Probability of no matching pairs of shoes There are 3 different pairs ( i.e. 6 units say aa, bb, cc) of shoes in a lot. Now three person come and pick the shoes randomly (each get 2 units). Let p be the probability that no one is able to wear shoes (i.e. no one gets a correct pair) , then the value of $\frac{13p}{4-p}$ is?

We use a technique that is definitely overkill for this problem, the Principle of Inclusion/Exclusion. But the idea might be useful for a larger problem, say $5$ people and $5$ pairs of shoes. 
Let our people be P, Q, and R. There are $\binom{6}{2}$ ways for P to pick two shoes. And there are $\binom{3}{1}$ ways to pick a matching pair. So the probability that P picks a matching pair is $\frac{\binom{3}{1}}{\binom{6}{2}}$.
This is $\frac{1}{5}$. 
If we find the sum $\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$, we will have counted twice the probability P and Q both choose a matching pair, also the same for Q and R, also for P and R. Given that P chose a matching pair, the probability Q does is $\frac{\binom{2}{1}}{\binom{4}{2}}=\frac{1}{3}$. 
So the probability P and Q choose a matching pair is $\frac{15}{5}\cdot\frac{1}{3}$. We subtract $3$ times this from $\frac{3}{5}$.
We have subtracted too much. We must add back the probability they all choose a matching pair, which is $\frac{1}{5}\cdot\frac{1}{3}$. 
So the probability there is at least one person with a matching pair of shoes is
$$3\cdot\frac{1}{5}-3\cdot \frac{1}{5}\cdot \frac{1}{3}+1\cdot\frac{1}{5}\cdot\frac{1}{3}.$$
This simplifies to $\frac{7}{15}$.
So the probability $p$ of no match is $1-\frac{7}{15}$. 
