Question

A car insurance company has determined that 7% of all drivers were involved in an accident last year. If 11 drivers are selected at random, what is the probability that 3 or more were involved in an accident last year?

Answer 1

Answer:

Therefore the probability that 3 or more were involved in an accident last year is 0.376.

Step-by-step explanation:

Binomial distribution:

A discrete random variable X having the set {0,1,2.....,n} as the spectrum, is said to have binomial distribution with parameter n= the number of trial and p = probability of getting successes in one trial if the p.m.f of X is give by

$P(X=x)=C(n,x)p^x(1-p)^{n-x}$   for x=0,1,2,....,n

               =0                                   elsewhere

where $C(n,x)= \frac{n!}{x!(n-x)!}$, n is a positive integer and 0<p<1.

Given that,

7% of all drivers were involved in an accident last year that was determined by a car insurance company.

p= 7%=0.07, n=11, x=3

P(X≥3)

= 1-P(X≤2)

= 1- P(X=0)-P(X=1)-P(X=2)

          $=1- C(11,0)(0.07)^0(1-0.07)^{11}- C(11,1)(0.07)^1(1-0.07)^{10}- C(11,2)(0.07)^2(1-0.07)^{9}$

$=1- C(11,0)(0.07)^0(0.93)^{11}- C(11,1)(0.07)^1(0.93)^{10}- C(11,2)(0.07)^2(0.93)^{9}$

=1-0.624

≈0.376

Therefore the probability that 3 or more were involved in an accident last year is 0.376.