Question

Flying over the western states with mountainous terrain in a small aircraft is 40% riskier than flying over similar distances in flatter portions of the nation, according to a General Accounting Office study completed in response to a congressional request. The accident rate for small aircraft in the 11 mountainous western states is 3.5 accidents per 100,000 flight operations. Find the probability of no accidents in 100,000 flight operations. (Use 4 decimal places.) Find the probability of at least 8 accidents in 180,000 flight operations. (Use 4 decimal places.)

Answer 1

This is a Binomial distribution with the parameters:

n = 100000, p = 3.5/100000

Let X denote the random variable for the number of accidents.

Probability that no accidents occur = P(X=0)

Now,

P(X=0) = ⁿC₀*(p^0)*((1-p)^(n-0)) = ¹⁰⁰⁰⁰⁰C₀*((3.5/100000)^0)*((1-(3.5/100000))^(100000-0)) = 0.0302

Next, we have n = 180000, p = 3.5/100000

We are asked to calculate P (X>=8)

Now,

P(X>=8) = 1 - P(X<8)

P(X < 8) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) +  P(X=5) + P(X=6) + P(X=7)

P(X=0) = ⁿC₀*(p^0)*((1-p)^(n-0)) = ¹⁸⁰⁰⁰⁰C₀*((3.5/180000)^0)*((1-(3.5/180000))^(180000-0)) = 0.0302

P(X=1) = ⁿC₁*(p^1)*((1-p)^(n-1)) = ¹⁸⁰⁰⁰⁰C₁*((3.5/180000)^1)*((1-(3.5/180000))^(180000-1)) = 0.105

Similarly,

P(X=2) = ¹⁸⁰⁰⁰⁰C₂*((3.5/180000)^2)*((1-(3.5/180000))^(180000-2)) = 0.184

P(X=3) = ¹⁸⁰⁰⁰⁰C₃*((3.5/180000)^3)*((1-(3.5/180000))^(180000-3)) = 0.215

P(X=4) = ¹⁸⁰⁰⁰⁰C₄*((3.5/180000)^4)*((1-(3.5/180000))^(180000-4)) = 0.188

P(X=5) = ¹⁸⁰⁰⁰⁰C₅*((3.5/180000)^5)*((1-(3.5/180000))^(180000-5)) = 0.132

P(X=6) = ¹⁸⁰⁰⁰⁰C₆*((3.5/180000)^6)*((1-(3.5/180000))^(180000-6)) = 0.077

P(X=7) = ¹⁸⁰⁰⁰⁰C₇*((3.5/180000)^7)*((1-(3.5/180000))^(180000-7)) = 0.038

So,

P(X < 8) = 0.0302 + 0.105 + 0.184 + 0.215 + 0.188 + 0.132 + 0.077 + 0.038 = 0.9692

So,

PX >= 8) = 1 - 0.9692 = 0.0308