Probability on rolling dice I am struggling to find answer for below Dice problem -
I roll a fair die 6 times.
(a) What is the probability that three of the six possible outcomes do not show up and each of the other three possible outcomes shows up two times?
(b) What is the probability that some outcome shows up at least three times?
Answer to the 1st question is given as $$\frac{\binom{6}{3}\binom{6}{2}\binom{6}{2}\binom{6}{2}}{6^6}$$
However, I failed to find any intuitive explanation to answer above 2 questions.
Can you please help me to explain them?
Any pointer will be highly appreciated.
 A: Part (a): The answer you have provided is incorrect. It should be
$$ \frac{\binom{6}{3} \binom{6}{2} \binom{4}{2} \binom{2}{2}}{6^6} .$$
Here is some explanation. There are $6^6$ total possible outcomes, so that is the denominator. The numerator should be the number of outcomes where only $3$ numbers come up during the $6$ rolls, with each number appearing twice.
There are $\binom{6}{3}$ ways to choose which $3$ outcomes show up on the dice. Say these outcomes are $a,b,c$. We want to allocate the letters $a,a,b,b,c,c$ to the six "slots"
$$\_, \_, \_, \_, \_, \_ .$$
There are $\binom{6}{2}$ ways to pick which slots to put the two $a$s in. Then there are $\binom{4}{2}$ ways to pick the slots for the $b$s and $\binom{2}{2}$ ways for the $c$s.
Part (b): Here is a hint. First, pick which outcome shows up three or more times. Then, compute the probability that this outcome appears exactly $3$ times. Then, compute the probabilities that this outcome appears exactly $4, 5, 6$ times. Finally, sum these probabilities.
To compute the probability that this outcome appears exactly $3$ times, think about first allocating slots for where the $3$ copies of that outcome appear, and then think about how to fill the three remaining slots.
(Edit: There is one tricky subtlety for the probability that the outcome appears exactly $3$ times. You have to make sure you are not overcounting the possibilities where there are two outcomes, call them $a$ and $b$, each of which appear $3$ times.)
A: (b) The probability of getting something that appears less that 3 times is $1-((\frac{1}{6})^6*6+(\frac16)^5(\frac56)*6*6+(\frac16)^4(\frac56)^2*6*15)=\frac{3685}{3888}$
