A bag contains 5 red marbles, 6 blue marbles and 3 green marbles. If three marbles are drawn out of the bag, what is the exact probability that all three marbles drawn will be red?
A bag contains 5 red marbles, 6 blue marbles and 3 green marbles. If three marbles are drawn out of the bag, what is the exact probability that all three marbles drawn will be red?
Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
ChapterCSR: Contents Of Student Resources
Section: Chapter Questions
Problem 11.22EP
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![### Probability Question
**Question:**
A bag contains 5 red marbles, 6 blue marbles, and 3 green marbles. If three marbles are drawn out of the bag, what is the **exact** probability that all three marbles drawn will be red?
**Answer:**
[Input box]
<button>Submit Answer</button>
**Attempts:** 2 out of 2 remaining
---
**Explanation:**
To solve this problem, we need to find the probability that all three drawn marbles are red. This involves calculating the number of favorable outcomes where all marbles are red and then dividing that by the total number of possible outcomes when drawing three marbles from the bag.
The total number of marbles in the bag is:
\[ 5 (\text{red}) + 6 (\text{blue}) + 3 (\text{green}) = 14 \text{ marbles} \]
The number of ways to draw 3 red marbles from the 5 red marbles can be calculated using the combination formula \( \binom{n}{k} \), which represents the number of ways to choose \( k \) items from \( n \) items without regard to order:
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = 10 \]
The total number of ways to draw 3 marbles from 14 marbles:
\[ \binom{14}{3} = \frac{14!}{3!(14-3)!} = 364 \]
Thus, the probability \( P \) that all three drawn marbles are red is:
\[ P = \frac{\binom{5}{3}}{\binom{14}{3}} = \frac{10}{364} = \frac{5}{182} \]
So the exact probability that all three marbles drawn will be red is:
\[ \frac{5}{182} \]
Feel free to input your answer and submit it using the "Submit Answer" button.
**Note:**
- Confirm that you understand how to use combinations to solve probability problems.
- Ensure all calculations are correct before submitting your answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48f1f7be-b80e-4859-806c-a5bfecf08176%2F39a0ce29-21f2-4c78-9c33-873beeb1ac71%2Fa93x2cb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability Question
**Question:**
A bag contains 5 red marbles, 6 blue marbles, and 3 green marbles. If three marbles are drawn out of the bag, what is the **exact** probability that all three marbles drawn will be red?
**Answer:**
[Input box]
<button>Submit Answer</button>
**Attempts:** 2 out of 2 remaining
---
**Explanation:**
To solve this problem, we need to find the probability that all three drawn marbles are red. This involves calculating the number of favorable outcomes where all marbles are red and then dividing that by the total number of possible outcomes when drawing three marbles from the bag.
The total number of marbles in the bag is:
\[ 5 (\text{red}) + 6 (\text{blue}) + 3 (\text{green}) = 14 \text{ marbles} \]
The number of ways to draw 3 red marbles from the 5 red marbles can be calculated using the combination formula \( \binom{n}{k} \), which represents the number of ways to choose \( k \) items from \( n \) items without regard to order:
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = 10 \]
The total number of ways to draw 3 marbles from 14 marbles:
\[ \binom{14}{3} = \frac{14!}{3!(14-3)!} = 364 \]
Thus, the probability \( P \) that all three drawn marbles are red is:
\[ P = \frac{\binom{5}{3}}{\binom{14}{3}} = \frac{10}{364} = \frac{5}{182} \]
So the exact probability that all three marbles drawn will be red is:
\[ \frac{5}{182} \]
Feel free to input your answer and submit it using the "Submit Answer" button.
**Note:**
- Confirm that you understand how to use combinations to solve probability problems.
- Ensure all calculations are correct before submitting your answer.
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