Q. 1 - A First Course in Probability

Question

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?

Step-by-Step Solution

Answer

If the West hand is chosen in a certain fashion, the result space is reduced to equally likely partner hands.

Make a list of possibilities.

Part $\left(a\right)$

$\left(a\right)$The probability of his  partner's having no aces is $P\left(E_0\mid W_0\right)=18.18\%.$

Part $\left(b\right)$

$\left(b\right)$The probability of his  partner's having $2$ or more aces is $P\left(E_2\cup E_3\cup E_4\mid W_0\right)=40.73\%.$

Part $\left(c\right)$

$\left(c\right)$The probability will be$P\left(E_0\mid W_1\right)=28.24\% ; P\left(E_2\cup E_3\cup E_4\mid W_1\right)=25.32\%$ if West had exactly $2$ ace.

Step: 1 Principles:

For outcome space reduction,the uniform random formula is

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

Step: 2 Probability no aces: (part a)

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

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

The possibilites are

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

Step: 3 Probability has two or more aces: (part b)

$E_0,E_1,E_2,E_3,E_4$ 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 $13-$card among  $39-$cards are not in west hand.

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

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

Step: 4 Possibilities if west had exactly two ace: (part c)

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

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