Question

1. sarah goes to a particular café every day for several months and notices that they are often out of certain items, so she starts collecting some data. she finds that when she goes to the café around 3 pm, there is a 0.08 probability that the café is out of regular coffee, a 0.11 probability that it is out of decaf coffee, and a 0.15 probability that it is out of tea. suppose the drink selections are available independent of one another. on a randomly selected day around 3 pm, what is the probability that all three drinks are available? a) less than 0.01 b) 0.304 c) 0.340 d) 0.696 e) greater than 0.99

Explanation:

Step1: Calculate the probability of each drink being available

Let \(P(R)\) be the probability that regular coffee is available, \(P(D)\) be the probability that decaf coffee is available, and \(P(T)\) be the probability that tea is available.
If the probability that the café is out of regular coffee is \(0.08\), then \(P(R)=1 - 0.08=0.92\).
If the probability that the café is out of decaf coffee is \(0.11\), then \(P(D)=1 - 0.11 = 0.89\).
If the probability that the café is out of tea is \(0.15\), then \(P(T)=1 - 0.15=0.85\).

Step2: Use the multiplication rule for independent events

Since the drink selections are independent, the probability that all three drinks are available is \(P(R\cap D\cap T)=P(R)\times P(D)\times P(T)\).
Substitute the values: \(P(R\cap D\cap T)=0.92\times0.89\times0.85\).
First, \(0.92\times0.89=(1 - 0.08)\times0.89=0.89-0.08\times0.89=0.89 - 0.0712 = 0.8188\).
Then, \(0.8188\times0.85=(0.8 + 0.0188)\times0.85=0.8\times0.85+0.0188\times0.85=0.68+0.01598 = 0.69598\approx0.696\).

Answer:

D. \(0.696\)