Stats: Help With Multiplication Rule

[shawnz1102]

I am having a difficult time grasping the multiplication rules - The rules are easy to remember but actually applying it to a problem is so confusing and frustrating as I don't know whether to use Rule #1 (independent events) or Rule #2 (Dependent events).

Here's a problem from my book that I just cannot solve:

An insurance company classifies drivers as low-risk, medium-risk, and high-risk. Of those insured, 60% are low risk, 30% are medium risk, and 10% are high risk. After a study, the company finds that during a 1-year period, 1% of the low risk drivers had an accident, 5% of the medium risk drivers had an accident, and 09% of the high risk drivers had an accident. If a drier is selected at random, find the probability that the driver will have had an accident during that year.

Ans: 0.03

What i tried doing was (0.01/0.6 * 0.05/0.3 * 0.09/0.1); looks like that's not the answer :(

Does anyone know how to solve this, and whether it's an independent or dependent event?

THANKS IN ADVANCE!

 

[Mark44]

I am having a difficult time grasping the multiplication rules - The rules are easy to remember but actually applying it to a problem is so confusing and frustrating as I don't know whether to use Rule #1 (independent events) or Rule #2 (Dependent events).

Here's a problem from my book that I just cannot solve:



What i tried doing was (0.01/0.6 * 0.05/0.3 * 0.09/0.1); looks like that's not the answer :(

Does anyone know how to solve this, and whether it's an independent or dependent event?

THANKS IN ADVANCE!

The expression you show above is wrong on two counts - you shouldn't be dividing by the group percentages, and you shouldn't be multiplying the three fractions.

This probability will be a weighted average of the accident rates in the three groups.
The probability is (accident rate in low-risk group * relative proportion of low-risk group) + (accident rate in medium-risk group * relative proportion of medium-risk group) + (accident rate in high-risk group * relative proportion of high-risk group)

 

[shawnz1102]

Thank you so much Mark!

When I encounter a problem like this, how do i know that it will be a weighted average problem?

 

[Mark44]

I wouldn't characterize it as a "weighted average problem"; I was just describing the expression I got. It's basically a probability problem with three mutually exclusive groups: low-risk, medium-risk, and high-risk drivers, and the relative probabilities of each group.

 

[shawnz1102]

Got it, thanks!