Questions: Use the data in the following table, which lists drive-thru order accuracy at popular fast food chains. Assume that orders are randomly selected from those included in the table. Drive-thru Restaurant A B C D Order Accurate 328 268 247 149 Order Not Accurate 31 58 34 12 If two orders are selected, find the probability that they are both from Restaurant D. a. Assume that the selections are made with replacement. Are the events independent? b. Assume that the selections are made without replacement. Are the events independent?

Solution Steps

To find the probability of selecting two orders from Restaurant D with replacement, we first calculate the probability of selecting one order from Restaurant D:

\[ P(D) = \frac{149}{1127} \approx 0.1323 \]

Since the selections are made with replacement, the probability of selecting two orders from Restaurant D is:

\[ P(X = 2) = P(D) \times P(D) = \left(\frac{149}{1127}\right)^2 \approx 0.0204 \]

Next, we calculate the probability of selecting two orders from Restaurant D without replacement using the hypergeometric distribution. The formula for the hypergeometric probability is:

\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]

Where:

• \(N = 1127\) (total orders)
• \(K = 149\) (orders from Restaurant D)
• \(n = 2\) (number of orders drawn)
• \(k = 2\) (number of orders from Restaurant D drawn)

Substituting the values, we have:

\[ P(X = 2) = \frac{\binom{149}{2} \binom{978}{0}}{\binom{1127}{2}} = \frac{11076}{634878} \approx 0.0203 \]

• With Replacement: The events are independent because the selection of the first order does not affect the probability of selecting the second order.

• Without Replacement: The events are not independent because the selection of the first order affects the probability of selecting the second order.

Final Answer