An insurance company writes policies for a large number of newly-licensed drivers each year. Suppose 40% of these are low-risk drivers, 40% are moderate risk, and 20% are high risk. The company has no way to know which group any individual driver falls in when it writes the policies. None of the low-risk drivers will have an at-fault accident in the next year, but 10% of the moderate-risk and 20% of the high-risk drivers will have such an accident. If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk

The probability that he or she is high-risk is 0.50

Step-by-step explanation:

P(Low risk) = 40% = 0.40

P( Moderate risk) = 40% = 0.40

P(High risk) = 20% = 0.20

P(At - fault accident | Low risk) = 0% = 0

P(At-fault accident | Moderate risk) = 10% = 0.10  

P(At-fault accident | High risk) = 20% = 0.20

If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk. Hence, We need to  calculate P( High risk | at-fault accident) = ?

Using Bayes' conditional probability theorem

P( High risk | at-fault accident) = ( P( High risk) * P(At-fault accident | High risk) ) /  { P( Low risk) * P(At-fault accident | Low risk) +P( Moderate risk) * P(At-fault accident | Moderate risk) +  P( High risk) * P(At-fault accident | High risk) }

P( High risk | at-fault accident)= (0.20 * 0.20) / ( 0.40 * 0 + 0.40 * 0.10 + 0.20 * 0.20 )

P( High risk | at-fault accident) = 0.04 / 0 + 0.04 + 0.04

P( High risk | at-fault accident) = 0.04 / 0.08

P( High risk | at-fault accident) = 0.50 .

two data sets with the same median will not have the same varaibility

step-by-step explanation:

51

step-by-step explanation:

73-22=51 the   mass number minus the amount of protons gives you the amount of neutrons