<s>[INST] Probability of a random selection of $2$ out of $5$ shoes 
From a set of five pairs of shoes, two of the shoes are selected at random. Find the probability of each of the following:
a) Both are from the same pair.

My answer: $\dfrac{\binom{10}{1}\binom{1}{1}}{\binom{10}{2}}$

b) One left shoe and one right shoe are selected.

My answer: $\dfrac{\binom{10}{1}\binom{5}{1}}{\binom{10}{2}}$
Note: Solve this problem by counting combinations (i.e $\binom{n}{k}$).
Can someone check if this is correct? If not, can you explain why it is incorrect?
 [/INST] The numerators in your solutions are incorrect. Considering the two different questions:


*

*Both shoes must be from the same pair. Using probability on the first and second draw, we find: $$\frac{10}{10} \cdot \frac{1}{9} = \frac{1}{9}$$ With your suggested approach, however, we find: $$\frac{10 \cdot 1}{45} = \frac{2}{9}$$ In the numerator, we must only mention the combinations of shoes we pick, not multiply the values for the two different draws. Instead, try writing it as follows: $$\frac{5 \choose 1}{10 \choose 2} = \frac{5}{45} = \frac{1}{9}$$ There are five pairs we can draw, and 45 possible pairs of draws.

*Similarly as for (a), the numerator is incorrect. Using probability on the first and second draw, we find: $$\frac{10}{10} \cdot \frac{5}{9} = \frac{5}{9}$$ Using combinations, you should arrive at the following: $$\frac{{5 \choose 1}{5 \choose 1}}{10 \choose 2} = \frac{25}{45} = \frac{5}{9}$$ In this way, we consider all pairs of draws in which one shoe is drawn from the left shoes and one shoe is drawn from the right shoes. 
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