Conditional Probability

[sunny1324]

I had an exam today and I'm pretty sure I did ok on it except one question.

The problem went something like this, I don't remember the except numbers but...

A store is selling 4 types of VCR; V1, V2, V3 and V4. 40% buy a V1, 30% buy a V2, 20% buy a V3, and 10% buy a V4. The store will fix your VCR if it breaks in less than a year. The probability that a V1 will break is 0.10, V2 0.08, V3 0.05 and a V4 is 0.02. A man buys a VCR. What is the probability that the VCR he buys breaks and is not a V1?


So the way I did it:

I created a tree...

-------- 0.4 V1 ------------ 0.10
---------0.3 V2---------------0.08
----------0.2 V3-----------0.05
------------0.1 V3 ----------0.02

[(0.3)(0.08) + (0.2)(0.05) + (0.1)(0.02)] / [(0.4)(0.10) + (0.3)(0.08) + (0.2)(0.05) + (0.1)(0.02)] = 0.4736


was I even close to getting it right?

 

[galactus]

A little thing I like to do with these is to build a table. Then, you can answer any question they through at you.

First, assume a nice number of VCR's to work with. Say, 1000.

Now, we want 'breaks' AND 'not a V1'

Can you see it better now?. Add them up.

 

[sunny1324]

So, the total that breaks is 250

and since it's not a V1, would you add up V2, V3, and V4? So that would be 600.


is the answer 250/600?

is that what you mean?