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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.)

 Apr 28, 2022

Best Answer 

 #1
avatar+2455 
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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:  \(\large{{1 \over 4} \times {1 \over 4} \times {3 \over 4} \times { 3 \over 4} \times {4 \choose 2}} = \color{brown}\boxed{7 \over 128}\)

 Apr 28, 2022
 #1
avatar+2455 
+1
Best Answer

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:  \(\large{{1 \over 4} \times {1 \over 4} \times {3 \over 4} \times { 3 \over 4} \times {4 \choose 2}} = \color{brown}\boxed{7 \over 128}\)

BuilderBoi Apr 28, 2022

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