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13 Oct 2024, 08:34
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What is the probability of rolling four fair six-sided dice and getting exactly two 6s and two 1s?
A. \(\frac{1}{6^3}\)
B. \(\frac{1}{6^2*3^2}\)
C. \(\frac{1}{2*6^3}\)
D. \(\frac{1}{3*6^3}\)
E. \(\frac{1}{6^4}\)
A. \(\frac{1}{6^3}\)
B. \(\frac{1}{6^2*3^2}\)
C. \(\frac{1}{2*6^3}\)
D. \(\frac{1}{3*6^3}\)
E. \(\frac{1}{6^4}\)
13 Oct 2024, 08:34
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Official Solution:
What is the probability of rolling four fair six-sided dice and getting exactly two 6s and two 1s?
A. \(\frac{1}{6^3}\)
B. \(\frac{1}{6^2*3^2}\)
C. \(\frac{1}{2*6^3}\)
D. \(\frac{1}{3*6^3}\)
E. \(\frac{1}{6^4}\)
The probability of obtaining two 6s and two 1s is \(\frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{4!}{2! * 2!}=\frac{1}{6^3}\). We multiply by \( \frac{4!}{2! * 2!}\) to account for all possible arrangements of (6, 6, 1, 1) across the four dice:
Each of these \( \frac{4!}{2! * 2!}=6\) cases has a probability of \(\frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{1}{6}\). Since there are 6 such arrangements, the total probability is:
\(\frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{4!}{2! * 2!}=\frac{1}{6^3}\)
Answer: A
What is the probability of rolling four fair six-sided dice and getting exactly two 6s and two 1s?
A. \(\frac{1}{6^3}\)
B. \(\frac{1}{6^2*3^2}\)
C. \(\frac{1}{2*6^3}\)
D. \(\frac{1}{3*6^3}\)
E. \(\frac{1}{6^4}\)
The probability of obtaining two 6s and two 1s is \(\frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{4!}{2! * 2!}=\frac{1}{6^3}\). We multiply by \( \frac{4!}{2! * 2!}\) to account for all possible arrangements of (6, 6, 1, 1) across the four dice:
| Die 1 | Die 2 | Die 3 | Die 4 |
| 6 | 6 | 1 | 1 |
| 6 | 1 | 6 | 1 |
| 6 | 1 | 1 | 6 |
| 1 | 6 | 6 | 1 |
| 1 | 6 | 1 | 6 |
| 1 | 1 | 6 | 6 |
Each of these \( \frac{4!}{2! * 2!}=6\) cases has a probability of \(\frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{1}{6}\). Since there are 6 such arrangements, the total probability is:
\(\frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{4!}{2! * 2!}=\frac{1}{6^3}\)
Answer: A
Slimeto21
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Schools: Goizueta '26 (A$$) Haas '26 (D) Kellogg '26 (D) Ross '26 (WL) Stanford '26 (D) Wharton '26 (D)
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15 Mar 2025, 19:16
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I did not quite understand the solution. Wondering how you use 4C2 to account for the all possible arrangements?
