Posted by
**Cari** on
.

70% of all burger chain stores decided to advertise in their local newspaper. Of those chain stores that advertised in thier local newspapers, 60% had an increase in sales. Of those chain stores that did not advertise, 25% had an increase in sales. What is the probability that a randomly selected store with an increase in sales advertised in its local newspaper?

is 28/33 correct?

0.7000 advertised in local newspapers.

0.4200 (60%) advertised and had increased sales

0.2800 advertised and did not have incresed sales

0.3000 did not advertise

0.2250 did not advertise and had no increase

0.0750 did not advertise and had an increase in sales.

The fraction that had an increase in sales was 0.075 + 0.4200 = 0.495. The fraction that had advertised was 0.42. All fractions refer to the total population.

I also get answer to be 420/495 = 28/33