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conditional prob

Posted on 2013-06-05
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Last Modified: 2013-06-06
Hi,

I know the answer but I cant work this out.  It should be a conditional prob question


For a particular petrol station 30% of customers buy ‘super’ and 60% buy unleaded and
10% diesel. When a customer buys super there is a 25% chance they will fill the tank.
Customers buying unleaded have a 20% chance they will fill the tank. Of those buying
diesel, 70% fill their tank.

Given that a car leaving the petrol station has a full tank, what is the probability that
the tank contains unleaded petrol?
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Question by:jagguy
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by:pony10us
ID: 39222419
First let me ask if I have the correct answer since you say you already know.  Is it 12% chance?

Let's say there are 10 people.

10 * 30% = 3 people buy super
10 * 60% = 6 people buy unleaded
10 * 10% = 1 person buys diesel

6 * 20% = 1.2 people that buy unleaded will fill their tank

1.2 / 10 = .12 (12%) probability that the car leaving will with a full tank will have unleaded.

Does this follow with what you have for an answer?
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by:TommySzalapski
ID: 39222420
Draw it out as a grid.

                  full         not full
super      .3*.25        .3*.75
unlead    
diesel

Now fill out the rest of the table.

If you add up the first column, you will have the total probability of a full tank. What percent of that is from unleaded?
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by:d-glitch
ID: 39222434
The best way to do these problems is to draw the full probability tree.

The top three-way branch is 30, 60, 10

Each of those branches twice into Fill/No_Fill

That gives you six possibilities on the bottom row.

Collect the Fill terms and scale the sum of the probabilities back to 100%.
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by:TommySzalapski
ID: 39222442
pony10us, that is the chance that any car will leave with a full tank of unleaded.

Given that a car's tank is full, the probability that it is unleaded is between 40 and 50%

I know the exact answer, but I don't want to post it until the asker does.
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Accepted Solution

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TommySzalapski earned 500 total points
ID: 39222464
Remember the probability of A given B is the probability that both are true divided by the probability that B is true

P(A|B) = P(A and B)/P(B)

So what is P(A and B) and what is P(B)?

Either the table or the tree will give you both.
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by:pony10us
ID: 39222595
Tommy,

Okay, I think I understand.  If I am correct then I may be closer to the answer.  I was only partially through the process apparently.

Been a long time since I did any math like this and it intrigued me.

:)
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by:awking00
ID: 39223133
From a layman's approach -
If 200 customers by petrol, 120 buy unleaded, 60 buy super, and 20 buy diesel.
The 120 unleaded customers fill up 24 times, the super customers 15 times, and the diesel customers 14 times. So a total of 53 customers have full tanks, 24 of which are unleaded.
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by:TommySzalapski
ID: 39223167
Yes. That is precisely the same thing that the table or tree would get you, except they use 1 (100%) instead of 200.
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by:awking00
ID: 39223191
Thanks, Tommy. I think it's just easier for us dummies to think in terms of 24, 15, and 14 people as opposed to .12, .075, and .07 people. :-)
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Author Comment

by:jagguy
ID: 39224232
the answer is .4528
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Author Closing Comment

by:jagguy
ID: 39224235
well done
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by:pony10us
ID: 39225729
jagguy

Exactly what I ended up with after Tommy corrected my logic.  :)

Thanks Tommy.

the formula I have on an excel spreadsheet is:

=(0.6*0.2)/((0.3*0.25)+(0.6*0.2)+(0.1*0.7))

I imagine there is a simpler method though.
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