Use the binomial probability to solve.
Basketball Suppose you make 90% of your free throws and you attempt 3 free throws. Use the Binomial Theorem to calculate each probability.
a. You do not make any of them.
- I have this one
b. You only make 1 of them.
- I have this one
c. You only make 2 of them.
- I don't understand this one
d. You make all of them.
- I don't understand this one
So....
c. .9*.9*(1-.9)=-0.09
d. .9^3=0.729
would that be correct?
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
c. .9 * .9 * (1-.9) = ?
d. .9^3 = ?
c. You only make 2 of them.
To calculate the probability of making 2 out of 3 free throws, we use the binomial probability formula:
P(X = k) = (nCk) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of getting exactly k successes
- n is the number of trials
- p is the probability of success on each trial
- k is the number of successes
In this case, n = 3 (number of free throws), p = 0.90 (probability of making a free throw), and k = 2 (number of successes).
Let's calculate the probability:
P(X = 2) = (3C2) * 0.90^2 * (1 - 0.90)^(3 - 2)
First, calculating (3C2):
(3C2) = 3! / (2!(3-2)!) = 3
Now, plugging the values into the formula:
P(X = 2) = 3 * 0.90^2 * (1 - 0.90)^(3 - 2)
= 3 * 0.90^2 * (0.10)^1
= 3 * 0.81 * 0.10
= 0.243
Therefore, the probability of making exactly 2 free throws out of 3 attempts is 0.243 or 24.3%.
d. You make all of them.
To calculate the probability of making all 3 free throws, we again use the binomial probability formula:
P(X = k) = (nCk) * p^k * (1 - p)^(n - k)
In this case, n = 3 (number of free throws), p = 0.90 (probability of making a free throw), and k = 3 (number of successes).
Let's calculate the probability:
P(X = 3) = (3C3) * 0.90^3 * (1 - 0.90)^(3 - 3)
First, calculating (3C3):
(3C3) = 3! / (3!(3-3)!) = 1
Now, plugging the values into the formula:
P(X = 3) = 1 * 0.90^3 * (1 - 0.90)^(3 - 3)
= 1 * 0.90^3 * (0.10)^0
= 1 * 0.90^3 * 1
= 0.729
Therefore, the probability of making all 3 free throws is 0.729 or 72.9%.
To solve these problems using the binomial probability, we need to understand a few key concepts:
1. Binomial distribution: The binomial distribution is a probability distribution that describes the number of successes (in this case, made free throws) in a fixed number of independent trials (in this case, the number of attempts).
2. Binomial formula: The binomial formula calculates the probability of getting exactly k successes in n independent trials, given the probability of success, p, in each trial. The formula is:
P(k successes) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- C(n, k) represents the number of ways to choose k successes from n trials (also known as the binomial coefficient).
- p is the probability of success in each trial.
- (1-p) is the probability of failure in each trial.
Now let's address each part of the question:
a. You do not make any of them.
In this case, the probability of making zero free throws is P(k=0).
Using the binomial formula, we substitute n=3 (number of trials) and p=0.9 (probability of success in each trial), we get:
P(k=0) = C(3, 0) * 0.9^0 * (1-0.9)^(3-0) = 1 * 1 * 0.1^3 = 0.1^3 = 0.001
Therefore, the probability of not making any free throws is 0.1% (or 0.001).
b. You only make 1 of them.
In this case, the probability of making exactly one free throw is P(k=1).
Using the binomial formula, we substitute n=3 (number of trials) and p=0.9 (probability of success in each trial), we get:
P(k=1) = C(3, 1) * 0.9^1 * (1-0.9)^(3-1) = 3 * 0.9 * 0.1^2 = 0.027
Therefore, the probability of making only one free throw is 2.7% (or 0.027).
c. You only make 2 of them.
In this case, the probability of making exactly two free throws is P(k=2).
Using the binomial formula, we substitute n=3 (number of trials) and p=0.9 (probability of success in each trial), we get:
P(k=2) = C(3, 2) * 0.9^2 * (1-0.9)^(3-2) = 3 * 0.9^2 * 0.1^1 = 0.243
Therefore, the probability of making only two free throws is 24.3% (or 0.243).
d. You make all of them.
In this case, the probability of making all three free throws is P(k=3).
Using the binomial formula, we substitute n=3 (number of trials) and p=0.9 (probability of success in each trial), we get:
P(k=3) = C(3, 3) * 0.9^3 * (1-0.9)^(3-3) = 1 * 0.9^3 * 0.1^0 = 0.729
Therefore, the probability of making all three free throws is 72.9% (or 0.729).
Remember, the binomial probability formula can be used to calculate probabilities for similar scenarios with different values of n and p.