Answer:
[tex]\mathbf{P(X \le 4 ) \simeq 0.0006722}[/tex]
Step-by-step 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;
[tex]X \sim Binom (15,0.7)[/tex]
Thus, the required probability that at most 4 involve in a single-vehicle is
[tex]P(X\le 4) \\ \\P(X \le 4) = P(X = 0) + P(X =1 ) + ... + P(X = 4)[/tex]
[tex]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}[/tex]
[tex]P(X \le 4 ) = (\dfrac{15!}{0!(15-0)!}) *0.7^0 *0.3^{15-0} + (\dfrac{15!}{1!(15-1)!}) *0.7^1 *0.3^{15-1} + (\dfrac{15!}{2!(15-2)!}) *0.7^2 *0.3^{15-2} + (\dfrac{15!}{3!(15-3)!}) *0.7^3 *0.3^{15-3} + (\dfrac{15!}{4!(15-4)!}) *0.7^4 *0.3^{15-4}[/tex]
[tex]P(X \le 4 ) =1.4348907 \times 10^{-8} +5.02211745 \times 10^{-7} + 8.20279183 \times 10^{-6} + 8.29393397 \times 10^{-5} + 5.80575378 \times 10^{-4}[/tex]
[tex]P(X \le 4 ) =6.7223407 \times 10^{-4}[/tex]
[tex]\mathbf{P(X \le 4 ) \simeq 0.0006722}[/tex]
