Answer:
p = 0.38, n = 20
The probability that he throws more than 10 strikes = 0.09233
Step-by-step explanation:
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = number of times Jack wants to bowl = 20
x = Number of successes required = number of strikes he intends to get
p = probability of success = probability that Jack throws a strike = 0.38
q = probability of failure = probability that Jack doesn't throw a strike = 0.62
P(X > x) = Σ ⁿCₓ pˣ qⁿ⁻ˣ (summing from x+1 to n)
P(X > 10) = Σ ²⁰Cₓ pˣ qⁿ⁻ˣ (summing from 11 to 20)
P(X > 10) = [P(X=11) + P(X=12) + P(X=13) + P(X=14) + P(X=15) + P(X=16) + P(X=17) + P(X=18) + P(X=19) + P(X=20)
P(X > 10) = 0.09233
There are binomial distribution cacalculators that can calculate all of this at once. Get one to minimize errors.
Sent a picture of the solution to the problem (s).
It’s 21 it technically just means your increasing the 15 by 50 percent
0.5^6 = 0.0156
Assuming an unbiased coin.
