<s>[INST] Did I answer this probability question correctly? Hello I have the following question to answer:
Imagine three boxes, each of which has three slips of paper in it each with a number
marked on it. The numbers for box A are 2, 4 and 9, for box B 1, 6 and 8, and for box C
3, 5 and 7. One slip is drawn, independently and with equal probability, from each box.
Compute P{A Slip > B slip} (a slip taken from A is greater than a slip from B)
My attempted solution:
First note the possible outcomes: (highlighted are outcomes that satisfy A>B)
{2,1} || {2,6} || {2,8}
{4,1} || {4,6} || {4,8}
{9,1} || {9,6} || {9,8}
Since there are 5 possible outcomes where A>B and since the events are independent from one another then the probability is simply:
(5/9) * P{B|A}, where B is a slip chosen from B that is smaller than A 
The probability P{B|A} is P{B|2} + P{B|4} + P{B|9} = 0+1/3+1 = 4/3
So finally P{Aslip>Bslip} = (5/9)(4/3) = 20/27 or 74%
 [/INST] There are $3$ equally likely possibilities for A's number, $2$, $4$, and $9$. 
If A got $2$, then the probability that A beat B is $\frac{1}{3}$, for B must get $1$. 
If A got $4$, then again the probability that A beat B is $\frac{1}{3}$.
If A got $9$, then the probability that A beat B is $1$, or more prettily $\frac{3}{3}$.
Thus the probability that A beats B is $\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{3}{3}$. 
Remark: Your counting argument gave the same result, but then got spoiled.
The problem will become interesting when you compute the probability that B beats C, and the probability that C beats A. 
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