In a game of bridge, West has no aces. What is the probability of his partner's having $\left(a\right)$no aces?$\left(b\right)$ $2$ or more aces? $\left(c\right)$ What would the probabilities be if West had exactly $1$ ace?

For outcome space reduction,the uniform random formula is

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

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" $$\begin{gathered} P\left(E_0\mid W_0\right)=\frac{\left(\begin{array}{l}35\\ 13\end{array}\right)}{\left(\begin{array}{l}39\\ 13\end{array}\right)} \\ P\left(E_0\mid W_0\right)\approx 18.18\mathrm{\%}. \end{gathered}$$

$E_0,E_1,E_2,E_3,E_4$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,

$\begin{array}{l}1=P\left(E_0\mid W_0\right)+P\left(E_1\mid W_0\right)+P\left(E_2\mid W_0\right)+P\left(E_3\mid W_0\right)+P\left(E_4\mid W_0\right)\\ \Downarrow \\ P\left(E_2\cup E_3\cup E_4\mid W_0\right)=1-P\left(E_0\mid W_0\right)-P\left(E_1\mid W_0\right)\\ P\left(E_1\mid W_0\right).\end{array}$

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

$\begin{array}{l}P\left(E_1\mid W_0\right)=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\times \left(\begin{array}{l}35\\ 12\end{array}\right)}{\left(\begin{array}{l}39\\ 13\end{array}\right)}\approx 41.09\mathrm{\%}\\ P\left(E_2\cup E_3\cup E_4\mid W_0\right)\approx 1-0.1818-0.4109=0.4073=40.73\mathrm{\%}.\end{array}$

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

$\begin{array}{l}P\left(E_0\mid W_1\right)=\frac{\left(\begin{array}{l}36\\ 13\end{array}\right)}{\left(\begin{array}{l}39\\ 13\end{array}\right)}\approx 28.45\mathrm{\%}\\ P\left(E_2\cup E_3\cup E_4\mid W_1\right)=1-P\left(E_0\mid W_1\right)-P\left(E_1\mid W_1\right).\end{array}$

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

$\begin{array}{l}P\left(E_1\mid W_1\right)=\frac{3\times \left(\begin{array}{l}36\\ 12\end{array}\right)}{\left(\begin{array}{l}39\\ 13\end{array}\right)}\approx 46.23\mathrm{\%}\\ P\left(E_2\cup E_3\cup E_4\mid W_1\right)\approx 1-0.2845-0.4623=0.2532=25.32\mathrm{\%}.\end{array}$