Probability question with 3 colored dice I have a question. A model exam paper has the following question.
Model paper question:
 1. Three fair, six-sided dice are rolled. One green, one red, and one blue. Find the probability that precisely two of the dice show the same number.
There are $6$ outcomes for the red die, $6$ outcomes for the green and $6$ outcomes for the blue. Hence by the Multiplication Principle there are $6\times 6 \times 6 = 216$ outcomes of the experiment.
Model paper answer:
There are $3$ possibilities for the die that is to show a different number and $6$ possible numbers for this die. The remaining two dice must have the same number, but different from the first die, whence there are $5$ possibilities for this number. Hence by the Multiplication Principle there are $3\times 6 \times 5 = 90$ outcomes where exactly two dice show the same number, and thus $P(\text{precisely }2\text{ the same}) = 90/216 = 5/12$.
However I don't understand the answer. I picture it in a different way.


*

*Only 2 dice can have the same number.

*One die will have $6$ outcomes. 2 other dice must have each $5$ outcomes because can't have the same number as the first die. 


So in this case. The probability of precisely 2 dice having same number is
$(6\times5\times5)/216 = 150/216$.
Why is my thinking wrong here?
 A: Your version has two errors.
$(1)\;$You didn't choose which die is the solo one. To fix it, you need a factor
of $3$.

$(2)\;$Of your two factors of $5$, the second one is incorrect. It should be $1$, not $5$, since there is no choice of color for the third die.

With those corrections, you get the product 
$$(3)(6)(5)(1) = 90$$
which matches the official answer.
A: The issue is that although the second and third dice each have $5$ outcomes, they have to be the same as each other. So there are only $5$ possible outcomes for the second and third dice combined, not $5\times5$.
Also, you need to mutliply by $3$ because there are three possible choices for the odd die out (red, green or blue).
The possible patterns for the three dice (red, green, then blue) are $ABB,BAB,BBA$. For each pattern, there are $6$ choices for $A$ and then $5$ choices for $B$, so each pattern has $30$ options. Therefore there are $90$ options overall.
A: One die has 6 outcomes. Next have 5 outcomes. But the third die has only 1 options either number shown on first die or number on second die.
So we have $6 × 5 × 1 = 30$.
But we have that two same number on either 12_ , _23, 1_3 positions. So we multiply by 3.
Now we have $30 × 3 = 90$
