91Ó°ÊÓ

For outcome space reduction,the uniform random formula is

$$\begin{gathered} P\left(A\right)=\frac{\mathrm{\#}\text{of equally likely events in}A}{\mathrm{\#}\text{of equally likely events}} \\ P(Aâˆ\pounds B)=\frac{\mathrm{\#}\text{of equally likely events in}A\text{if}B\text{occurred}}{\mathrm{\#}\text{of equally likely events in}B}. \end{gathered}$$

East-west possible partner - $\left(\begin{array}{l}52\\ 13\end{array}\right)$

The numbe of east hand - localid="1649492092335" $\left(\begin{array}{l}39\\ 13\end{array}\right)$,the number of localid="1649492081417" $13-$card among localid="1649492087719" $39-$cards are not in west hand.

The possibilites are

localid="1649491553162" $$\begin{gathered} P\left(E_0âˆ\pounds W_0\right)=\frac{\left(\begin{array}{l}35\\ 13\end{array}\right)}{\left(\begin{array}{l}39\\ 13\end{array}\right)} \\ P\left(E_0âˆ\pounds W_0\right)≈18.18\mathrm{\%}. \end{gathered}$$

$E_0,E_1,E_2,E_3,E_4$are five mutually exclusive events to make the whole outcome space.

The conditional probability $P\left(\raisebox{1ex}{$.$}\!\left/ \!\raisebox{-1ex}{$W_0$}\right.\right)$is,

$\begin{array}{l}1=P\left(E_0âˆ\pounds W_0\right)+P\left(E_1âˆ\pounds W_0\right)+P\left(E_2âˆ\pounds W_0\right)+P\left(E_3âˆ\pounds W_0\right)+P\left(E_4âˆ\pounds W_0\right)\\ ⇓\\ P\left(E_2∪E_3∪E_4âˆ\pounds W_0\right)=1−P\left(E_0âˆ\pounds W_0\right)−P\left(E_1âˆ\pounds W_0\right)\\ P\left(E_1âˆ\pounds W_0\right).\end{array}$

The numbe of east hand - $\left(\begin{array}{l}39\\ 13\end{array}\right)$,the number of role="math" localid="1649492060783" $13-$card among $39-$cards are not in west hand.

The possibilities of $\left(\begin{array}{l}35\\ 12\end{array}\right)$are

$\begin{array}{l}P\left(E_1âˆ\pounds W_0\right)=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)×\left(\begin{array}{l}35\\ 12\end{array}\right)}{\left(\begin{array}{l}39\\ 13\end{array}\right)}≈41.09\mathrm{\%}\\ P\left(E_2∪E_3∪E_4âˆ\pounds W_0\right)≈1−0.1818−0.4109=0.4073=40.73\mathrm{\%}.\end{array}$

West has one ace - $W_1$,the possibilties are

$\begin{array}{l}P\left(E_0âˆ\pounds W_1\right)=\frac{\left(\begin{array}{l}36\\ 13\end{array}\right)}{\left(\begin{array}{l}39\\ 13\end{array}\right)}≈28.45\mathrm{\%}\\ P\left(E_2∪E_3∪E_4âˆ\pounds W_1\right)=1−P\left(E_0âˆ\pounds W_1\right)−P\left(E_1âˆ\pounds W_1\right).\end{array}$

and P$\left(\raisebox{1ex}{$E_1$}\!\left/ \!\raisebox{-1ex}{$W_1$}\right.\right)$.

There are three choice of ace,the remaining $12$cards of $\left(\begin{array}{l}36\\ 12\end{array}\right)$is

$\begin{array}{l}P\left(E_1âˆ\pounds W_1\right)=\frac{3×\left(\begin{array}{l}36\\ 12\end{array}\right)}{\left(\begin{array}{l}39\\ 13\end{array}\right)}≈46.23\mathrm{\%}\\ P\left(E_2∪E_3∪E_4âˆ\pounds W_1\right)≈1−0.2845−0.4623=0.2532=25.32\mathrm{\%}.\end{array}$