Each of the digits 0-9 is written on a slip of paper, and the slips are placed in a hat. If 2 slips of paper are selected at random, determine the probability that the 2 numbers selected are greater than 3. The problem is to be done without replacement. Use combinations to determine the probability. The probability is (Type an integer or a simplified fraction.)

Algebra and Trigonometry (6th Edition)
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ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Problem Statement

Each of the digits 0-9 is written on a slip of paper, and the slips are placed in a hat. If 2 slips of paper are selected at random, determine the probability that the 2 numbers selected are greater than 3.

The problem is to be done without replacement. Use combinations to determine the probability.

The probability is ____ . (Type an integer or a simplified fraction.)

---

**Explanation:**

To solve this problem, we need to calculate the probability that both numbers selected from the hat are greater than 3. Let's break down the steps:

1. **Identify the total number of possible outcomes:**
   - There are 10 digits (0-9) written on slips and placed in a hat.
   - We are selecting 2 slips out of these 10, so the number of possible combinations (total outcomes) is given by the combination formula \( \binom{n}{k} \), where \( n \) is the total number of items, and \( k \) is the number of items to choose.
   - Total combinations = \( \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \).

2. **Identify the favorable number of outcomes:**
   - We need to find the number of combinations that yield numbers greater than 3. The digits greater than 3 are 4, 5, 6, 7, 8, and 9 (a total of 6 digits).
   - We are selecting 2 slips out of these 6, so the number of favorable combinations is \( \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \).

3. **Calculate the probability:**
   - The probability of selecting 2 numbers both greater than 3 is the number of favorable outcomes divided by the total number of possible outcomes:
   \[
   \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{15}{45} = \frac{1}{3}
   \]

So the probability is \( \frac{1}{3} \).
Transcribed Image Text:### Problem Statement Each of the digits 0-9 is written on a slip of paper, and the slips are placed in a hat. If 2 slips of paper are selected at random, determine the probability that the 2 numbers selected are greater than 3. The problem is to be done without replacement. Use combinations to determine the probability. The probability is ____ . (Type an integer or a simplified fraction.) --- **Explanation:** To solve this problem, we need to calculate the probability that both numbers selected from the hat are greater than 3. Let's break down the steps: 1. **Identify the total number of possible outcomes:** - There are 10 digits (0-9) written on slips and placed in a hat. - We are selecting 2 slips out of these 10, so the number of possible combinations (total outcomes) is given by the combination formula \( \binom{n}{k} \), where \( n \) is the total number of items, and \( k \) is the number of items to choose. - Total combinations = \( \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \). 2. **Identify the favorable number of outcomes:** - We need to find the number of combinations that yield numbers greater than 3. The digits greater than 3 are 4, 5, 6, 7, 8, and 9 (a total of 6 digits). - We are selecting 2 slips out of these 6, so the number of favorable combinations is \( \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \). 3. **Calculate the probability:** - The probability of selecting 2 numbers both greater than 3 is the number of favorable outcomes divided by the total number of possible outcomes: \[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{15}{45} = \frac{1}{3} \] So the probability is \( \frac{1}{3} \).
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