Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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![**Problem Statement:**
A die is thrown twice. What is the probability that a 3 will result the first time and a 5 the second time? (Assume the die is six-sided with each side numbered one through six. Enter your probabilities as a fraction.)
**Answer Box:** [Text box for entering the probability fraction]
**Explanation:**
When a six-sided die is thrown, each number (1 through 6) has an equal chance of appearing. Therefore, for any single roll, the probability of rolling a specific number, such as 3, is \( \frac{1}{6} \).
The same logic applies to rolling a 5 on the second throw. Both events are independent, meaning the outcome of the first roll does not affect the second.
To find the probability of two independent events both occurring, you multiply their probabilities:
\[ P(\text{rolling a 3 first and a 5 second}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \]
So, the probability is \(\frac{1}{36}\).](https://content.bartleby.com/qna-images/question/8533e211-c3b0-4c2c-a0a5-c3ff9e21dbc2/8eabad97-d40e-47a3-b4e8-4540796585ef/qf1snzo_processed.png)
Transcribed Image Text:**Problem Statement:**
A die is thrown twice. What is the probability that a 3 will result the first time and a 5 the second time? (Assume the die is six-sided with each side numbered one through six. Enter your probabilities as a fraction.)
**Answer Box:** [Text box for entering the probability fraction]
**Explanation:**
When a six-sided die is thrown, each number (1 through 6) has an equal chance of appearing. Therefore, for any single roll, the probability of rolling a specific number, such as 3, is \( \frac{1}{6} \).
The same logic applies to rolling a 5 on the second throw. Both events are independent, meaning the outcome of the first roll does not affect the second.
To find the probability of two independent events both occurring, you multiply their probabilities:
\[ P(\text{rolling a 3 first and a 5 second}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \]
So, the probability is \(\frac{1}{36}\).
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