Question

A car insurance company has determined that 9% of all drivers were involved in a car accident last year. among 14 drivers living on one particular street, 3 were involved in a car accident last year. If 14 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?
a) 0.094.
b) 0.906.
c) 0.126.
d) 0.393.

Answer 1

The probability of getting 3 or more who were involved in a car accident last year is 0.126.

Given 9% of the drivers were involved in a car accident last year.

We have to find the probability of getting 3 or more who were involved in a car accident last year if 14 were selected randomly.

We have to use binomial theorem which is as under:

n$C_{r}p^{r} (1-p)^{n-r}$

where p is the probability an r is the number of trials.

Probability that 3 or more involved in a car accident last year if 14 are randomly selected=1-[P(X=0)+P(X=1)+P(X=2)]

=1-{$14C_{0}0.9^{0} (0.1)^{14} +14C_{1} (0.9)^{1} (0.1)^{13} +14C_{2} (0.9)^{2} (0.1)^{12}$}

=1-{1*0.2670+14!/13!*0.9*0.29+14!/2!12!*0.0081*0.2358}

=1-{0.2670+0.3654+0.2358}

=1-0.8682

=0.1318

Among the options given the nearest is 0.126.

Hence the probability that 3 or more are involved in the accident is 0.126.

Learn more about probability at brainly.com/question/24756209

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