Question 260099: Can someone please help me understand this.
Based on past records, a car insurance company has determined that 9% of all drivers were involved in a car accident last year.
a. If 12 drivers were randomly selected, what is the probability of getting 0 who were involved in a car accident last year?
b. If 12 drivers were randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Based on past records, a car insurance company has determined that 9% of all drivers were involved in a car accident last year.
a. If 12 drivers were randomly selected, what is the probability of getting 0 who were involved in a car accident last year?
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This is a binomial problem with n = 12 ; p = 0.09
P(x=0) = 12C0(0.09)^0(0.91)^12 = 0.91^12 = 0.3225
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b. If 12 drivers were randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?
P(x>=3) = 1-P(0<= x <=2) = 1 - binomcdf(12,0.09,2) = 0.0866..
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Cheers,
Stan H.
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