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
Yes, your answer of 28/33 is correct.
To calculate the probability that a randomly selected store with an increase in sales advertised in its local newspaper, you need to find the ratio of the number of stores that both advertised and had an increase in sales to the total number of stores that had an increase in sales.
From the information given:
- 70% of all burger chain stores advertised in their local newspaper, which means 0.7 is the probability that a randomly selected store advertised.
- Of those chain stores that advertised, 60% had an increase in sales, which means 0.6 is the probability that a store had an increase in sales given that it advertised.
- Of those chain stores that did not advertise, 25% had an increase in sales, which means 0.25 is the probability that a store had an increase in sales given that it did not advertise.
To find the probability that a randomly selected store with an increase in sales advertised, we can use the law of total probability. Let A be the event that a store advertised, and B be the event that a store had an increase in sales. We are interested in calculating P(A|B), the conditional probability of A given B.
Using Bayes' theorem, P(A|B) = P(B|A) * P(A) / P(B)
P(B|A) is the probability of having an increase in sales given that the store advertised, which is 0.6.
P(A) is the probability that a store advertised, which is 0.7.
P(B) is the probability that a store had an increase in sales, which is obtained by summing the probabilities of having an increase in sales for both stores that advertised and stores that did not advertise: P(B) = P(B and A) + P(B and not A) = 0.6 * 0.7 + 0.25 * 0.3.
Therefore, P(A|B) = 0.6 * 0.7 / (0.6 * 0.7 + 0.25 * 0.3) = 0.42 / (0.42 + 0.075) = 0.42 / 0.495 = 28/33.
So, the probability that a randomly selected store with an increase in sales advertised in its local newspaper is indeed 28/33.