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?

Answers

Answer 1

Answer:

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

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$C_(n,x) = (n!)/(x!(n-x)!)$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

7% of all drivers were involved in an accident last year, hence $p = 0.07$ .

11 drivers are selected at random, hence $n = 11$ .

The probability is:

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 0) = C_(11,0).(0.07)^(0).(0.93)^(11) = 0.4501$

$P(X = 1) = C_(11,1).(0.07)^(1).(0.93)^(10) = 0.3727$

$P(X = 2) = C_(11,2).(0.07)^(2).(0.93)^(9) = 0.1403$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.4501 + 0.3727 + 0.1403 = 0.9631$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.9631 = 0.0369$

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 brainly.com/question/24863377

Answer 2

Answer:

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 $C(n,x)= (n!)/(x!(n-x)!)$ , 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.

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Answers

Answer:

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Step-by-step explanation:

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Answers

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Answers

Answer:

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Answers

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Answers

Answer:

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