Solving Probability Questions: Big Burger Chain Stores

[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?

 

[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.

 

[mkir]

So for b), would

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

be correct?

 

[regor60]

So for b), would

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

be correct?

yes it is

 

[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.