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%

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.