# Probablity Question

1. Feb 25, 2009

### mkir

Hi, I need some help with this question.

Seventy percent of all Big Burger chain stores decided to advertise in their local newspapers. Of those chain stores that advertised in their local newspapers, 60% had an increase in sales. Of those chain stores that did not advertise in their local newspapers, 25% had an increase in sales.
(a) What is the probability that a randomly selected store has an increase in sales.
(b) What is the probability that a randomly selected store with an increase in sales advertised in its local newspaper?

for a)
I did
(.7)(.6) + (.3)(.25) = 0.495

for b)
A = Increased sales
B = Advertised

P(A|B) = 0.495/0.7 = 0.707

Did I do this right?

2. Feb 26, 2009

### jambaugh

a.) Looks right.
b.) It looks like you have A & B reversed.
The question asks for the probability one advertised given one had an increase in sales. So that should be P(B|A) as you defined them.

3. Feb 26, 2009

### mkir

So for b), would

P(B|A) = (.7)(.6)/(.495) = 0.848

be correct?

4. Feb 27, 2009

### regor60

yes it is

5. Feb 27, 2009

### HallsofIvy

Another way to do (b) is to imagine that there are 1000 stores. Then 70% of them, or 700, advertise. Of those, 60%, or 420, have an increase in sales. 300 stores do not advertise and 25% of them, 75, also have an increase in sales. So a total of 420+ 75= 495 stores have an increase in sales and 420 advertised: that is, 420/495= .848 (approximately) or 84.8% of the stores that had an increase in sales advertised.

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