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Here is my solution:

There are $\binom{6}{2}$ ways for two numbers to appear. On each of the 4 rolls of the die, there are two possible outcomes giving $\binom{6}{2}$$\cdot$$2^4$ possible outcomes which two of the six numbers appear. This probability is $\binom{6}{2}(\cdot2^4-2)\over 6^4$

Is this correct?

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Your denominator is correct: six outcomes and four trials gives $6^4$ possible sequences.

For the numerator, ask how many sequences do contain exactly two 5's. We can places the 5's in the sequence in $\binom{4}{2}$ ways. After that, we can fill in the remaining two positions with any non-5 roll. This gives a total of $\binom{4}{2} \cdot 5^2$ sequences containing exactly two 5's.

Putting it all together, we have a probability of $$ \frac{6^4 - \binom{4}{2} \cdot 5^2}{6^4}, $$ where we are subtracting the number of "bad" outcomes from the total number of outcomes in the numerator.

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We can work out the probability that 2 fives appear exactly 2 times and then take the compliment.

The probability that 2 chosen roles of the 4 roles are 5 and the other 2 are not is: $$ \left (\frac{1}{6} \right ) ^2 \left(\frac{5}{6}\right) ^2 $$ Then there are $ {4 \choose 2} $ ways of choosing which are the 2 chosen roles. from this we deduce your final answer to be:

$$ 1- {4 \choose 2} \left (\frac{1}{6} \right ) ^2 \left(\frac{5}{6}\right) ^2 $$

This does give a different value than your answer, but I'm not really sure where your method went wrong. I didn't really understand the wording at the beginning about 2 numbers appearing. Hope this helps and feel free to help me understand the wording! :)

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