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%
 A: 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. 
A: First, calculate the total possible ways you can draw 3 slips. There are 3 choices for $A$, 3 for $B$, and 3 for box $C$ giving a total of $3(3)(3)=27$ total possible ways to draw slips. 
Notice what you draw from the 3rd box doesn't matter at all. So we can calculate the ways to draw slips such that # slip $A>$ # slip $B$ and just know that for each way we could do that there are 3 choices for $C$. 
Because the events are independent, if we draw $1$ from $B$ there are 3 choices that I can draw from $A$ that satisfy what we want and of course 3 choices for $C$. If we draw $6$ there is only one choice, $9$, that we can draw from $A$ to get what we want (and $3$ choices for $C$). Similarly, if we draw $8$ from $B$. Then there are $3(3)+3(1)+3(1)=15$ total ways to get the desired outcome. That gives a probability of $\frac{15}{27}=\frac{5}{9}=55.\overline{5}\%$ probability of the desired event occurring.
