A card from a pack of 52 playing cards is lost. From the remaining cards, 2 cards are drawn at random without replacement, and are found to be both…

$\displaystyle E_1$ : Lost card is an ace$\displaystyle E_2$ : Lost card is not an ace$\displaystyle A$ : $\displaystyle 2$ ace cards are drawn\[P(E_1) = \frac{1}{13} \qquad P(E_2) = \frac{12}{13}\]\[P(A/E_1) = \frac{{}^{3}C_2}{{}^{51}C_2} \qquad P(A/E_2) = \frac{{}^{4}C_2}{{}^{51}C_2}\]\[P(E_1/A) = \frac{P(E_1)\,P(A/E_1)}{P(E_1)\,P(A/E_1) + P(E_2)\,P(A/E_2)}\]\[= \frac{\dfrac{1}{13}\dfrac{{}^{3}C_2}{{}^{51}C_2}}{\dfrac{1}{13}\cdot\dfrac{{}^{3}C_2}{{}^{51}C_2} + \dfrac{12}{13}\cdot\dfrac{{}^{4}C_2}{{}^{51}C_2}} \;=\; \frac{3}{75} \ \text{ or } \ \frac{1}{25}\]