Question

Calculate the following probabilities and indicate which probability distribution model is appropriate in each case. A very good darts player can hit the bull’s eye (red circle in the center of the dart board) 65% of the time. What is the probability that the (a) hits the bullseye for the 10th time on the 15th try? (b) hits the bullseye 10 times in 15 tries? (c) hits the first bullseye on the third try?

Answer 1

Answer:

a) P(10,10)=0.125 (Poisson distribution)

b) P(X=10)=0.212 (Binomial distribution)

c) P(first bullseye on the third try)=0.08 (Binomial experiment)

Step-by-step explanation:

a) Given the event:

X=The player hits the bull’s eye

p=0.65 (probability of success)

We use Poisson Distribution:

P(k,α)=$\frac{e^{-\alpha} \alpha^k }{k!}$

where:

α=n×p=9.75≈10 (n=15 tries)

and k=10 (times that hits the bullseye)

So,

P(10,10)=$\frac{e^{-10} 10^10 }{10!}$=0.125

b) Binomial distribution

P(X)=$\frac{n!}{x!(n-x)!} p^{x}(1-p)^{n-x}$

where:

p=0.65 (probability of success)

n=15 (experiments)

P(X=10)=$\frac{15!}{10!(15-10)!} (0.65)^{10}(1-0.65)^{15-10}$=0.212

c) Binomial experiment:

This experiment consists of miss bull's eye twice and hit it on the third try.

We check the attached tree diagram and we follow the red path to get the probability:

note that p= 0.65 and q=1-p=1-0.65=0.35

P(first bullseye on the third try)=q×q×p=0.35×0.35×0.65=0.080