketball player, Player A made 80% of her free throws last season and that she continues to shoot free throws at the same rate. Assume that free throw attempts are independent. Let the random variable X be the number of free throws that player A makes in her next 10 attempts.
a. what is the distribution
b. probability player A makes all 10
c.probabiliity makes at least 8
d.makes between 5 and 9
.869
a. The distribution of X, the number of free throws Player A makes in her next 10 attempts, follows a binomial distribution since each attempt is independent and has the same probability of success (making the free throw) of 80%.
b. To find the probability that Player A makes all 10 free throws, we can use the binomial probability formula:
P(X = k) = nCk * p^k * q^(n-k)
where:
- n is the number of trials/attempts (10 in this case)
- k is the number of successful outcomes (all 10 in this case)
- p is the probability of a successful outcome (making a free throw, which is 0.8)
- q is the probability of an unsuccessful outcome (missing a free throw, which is 1 - p = 0.2)
Using these values in the formula:
P(X = 10) = 10C10 * 0.8^10 * 0.2^(10-10)
= 1 * 0.8^10 * 0.2^0
= 0.8^10
= 0.1074
Therefore, the probability that Player A makes all 10 free throws is approximately 0.1074 or 10.74%.
c. To find the probability that Player A makes at least 8 free throws, we need to calculate the probabilities for making 8, 9, and 10 free throws, and sum them together.
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
Using the same binomial probability formula as before:
P(X = 8) = 10C8 * 0.8^8 * 0.2^(10-8)
= 45 * 0.8^8 * 0.2^2
= 0.301989888
P(X = 9) = 10C9 * 0.8^9 * 0.2^(10-9)
= 10 * 0.8^9 * 0.2^1
= 0.268435456
P(X = 10) = 0.1074 (as calculated in part b)
Hence,
P(X ≥ 8) = 0.301989888 + 0.268435456 + 0.1074
≈ 0.677825344
Therefore, the probability that Player A makes at least 8 free throws is approximately 0.6778 or 67.78%.
d. To find the probability that Player A makes between 5 and 9 free throws, we need to calculate the probabilities for making 5, 6, 7, 8, and 9 free throws, and sum them together.
P(5 ≤ X ≤ 9) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)
Using the same binomial probability formula:
P(X = 5) = 10C5 * 0.8^5 * 0.2^(10-5)
= 252 * 0.8^5 * 0.2^5
= 0.0264241152
P(X = 6) = 10C6 * 0.8^6 * 0.2^(10-6)
= 210 * 0.8^6 * 0.2^4
= 0.088080384
P(X = 7) = 10C7 * 0.8^7 * 0.2^(10-7)
= 120 * 0.8^7 * 0.2^3
= 0.201326592
P(X = 8) = 0.301989888 (calculated in part c)
P(X = 9) = 0.268435456 (calculated in part c)
Hence,
P(5 ≤ X ≤ 9) = 0.0264241152 + 0.088080384 + 0.201326592 + 0.301989888 + 0.268435456
≈ 0.8862564352
Therefore, the probability that Player A makes between 5 and 9 free throws is approximately 0.8863 or 88.63%.
To answer these questions, we need to understand the concept of a binomial distribution. The binomial distribution is used when there are two possible outcomes (success or failure) for a fixed number of independent trials.
In this case, the random variable X represents the number of successful free throws Player A makes in her next 10 attempts. The probability of making each free throw is 80% (or 0.8), and the probability of missing each free throw is 20% (or 0.2).
a. The distribution of X follows a binomial distribution with parameters n (number of trials) and p (probability of success). In this case, n = 10 and p = 0.8. Therefore, X ~ B(10, 0.8).
b. To find the probability that Player A makes all 10 free throws, we calculate P(X = 10) using the binomial distribution formula:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
P(X = 10) = (10 choose 10) * 0.8^10 * (1 - 0.8)^(10 - 10)
= (1) * 0.8^10 * 0.2^0
= 0.8^10
≈ 0.1074
Therefore, the probability that Player A makes all 10 free throws is approximately 0.1074 or 10.74%.
c. To find the probability that Player A makes at least 8 free throws, we calculate P(X ≥ 8) using the binomial distribution formula and summing up the probabilities for X = 8, 9, and 10:
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
P(X ≥ 8) = (10 choose 8) * 0.8^8 * 0.2^(10 - 8) + (10 choose 9) * 0.8^9 * 0.2^(10 - 9) + (10 choose 10) * 0.8^10 * 0.2^(10 - 10)
Performing this calculation, we find:
P(X ≥ 8) ≈ 0.3751
Therefore, the probability that Player A makes at least 8 free throws is approximately 0.3751 or 37.51%.
d. To find the probability that Player A makes between 5 and 9 free throws (inclusive), we calculate P(5 ≤ X ≤ 9) using the binomial distribution formula and summing up the probabilities for X = 5, 6, 7, 8, and 9:
P(5 ≤ X ≤ 9) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)
P(5 ≤ X ≤ 9) = (10 choose 5) * 0.8^5 * 0.2^(10 - 5) + (10 choose 6) * 0.8^6 * 0.2^(10 - 6) + (10 choose 7) * 0.8^7 * 0.2^(10 - 7) + (10 choose 8) * 0.8^8 * 0.2^(10 - 8) + (10 choose 9) * 0.8^9 * 0.2^(10 - 9)
Performing this calculation, we find:
P(5 ≤ X ≤ 9) ≈ 0.3754
Therefore, the probability that Player A makes between 5 and 9 free throws (inclusive) is approximately 0.3754 or 37.54%.
You can find the probabilities by using a binomial probability table or do these problems by hand using the following:
P(x) = (nCx)(p^x)[q^(n-x)]
n = 10
p = .80
q = 1 - p
I'll give you some hints:
b. Find P(10).
c. Find P(8), P(9), and P(10), then add together for total probability.
d. I'll let you determine this one.