Suppose a certain combination lock requires three selections of numbers, each from 0 through 34. What is the probability that a randomly selected combination has a repeated number? Give your answer as a decimal probability, not a percentage. Give your answer to exactly two decimal points.

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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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**Problem Statement:**

Suppose a certain combination lock requires three selections of numbers, each from 0 through 34. What is the probability that a randomly selected combination has a repeated number? Give your answer as a decimal probability, not a percentage. Give your answer to exactly two decimal points.

**Your Answer:**

[Text box for answer entry]

---

**Explanation:**

This problem asks for the probability of a repeated number in a three-number sequence, each selected from the range 0 to 34. To solve this:

1. **Total Possible Combinations:** Each of the three selections can be any number from 0 to 34. Therefore, the total number of combinations is \(35 \times 35 \times 35 = 42,875\).

2. **Combinations with No Repeats:** For no repeated numbers, the first number can be any of the 35, the second any of the remaining 34, and the third any of the remaining 33. Thus, there are \(35 \times 34 \times 33 = 39,270\) combinations with no repeats.

3. **Combinations with Repeats:** Subtract the no-repeat combinations from the total combinations: \(42,875 - 39,270 = 3,605\).

4. **Probability Calculation:** The probability of a repeated number is the number of combinations with repeats divided by the total number of combinations: \(\frac{3,605}{42,875}\).

5. **Decimal Probability:** Perform the division to obtain the decimal probability, rounding to two decimal points.
Transcribed Image Text:**Problem Statement:** Suppose a certain combination lock requires three selections of numbers, each from 0 through 34. What is the probability that a randomly selected combination has a repeated number? Give your answer as a decimal probability, not a percentage. Give your answer to exactly two decimal points. **Your Answer:** [Text box for answer entry] --- **Explanation:** This problem asks for the probability of a repeated number in a three-number sequence, each selected from the range 0 to 34. To solve this: 1. **Total Possible Combinations:** Each of the three selections can be any number from 0 to 34. Therefore, the total number of combinations is \(35 \times 35 \times 35 = 42,875\). 2. **Combinations with No Repeats:** For no repeated numbers, the first number can be any of the 35, the second any of the remaining 34, and the third any of the remaining 33. Thus, there are \(35 \times 34 \times 33 = 39,270\) combinations with no repeats. 3. **Combinations with Repeats:** Subtract the no-repeat combinations from the total combinations: \(42,875 - 39,270 = 3,605\). 4. **Probability Calculation:** The probability of a repeated number is the number of combinations with repeats divided by the total number of combinations: \(\frac{3,605}{42,875}\). 5. **Decimal Probability:** Perform the division to obtain the decimal probability, rounding to two decimal points.
Suppose 12 customers at a restaurant ordered an appetizer, 9 customers ordered a main dish, and 3 customers ordered an appetizer and a main dish. How many customers ordered an appetizer or a main dish?

Your Answer:

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Transcribed Image Text:Suppose 12 customers at a restaurant ordered an appetizer, 9 customers ordered a main dish, and 3 customers ordered an appetizer and a main dish. How many customers ordered an appetizer or a main dish? Your Answer: [Answer box for students to fill in]
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