Question

At a local coffee shop, the manager has determined that 56% of drink orders are for specialty espresso drinks and 44% are for plain coffee. The manager also noted that 40% of customers order food. For customers who purchase the specialty espresso drinks, 35% also purchase a food item, and for customers who purchase plain coffee, 30% also purchase a food item. What is the probability that a randomly chosen customer will purchase a specialty espresso drink and a food item?

0.13
0.14
0.20
0.22

Answer 1

Let's use conditional probability to solve this problem. We want to find the probability of a customer purchasing a specialty espresso drink and a food item, so we can use the following formula:

P(specialty espresso drink and food) = P(food|specialty espresso drink) * P(specialty espresso drink)

We know that P(specialty espresso drink) = 0.56 and P(food|specialty espresso drink) = 0.35, so we can substitute these values into the formula:

P(specialty espresso drink and food) = 0.35 * 0.56

P(specialty espresso drink and food) = 0.196

Therefore, the probability that a randomly chosen customer will purchase a specialty espresso drink and a food item is 0.196 or approximately 0.20 (rounded to two decimal places).

So, the answer is option C: 0.20.