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)=[tex]\frac{e^{-\alpha} \alpha^k }{k!}[/tex]
where:
α=n×p=9.75≈10 (n=15 tries)
and k=10 (times that hits the bullseye)
So,
P(10,10)=[tex]\frac{e^{-10} 10^10 }{10!}[/tex]=0.125
b) Binomial distribution
P(X)=[tex]\frac{n!}{x!(n-x)!} p^{x}(1-p)^{n-x}[/tex]
where:
p=0.65 (probability of success)
n=15 (experiments)
P(X=10)=[tex]\frac{15!}{10!(15-10)!} (0.65)^{10}(1-0.65)^{15-10}[/tex]=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
