If $3$ dice are rolled, what is the probability that two show the same number, but the third is different?
Is my approach correct?
$$\left(\frac6{36}\times\frac5{36}\right)$$
There are $6$ possible ways that both can be the same number for the first 2 dice, and there are $5$ different ways for the third to be a different one.
On
A way to approach problems like this is in steps. The probability of rolling 3 dice and getting no dice in common is $\frac{C(6,3)}{6^3}$.
So rolling 3 dice and getting at least two dice in common is $1 - \frac{C(6,3)}{6^3} = \frac{6^3-C(6,3)}{6^3} $.
However this over counts as it includes the case where the three are identical. So we can subtract those 6 cases. $\frac{(6^3-C(6,3)) - 6}{6^3} $
OK you want to know the changes that, if you throw three dice, you will get two showing the same number and the third showing a different number? Well there are $6 \times 6 \times 6=216$ ways that the three dice could fall. Of those, there are 6 in which all three are equal (to 1,2,3,4,5 or 6). The number of ways for them to be totally unequal is $6 \times 5 \times 4=120$. So that leaves us with $216-120-6=90$ ways to have exactly two equal to each other. Another way to think of this is to assume that we have thrown the first dice. There are six ways to do this. Now there are two possibilities for the second: either it is the same or different. Let us say it is the same then there are five ways for the last dice to be different this gives us 30 options. If the second dice is different to the first, there are five ways for this to happen and now two ways for the third dice to be equal to one of the first two (it can be equal to the first or the second). This gives us a total of 60 options (if the second dice is different to the first) and 30 if it is the same. Adding these two together gives us 90 options. So the probability of having two of the dice equal to each other and the other different is $90/216 = 15/36 = 5/12$. It's just under half so actually quite likely. Having three equal is unlikely $1/36$ and three different is the most likely $120/216=20/36=5/9$ is more than $50$ percent.