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?

Using the binomial distribution, it is found that there is a 0.0369 = 3.69% probability that 3 or more were involved in an accident last year.

For each driver, there are only two possible outcomes, either they were involved in an accident, or they were not. The probability of a driver being involved in an accident is independent of any other driver, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

The parameters are:

In this problem:

The probability is:

In which:

Then

0.0369 = 3.69% probability that 3 or more were involved in an accident last year.

To learn more about the binomial distribution, you can check link

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

  for x=0,1,2,....,n

               =0                                   elsewhere

where , 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)

=1-0.624

≈0.376

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

21/58.

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b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5