I WILL GIVE I need help figuring this out, no links thank you :)

0.12 accidents.
Step-by-step explanation:
The expected value is:
EV = ∑pₙ*xₙ
where xₙ is the event number n, pₙ is the probability of event xₙ
in this case we have:
x₀ = 0 accidents and p₀ = 0.935
x₁ = 1 accident, and p₁ = 0.03
x₂ = 2 accidents, and p₂ = 0.02
x₃ = 3 accidents, and p₃ = 0.01
x₄ = 4 accidents, and p₄ = 0.005
Then the expected value is:
EV = (0*0.935 + 1*0.03 + 2*0.02 + 3*0.01 + 4*0.05) = 0.12 accidents.

1.3 and 1 1/3 are both closest to 1. hope this ; )