Q 3.13 - A First Course in Probability

Question

Suppose that an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each. We are interested in determining p, the probability that each hand has an ace. Let Ei be the event that I the hand has exactly one ace. Determine p = P(E1E2E3E4) by using the multiplication rule. 

Step-by-Step Solution

Answer

The probability that each hand has an ace P Is 0.105. 

Step 1 Given Information

Given that an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each. 

We have to determine p, the probability that each hand has an ace. 

Step 2 Explanation-1

Given that, an ordinary deck of 52 cards (Deck of cards containing 4 aces) is randomly divided into 4 hands of 13 cards each.

Let the four events be$E_1,E_2,E_3$, and $E_4$

$P\left(E_1\right)=\text{ Probability that I" hand has one Ace }$

$P\left(E_2\right)={\text{ Probability that II }}^{\text{ad }}\text{ hand has one Ace }$

$P\left(E_3\right)={\text{ Probability that III }}^{\text{rd }}\text{ hand has one Ace }$

$P\left(E_4\right)={\text{ Probability that IV IV }}^{\text{th }}\text{ hand has one Ace }$

Thus,

$P\left(E_1{E}_2{E}_3{E}_4\right)=P\left(E_1\right)·P\left(E_2\mid E_1\right)·P\left(E_3\mid E_1{E}_2\right)·P\left(E_4\mid E_1{E}_2{E}_3\right)$

Consider,

$P\left(E_1\right)=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\times \left(\begin{array}{l}48\\ 12\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)}$

          

         $=0.438847$

Step 3 Explanation-2

Here $\left(\begin{array}{l}4\\ 1\end{array}\right)$is exactly one ace from 4 aces $\left(\begin{array}{c}48\\ 12\end{array}\right)$ is  remaining 12 cards  from 48 cards which does not have an ace $\left(\begin{array}{l}52\\ 13\end{array}\right)$and  is sample space.

similarly,

$P\left(E_2\mid E_1\right)=\frac{\left(\begin{array}{l}3\\ 1\end{array}\right)\times \left(\begin{array}{l}36\\ 12\end{array}\right)}{\left(\begin{array}{l}39\\ 13\end{array}\right)}$

                  $=462304$

Step 4 Explanation-3

After$1^{\text{st }}$ hand, total 39 cards are remaining with 3 aces and 36 cards which do not have an ace.

Similarly

$P\left(E_3\mid E_1{E}_2\right)=\frac{\left(\begin{array}{l}2\\ 1\end{array}\right)\times \left(\begin{array}{l}24\\ 12\end{array}\right)}{\left(\begin{array}{l}26\\ 13\end{array}\right)}$

 

                     $=0.52$

Step 5 Explanation-4

After $2^{st}$ the hand, a total of 26 cards are remaining with 2 aces and 24 cards which do not have an ace. 

$P\left(E_4\mid E_1{E}_2{E}_3\right)=\frac{1\times \left(\begin{array}{l}12\\ 12\end{array}\right)}{\left(13\right)}$

                            $=1$

Step 6 Explanation-5

After 3 hands are distributed last hand has exactly 1 ace and 12 non ace cards

so,

$p=P\left(E_1{E}_2{E}_3{E}_4\right)$

$=P\left(E_1\right)P\left(E_2\right)P\left(E_3\right)P\left(E_4\right)$

$=0.438847\times 0.462304\times 0.52\times 1$

$=0.105498$

$=0.105$

Step 7 Final Answer

The probability that each hand has an ace P Is 0.105.