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[tex]C_{r}p^{r} (1-p)^{n-r}[/tex]
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-{[tex]14C_{0}0.9^{0} (0.1)^{14} +14C_{1} (0.9)^{1} (0.1)^{13} +14C_{2} (0.9)^{2} (0.1)^{12}[/tex]}
=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.
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