A car insurance company has determined that 9% of all drivers were involved in a car accident last year. Among the 10 drivers living on one particular street, 3 were involved in a car accident last year. If 10 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?

The correct 0.05404

Step-by-step explanation:

Given:

Binomial distribution = x

n = 10

and p = 0.09

solution:

P(X=x) =1 0Cx*(0.09^x)*((1-0.09)^(10-x)) for x=0,1,2,...,10

So the probability is calculated by the Formula:

P(X>=3) = 1-P(X=0)-P(X=1)-P(X=2)

putting the given values in the formula

= 1-10C0*(0.09^0)*((1-0.09)^(10-0))-...-10C2*(0.09^2)*((1-0.09)^(10-2))

= 0.0540400

Thus, the correct 0.05404

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

any options?

step-by-step explanation: