2. 32% of drivers involved in an accident in one year will not be involved in an accident in the following year.
10% of drivers are not involved in an accident will be involved in an accident in the following year.
a. If 8% of all drivers had an accident one year, what is the probability that a driver, picked at random, has an accident in the following year?

Question

2.	32% of drivers involved in an accident in one year will not be involved in an accident in the following year.
10% of drivers are not involved in an accident will be involved in an accident in the following year.
a.	If 8% of all drivers had an accident one year, what is the probability that a driver, picked at random, has an accident in the following year?

turingtest · Answer

let $A_1$ be the event that a driver has an accident in some year, and $A_2$ be the event that they have one the following year

turingtest · Answer

we are told that $$P(A_2^c|A_1)=0.32$$and$$P(A_2|A_1^c)=0.1$$

turingtest · Answer

since we can start counting at any year as year 1, we can also say that for a random person, $P(A_1)=0.08$

turingtest · Answer

this gives the following diagram|dw:1396035258466:dw|

turingtest · Answer

the probability is then the sum of the disjoint paths|dw:1396035475107:dw|

turingtest · Answer

am i making any sense?

rocklee88 · Answer

Well, our doctor in the university has another way of solving this kind of scenarios..

Please see the posted file for a similar question which is answered by the doctor.

I tried to solve the question with same format, but not sure if my answer is correct or not..
I will post both files now.

rocklee88 · Answer

Here is the similar question:

rocklee88 · Answer

Here is my answer for the first question posted, answerd it with same format as doctor did for the similar one.

rocklee88 · Answer

Of course, I appreciate your effort though,
So, what do you think?

turingtest · Answer

It's very strange to me to see this problem solved with what appears to be graph theory, but the answer you got is different than mine, which is $$P(A_2)=P(A_1)P(A_2|A_1)+P(A_1^c)P(A_2|A^c_1)=0.08\cdot0.68+0.92\cdot0.1=0.1416$$

turingtest · Answer

I don't understand why in your diagram you have 10% as the chance of an accident in year 1

rocklee88 · Answer

I am not sure if my answer is true or not, as I said earlier, I answered the question, the same way as doctor did with the similar question.

The doctor asked us in class to solve the question by ourselves first, and my answer was similar to yours in your diagram. He did that to see how we would think then he showed us how to solve it in the way you saw in first posted pic.

any way in your second diagram, you wrote "not A1and then not A1 and not A2"!!
while the question says: "10% of drivers are not involved in an accident will be involved in an accident in the following year"
I think here is your mistake.

turingtest · Answer

|dw:1396104335909:dw|you disagree with this diagram?

rocklee88 · Answer

In the second diagram, you marked with black that "A1 then A1 and not A2" ,well that is ok, but then you chose "not A2 and then not A1 and not A2" <- I think here is the mistake, and where you calculated wrongly.

turingtest · Answer

oh right, I see

turingtest · Answer

but in my calculation I followed the right paths

turingtest · Answer

|dw:1396105446898:dw|

rocklee88 · Answer

^_^" I don't know man, maybe your calculation is correct, maybe I am not familiar with your format.
Anyway tomorrow I have math class, I will check with the doctor ^_^"
,and when I get the answer I will post it here.

turingtest · Answer

I'd appreciate that, as I'm still taking probability, too. Good luck!