Answer:
a) Probability of no broken bats in a game P(X=0) = 0.3678
b) Probability of at-least broken bats in a game P(X≥2) = 0.2644
Explanation:
we will use Poisson distribution
P(X=x) = e^(-λ) λ^ r/r!
a) Probability of no broken bats in a game P(X=0) = e^-1
= 0.3678
b) Probability of at-least two broken bats in a game
P(X≥2) = 1-([p(x=0)+P(x=1)]
= 1- (e^(-1)+e^(-1))
= 1-(0.3678+0.3678)
P(X≥2)= 1- 0.7356
P(X≥2)= 0.2644