Math Problem Statement

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

Let's analyze the table you provided and the questions that follow.

The table gives the number of accurate and inaccurate orders for four different drive-thru restaurants (A, B, C, and D).

Drive-thru RestaurantABCD
Order Accurate315276230141
Order Not Accurate35523615

We are asked to find the probability that two randomly selected orders are both from Restaurant D, and answer two sub-questions:

  1. For selections with replacement (and whether the events are independent).
  2. For selections without replacement (and whether the events are independent).

Total Orders for Each Restaurant

  • Restaurant D accurate orders: 141
  • Restaurant D inaccurate orders: 15
  • Total orders from Restaurant D = 141 + 15 = 156

The total number of orders across all restaurants is:

315+35+276+52+230+36+141+15=1100315 + 35 + 276 + 52 + 230 + 36 + 141 + 15 = 1100

Part a: With Replacement

If two orders are selected with replacement, the probability that both are from Restaurant D is:

P(both from D)=P(first from D)×P(second from D)=(1561100)×(1561100)P(\text{both from D}) = P(\text{first from D}) \times P(\text{second from D}) = \left(\frac{156}{1100}\right) \times \left(\frac{156}{1100}\right)

P(both from D)=(1561100)2=2433612100000.0201P(\text{both from D}) = \left(\frac{156}{1100}\right)^2 = \frac{24336}{1210000} \approx 0.0201

Since we are selecting with replacement, the two selections are independent events.

Part b: Without Replacement

If two orders are selected without replacement, the probability that both are from Restaurant D is:

P(first from D)=1561100P(\text{first from D}) = \frac{156}{1100} P(second from D)=1551099P(\text{second from D}) = \frac{155}{1099}

Thus, the probability that both orders are from D is:

P(both from D)=1561100×1551099=156×1551100×1099=2418012089000.0200P(\text{both from D}) = \frac{156}{1100} \times \frac{155}{1099} = \frac{156 \times 155}{1100 \times 1099} = \frac{24180}{1208900} \approx 0.0200

Since we are selecting without replacement, the two selections are not independent, because the second probability changes based on the outcome of the first selection.


Let me know if you'd like further clarification or have more questions! Here are five related questions for further exploration:

  1. What is the probability of selecting two orders from Restaurant A with replacement?
  2. What is the probability of selecting one accurate order and one inaccurate order from Restaurant D with replacement?
  3. How would the probability change if three orders were selected from Restaurant D with replacement?
  4. Can you calculate the probability of selecting one order from Restaurant A and one from Restaurant B without replacement?
  5. What would be the probability of selecting one order from Restaurant D and another from any other restaurant without replacement?

Tip:

When events are independent, the outcome of one event doesn't affect the probability of the next. When they are dependent, the outcome of one event influences the probability of subsequent events.

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Math Problem Analysis

Mathematical Concepts

Probability
Independent Events
Dependent Events

Formulas

P(both from D with replacement) = (156/1100) * (156/1100)
P(both from D without replacement) = (156/1100) * (155/1099)

Theorems

Law of Multiplication for Independent Events
Law of Multiplication for Dependent Events

Suitable Grade Level

Grades 10-12