Subject: Mathematics — Probability
Question: Consider three sets $E_{1} = \{ 1,2,3\}$,
$F_{1} = \{ 1,3,4\}$ and $G_{1} = \{ 2,3,4,5\}$. Two elements are chosen
at random, without replacement, from the set $E_{1}$, and let $S_{1}$
denote the set of these chosen elements.
Let $E_{2} = E_{1} - S_{1}$ and $F_{2} = F_{1} \cup S_{1}$. Now two
elements are chosen at random, without replacement, from the set
$F_{2}$, and let $S_{2}$ denote the set of these chosen elements.
Let $G_{2} = G_{1} \cup S_{2}$. Finally, two elements are chosen at
random, without replacement, from the set $G_{2}$ and let $S_{3}$
denote the set of these chosen elements.
Let $E_{3} = E_{2} \cup S_{3}$. Given that $E_{1} = E_{3}$, let $p$ be
the conditional probability of the event $S_{1} = \{ 1,2\}$. Then find
the value of $p$.
Answer: $\frac{1}{5}$
Solution: [IMAGE] $$P = \frac{P\left( S_{1} \cap \left( E_{1} = E_{3} \right) \right)}{P\left( E_{1} = E_{3} \right)} = \frac{P\left( A_{1,2} \right)}{P(A)}$$
$$P(A) = P\left( A_{1,2} \right) + P\left( A_{1,3} \right) + P\left( A_{2,3} \right)$$
$$P\left( A_{1,2} \right) = \frac{1}{3} \times \frac{1 \times \binom{3}{1}}{\binom{4}{2}} \times \frac{1}{\binom{5}{2}}$$
$$P\left( A_{1,2} \right) = \frac{1}{3} \times \frac{1}{2} \times \frac{1}{10} = \frac{1}{60}$$
$$P\left( A_{1,3} \right) = \frac{1}{3} \times \frac{1 \times \binom{2}{1}}{\binom{3}{2}} \times \frac{1}{\binom{5}{2}}$$
$$P\left( A_{1,3} \right) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{10} = \frac{1}{45}$$
$$P\left( A_{2,3} \right) = \frac{1}{3}\left\lbrack \frac{\binom{3}{2} \times 1}{\binom{4}{2}} \times \frac{1}{\binom{4}{2}} + \frac{1 \times \binom{3}{1}}{\binom{4}{2}} \times \frac{1}{\binom{5}{2}} \right\rbrack$$
$$P\left( A_{2,3} \right) = \frac{1}{3}\left( \frac{1}{12} + \frac{1}{20} \right) = \frac{2}{45}$$
$$P(A) = \frac{1}{60} + \frac{1}{45} + \frac{2}{45} = \frac{1}{12}$$
$$\frac{P\left( A_{1,2} \right)}{P(A)} = \frac{1}{5}$$