a basketball player shoots 70% of freethrows he has a 1 an 1 what is the probability of him shooting o,1,2 points
Pr(0)=.3
Pr(1)= .7*.3
pr(2)=.7*.7
To find the probability of a basketball player shooting 0, 1, or 2 points, given that he shoots 70% of free throws, we can use the concept of binomial probability.
The probability of shooting a free throw successfully can be represented as p = 0.7 (70% or 0.7 out of 1).
Now, let's find the probabilities for each scenario:
1. Probability of shooting 0 points (missing both free throws):
To calculate this probability, we use the binomial formula: P(X = k) = nCk * p^k * (1-p)^(n-k), where n is the total number of attempts and k is the number of successful attempts we want (in this case, 0).
In this scenario, n = 2 (since the player attempts 2 free throws). So, we need to find P(X = 0).
P(X = 0) = 2C0 * 0.7^0 * (1-0.7)^(2-0) = 1 * 1 * 0.3^2 = 0.09 (or 9%)
2. Probability of shooting 1 point (making the first free throw but missing the second):
Again, we use the binomial formula: P(X = k) = nCk * p^k * (1-p)^(n-k), where n = 2 and k = 1 (since we want to calculate the probability of making exactly one free throw).
P(X = 1) = 2C1 * 0.7^1 * (1-0.7)^(2-1) = 2 * 0.7 * 0.3 = 0.42 (or 42%)
3. Probability of shooting 2 points (making both free throws):
Using the same binomial formula: P(X = k) = nCk * p^k * (1-p)^(n-k), but this time we want k = 2 (both free throws made).
P(X = 2) = 2C2 * 0.7^2 * (1-0.7)^(2-2) = 1 * 0.7^2 * 0.3^0 = 0.49 (or 49%)
Therefore, the probabilities of shooting 0, 1, or 2 points are approximately 9%, 42%, and 49%, respectively.