4 12-sided dice are rolled. What is the probability that the number of dice showing a two digit number is equal to 2? Express your answer as a common fraction. (Assume that the numbers on the 12 sides are the numbers from 1 to 12 expressed in decimal.)
+2666 3 numbers have 2 digits (10, 11, 12)
The probability of getting a 2 digit number is \({1 \over 4}\)
We need 2 rolls to satisfy this condition, so the probability of this is \({1 \over 4} ^2\)
We also need 2 rolls to not have 2 digits. The probability of this is \({3 \over 4}^2\)
There are \(4 \choose 2 \) ways to choose which rolls are \(\ge 10 \).
Thus, the probability is: \(\large{{1 \over 4} \times {1 \over 4} \times {3 \over 4} \times { 3 \over 4} \times {4 \choose 2}} = \color{brown}\boxed{7 \over 128}\)
+2666 3 numbers have 2 digits (10, 11, 12)
The probability of getting a 2 digit number is \({1 \over 4}\)
We need 2 rolls to satisfy this condition, so the probability of this is \({1 \over 4} ^2\)
We also need 2 rolls to not have 2 digits. The probability of this is \({3 \over 4}^2\)
There are \(4 \choose 2 \) ways to choose which rolls are \(\ge 10 \).
Thus, the probability is: \(\large{{1 \over 4} \times {1 \over 4} \times {3 \over 4} \times { 3 \over 4} \times {4 \choose 2}} = \color{brown}\boxed{7 \over 128}\)
