At least two people roll the same number?
- Thread starter amy-lee
- Start date
Hello, amy-lee!
The opposite of "at least two roll the same number" is "no one rolls the same number."
We will find the probability that all four players roll differerent numbers
\(\displaystyle \text{Player 1 can roll any number; it doesn't matter.}\)
\(\displaystyle \text{Player 2 must roll a number different from 1: }\
= \frac{5}{6}\)
\(\displaystyle \text{Player 3 must roll a number different from 1 and 2: }\ = \frac{4}{6}\)
\(\displaystyle \text{Player 4 must roll a number different from 1, 2, and 3: }\ = \frac{3}{6}\)
\(\displaystyle \text{The probability that they roll different numbers is: }\:\frac{5}{6}\cdot\frac{4}{6}\cdot\frac{3}{6} \:=\:\frac{5}{18}\)
\(\displaystyle \text{Therefore, the probability that at least two roll the same number is: }\;1 - \frac{5}{18} \;=\;\boxed{\frac{13}{18}}\)
The opposite of "at least two roll the same number" is "no one rolls the same number."
We will find the probability that all four players roll differerent numbers
\(\displaystyle \text{Player 1 can roll any number; it doesn't matter.}\)
\(\displaystyle \text{Player 2 must roll a number different from 1: }\
= \frac{5}{6}\)\(\displaystyle \text{Player 3 must roll a number different from 1 and 2: }\
= \frac{4}{6}\)\(\displaystyle \text{Player 4 must roll a number different from 1, 2, and 3: }\
= \frac{3}{6}\)\(\displaystyle \text{The probability that they roll different numbers is: }\:\frac{5}{6}\cdot\frac{4}{6}\cdot\frac{3}{6} \:=\:\frac{5}{18}\)
\(\displaystyle \text{Therefore, the probability that at least two roll the same number is: }\;1 - \frac{5}{18} \;=\;\boxed{\frac{13}{18}}\)