How to find the the probability that both of them draw exactly one common card? Harry and Ron play a game of cards. They only have half a standard pack of cards. Harry draws two cards and replaces them back. Ron again repeats the same process. How to find the the probability that both of them draw exactly one common card?
 A: Ron must choose one card from the two cards drawn by Harry ($2$ possibilities) and one card from the $n-2=24$ cards not drawn by Harry ($n-2$ possibilities) and their order ($2$ possibilities) hence the probability is $2\cdot2\cdot(n-2)\cdot(n(n-1))^{-1}$.
A: We find the probability of no match, and of two matches, and subtract the sum of these from $1$.
For the probability of no match, imagine that Harry drew $2$ cards, and wrote down what they were.  Now Ron draws $2$ cards. There are $\binom{26}{2}$ ways for Ron to draw $2$ cards from $26$, all equally likely. There are $\binom{24}{2}$ ways to draw $2$ cards that are both different from what Harry wrote down. So the probability that neither of Ron's cards matches one of Harry's is
$$\frac{\binom{24}{2}}{\binom{26}{2}}.$$
The probability of $2$ matches is  simpler to find, since only $1$ pair from the $\binom{26}{2}$ is the same as the pair Harry wrote down. For fun, we will write $\binom{2}{2}$ instead of $1$.
Thus the probability of no match or two matches is
$$\frac{\binom{24}{2}+\binom{2}{2}}{\binom{26}{2}}.$$
Calculate. We get $\frac{277}{325}$. Thus the probability of exactly $1$ match is $\frac{48}{325}$.

Or else, more simply, we can calculate directly. There are $\binom{2}{1}$ ways to choose the card that will match one of Harry's.  For each such choice, there are $\binom{24}{1}$ ways to choose the card that will not match any of Harry's. So the required probability is
$$\frac{\binom{2}{1}\binom{24}{1}}{\binom{26}{2}}.$$
