Question

Amarillo Slim, a professional dart player, has an 82% chance of hitting the bull’s-eye on a dartboard with any throw. Suppose that he throws 15 darts, one at a time, at the dartboard. Find the probability that Slim hits the bull’s-eye exactly nine times.

Answer 1

Answer:

0.0285 = 2.85% probability that Slim hits the bull’s-eye exactly nine times.

Step-by-step explanation:

For each throw, there are only two possible outcomes, either he hits the bull’s-eye, or he does not. Each throw is independent of other throws. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

82% chance of hitting the bull’s-eye on a dartboard with any throw.

This means that $p = 0.82$

He throws 15 darts

This means that $n = 15$

Find the probability that Slim hits the bull’s-eye exactly nine times.

This is P(X = 9).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{15,9}.(0.82)^{9}.(0.18)^{15} = 0.0285$

0.0285 = 2.85% probability that Slim hits the bull’s-eye exactly nine times.