Probability Part 1

Let $S$ and $N S$ be the respective events of choosing a spade card and a card which is not a spade. $P(S)=\;\frac{{}^{13}C_{1}}{{}^{52}C_{1}}=\;\frac{1}{4}$ $P(N S)=\;\frac{{}^{39}C_{1}}{{}^{52}C_{1}}=\;\frac{3}{4}$ Let $A$ be the event when both the chosen cards are spade. Let $\overline{A}$ be the event when both the chosen cards are not spade.We have to find the probability of missing card not being spade when both the chosen cards are spade $=P(\overline{S}\mid A)$ By using Bayes theorem, $P(\overline{S}\mid A)=\;\frac{P(\overline{S}\cap A)}{P(A)}$ $=\;\frac{P(A\mid\overline{S})P(\overline{S})}{P(A\mid S)P(S)+P(A\mid\overline{S})P(\overline{S})}$ $=\;\frac{\left(\frac{{}^{13}C_{2}}{{}^{51}C_{2}}\right)\left(\frac{{}^{39}C_{1}}{{}^{52}C_{1}}\right)}{\left(\;\frac{{}^{13}C_{1}}{{}^{52}C_{1}}\right)\left(\;\frac{{}^{12}C_{2}}{{}^{51}C_{2}}\right)+\left(\;\frac{{}^{13}C_{2}}{{}^{51}C_{2}}\right)\left(\;\frac{{}^{39}C_{1}}{{}^{52}C_{1}}\right)}$ $=\;\frac{\left(\frac{13\cdot12}{51\cdot50}\right)\left(\frac{3}{4}\right)}{\left(\;\frac{1}{4}\right)\left(\;\frac{12\cdot11}{51\cdot50}\right)+\left(\;\frac{3}{4}\right)\left(\;\frac{13\cdot12}{51\cdot50}\right)}$ $=\;\frac{13\times12\times3}{(12\times11)+(3\times12\times13)}$ $=\;\frac{39}{50}$