Question

Suppose that 7 in every 10 auto accidents involve a single vehicle. If 15 auto accidents are randomly selected, compute the probability that at most 4 of them involve a single vehicle.

Answer 1

`\mathbf{P(X \le 4 ) \simeq 0.0006722}`

Explanation:

From the information given:

p = x/n

p = 7/10

p = 0.7

sample size n = 15

Suppose X be the number of accidents involved by a single-vehicle.

Then;

`X \sim Binom (15,0.7)`

Thus, the required probability that at most 4 involve in a single-vehicle is

`P(X\le 4) \\ \\P(X \le 4) = P(X = 0) + P(X =1 ) + ... + P(X = 4)`

`P(X \le 4 ) = (^(15)_0) *0.7^0 *0.3^(15-0) + (^(15)_1) *0.7^1 *0.3^(15-1) + (^(15)_2) *0.7^2 *0.3^(15-2) + (^(15)_3) *0.7^3 *0.3^(15-3) + (^(15)_4) *0.7^4 *0.3^(15-4)`

`P(X \le 4 ) = ((15!)/(0!(15-0)!)) *0.7^0 *0.3^(15-0) + ((15!)/(1!(15-1)!)) *0.7^1 *0.3^(15-1) + ((15!)/(2!(15-2)!)) *0.7^2 *0.3^(15-2) + ((15!)/(3!(15-3)!)) *0.7^3 *0.3^(15-3) + ((15!)/(4!(15-4)!)) *0.7^4 *0.3^(15-4)`

`P(X \le 4 ) =1.4348907 * 10^(-8) +5.02211745 * 10^(-7) + 8.20279183 * 10^(-6) + 8.29393397 * 10^(-5) + 5.80575378 * 10^(-4)`

`P(X \le 4 ) =6.7223407 * 10^(-4)`

`\mathbf{P(X \le 4 ) \simeq 0.0006722}`