Elias has a jar of 15 black marbles, 8 red marbles, and 9 white marbles. He will randomly choose a marble, record its color, and replace it in the jar. He will then randomly choose a second marble and record its color. 15/24 What is the probability of choosing a black marble and then a red marble? 9/128 135/1024 15/128
Elias has a jar of 15 black marbles, 8 red marbles, and 9 white marbles. He will randomly choose a marble, record its color, and replace it in the jar. He will then randomly choose a second marble and record its color. 15/24 What is the probability of choosing a black marble and then a red marble? 9/128 135/1024 15/128
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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![### Probability Exercise: Drawing Marbles
Elias has a jar containing 15 black marbles, 8 red marbles, and 9 white marbles. He will perform the following experiment twice:
1. Randomly choose a marble from the jar.
2. Record its color.
3. Replace the marble back into the jar.
He will then choose a second marble and record its color.
#### Question:
What is the probability of choosing a black marble first and then a red marble on the second draw?
### Answer Options:
- \( \frac{15}{24} \)
- \( \frac{9}{128} \)
- \( \frac{135}{1024} \)
- \( \frac{15}{128} \)
To find the correct answer, it is important to note:
1. The total number of marbles is \( 15 + 8 + 9 = 32 \).
2. The probability of drawing a black marble first is \( \frac{15}{32} \).
3. Since the marble is replaced, the second draw also consists of 32 marbles.
4. The probability of drawing a red marble next is \( \frac{8}{32} \).
Thus, the combined probability:
\[ P(\text{Black, then Red}) = P(\text{Black}) \times P(\text{Red}) \]
\[ P(\text{Black, then Red}) = \frac{15}{32} \times \frac{8}{32} = \frac{120}{1024} = \frac{15}{128} \]
### Final Answer:
\[ P(\text{Black, then Red}) = \frac{15}{128} \]
Hence, the correct option is:
- \( \boxed{\frac{15}{128}} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b5161d0-8c1c-4aef-9d6c-2e647d1b4222%2Fb36f1304-4a4e-4dda-b17f-21be68d2bfa6%2Fdrauday_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability Exercise: Drawing Marbles
Elias has a jar containing 15 black marbles, 8 red marbles, and 9 white marbles. He will perform the following experiment twice:
1. Randomly choose a marble from the jar.
2. Record its color.
3. Replace the marble back into the jar.
He will then choose a second marble and record its color.
#### Question:
What is the probability of choosing a black marble first and then a red marble on the second draw?
### Answer Options:
- \( \frac{15}{24} \)
- \( \frac{9}{128} \)
- \( \frac{135}{1024} \)
- \( \frac{15}{128} \)
To find the correct answer, it is important to note:
1. The total number of marbles is \( 15 + 8 + 9 = 32 \).
2. The probability of drawing a black marble first is \( \frac{15}{32} \).
3. Since the marble is replaced, the second draw also consists of 32 marbles.
4. The probability of drawing a red marble next is \( \frac{8}{32} \).
Thus, the combined probability:
\[ P(\text{Black, then Red}) = P(\text{Black}) \times P(\text{Red}) \]
\[ P(\text{Black, then Red}) = \frac{15}{32} \times \frac{8}{32} = \frac{120}{1024} = \frac{15}{128} \]
### Final Answer:
\[ P(\text{Black, then Red}) = \frac{15}{128} \]
Hence, the correct option is:
- \( \boxed{\frac{15}{128}} \)
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