a basketball player traditionally makes 85% of her free throws. Suppose she shoots 10 baskets and count the number she makes. What is the probability she makes less than 9 baskets.
To answer this question, we can use the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
P(X=k) is the probability of getting exactly k successes
n is the number of trials (in this case, shooting 10 baskets)
k is the number of successes (in this case, making baskets)
p is the probability of success (in this case, 85% or 0.85)
(1-p) is the probability of failure (1 - 0.85 = 0.15)
To find the probability she makes less than 9 baskets, we need to sum the probabilities of making 0, 1, 2, 3, 4, 5, 6, 7, 8 baskets:
P(X < 9) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8)
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
P(X=0) = (10 choose 0) * 0.85^0 * (1-0.85)^(10-0)
= 1 * 1 * 0.15^10
= 0.15^10
P(X=1) = (10 choose 1) * 0.85^1 * (1-0.85)^(10-1)
= 10 * 0.85 * 0.15^9
P(X=2) = (10 choose 2) * 0.85^2 * (1-0.85)^(10-2)
= 45 * 0.85^2 * 0.15^8
P(X=3) = (10 choose 3) * 0.85^3 * (1-0.85)^(10-3)
= 120 * 0.85^3 * 0.15^7
P(X=4) = (10 choose 4) * 0.85^4 * (1-0.85)^(10-4)
= 210 * 0.85^4 * 0.15^6
P(X=5) = (10 choose 5) * 0.85^5 * (1-0.85)^(10-5)
= 252 * 0.85^5 * 0.15^5
P(X=6) = (10 choose 6) * 0.85^6 * (1-0.85)^(10-6)
= 210 * 0.85^6 * 0.15^4
P(X=7) = (10 choose 7) * 0.85^7 * (1-0.85)^(10-7)
= 120 * 0.85^7 * 0.15^3
P(X=8) = (10 choose 8) * 0.85^8 * (1-0.85)^(10-8)
= 45 * 0.85^8 * 0.15^2
Now we can calculate the sum:
P(X < 9) = 0.15^10 + 10 * 0.85 * 0.15^9 + 45 * 0.85^2 * 0.15^8 + 120 * 0.85^3 * 0.15^7 + 210 * 0.85^4 * 0.15^6 + 252 * 0.85^5 * 0.15^5 + 210 * 0.85^6 * 0.15^4 + 120 * 0.85^7 * 0.15^3 + 45 * 0.85^8 * 0.15^2
Calculating this sum, we find that the probability she makes less than 9 baskets is approximately 0.969