We roll five six-sided fair dice. a) What is the probability that we get a three dice of one face and two of a second face? (For example, an outcome like 55533.) Explain your counting process. b) What is the probability that the dice all show different faces? Explain your counting process.
a) I worked the problem, but it felt so simple that I don't trust it. For $5$ rolls, you have $6$ options, so the number of possible outcomes is $6^5$. For the first roll, you have $6C1$ choices, for the second and third roll, $1C1$ choice. For the fourth roll you have $5C1$ choices, and then $1C1$ for the fifth roll, Making the probability $(6*5)/6^5$?
b) This one I'm more confident in. Number of possible out comes is still $6^5$, with us having one less option with each roll. $(6*5*4*3*2)/6^5$
For a) think about this: how many pairs there are to make a roll of the kind $aaabb$, that is, how many pairs of $ab$? There are $6\cdot 5=30$ kinds of these pairs. However each throw of the kind $aaabb$ can appear in any order (by example as $ababa$ or $abbaa$), that is, for each pair $ab$ there are $\binom52=10$ different ways that a throw can be ordered.
Thus the total probability is $30\cdot10\cdot \frac1{6^5}$
For the part b) it happen similarly: you want to choose $5$ different numbers of $6$, and these numbers can appear in any order, that is, it can appear as $abcde$ or $acdeb$. How many distinct groups of $5$ numbers, taken from $6$ exists? there are $\binom65=6$, and each one can appear in $5!$ different ways, then the probability is $6\cdot 5!\cdot\frac1{6^5}$.
In part b) you have the same result than me but it is not clear if your reasoning is correct or not.