Question

A car insurance company has determined that 66​% of all drivers were involved in a car accident last year. Among the 1010 drivers living on one particular​ street, 3 were involved in a car accident last year. If 1010 drivers are randomly​ selected, what is the probability of getting 3 or more who were involved in a car accident last​ year? Round to three decimal places.

Answer 1

Answer: Our required probability is 0.618.

Step-by-step explanation:

Since we have given that

Probability of all drivers were involved in a car accident last year = 66%

Number of drivers = 10

Number of drivers involved in a car accident = 3

So, probability of drivers involved in car accident is given by

$\dfrac{3}{10}\\\\=0.3$

Probability of getting 3 or more who were involved in a car accident is given by

P(X≥3)=1-P(X=0)-P(X=1)-P(X=2)

Here,

$P(X=0)=^{10}C_0(0.7)^{10}=0.028\\\\P(X=1)=^{10}C_1(0.7)^9(0.3)=0.121\\\\P(X=2)=^{10}C_2(0.7)^8(0.3)^2=0.233$

So, our equation becomes

P(X≥3)=1-P(X=0)-P(X=1)-P(X=2)

$P(X\geq 3)=1-0.028-0.121-0.233=0.618$

Hence, our required probability is 0.618.