A car insurance company has determined that 7% of all drivers were involved in a car...

A car insurance company has determined that 7% of all drivers were involved in a car accident last year. Among the 11 drivers living on one particular street, 3 were involved in a car accident last year. If 11 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year? Group of answer choices 0.0317 0.0370 0.9683 0.481

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Expert Solution

Solution

Given that ,

p = 0.07

q = 1 - p = 1 - 0.07 = 0.93

n = 11

Using binomial probability formula ,

P(X = x) = (n C x) * p x * (1 - p)n - x

P(X 3 ) = 1 - P( x <3)

= 1 - P(X = 0) - P(X = 1) - P(X = 2)

= 1 - (11 C 0) * 0.07 0 * (0.93)11 - (11 C 1) * 0.07 1 * (0.93)10 - (11 C 2)2 * 0.07 2 * (0.93)9

=1-0.9630

probability=0.0370

orchestra answered 1 year ago

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