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.

Short answer

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

Step-by-step solution

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.

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$E_4$

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

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

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

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

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,

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

$=0.438847$

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,

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

$=462304$

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

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

$=0.52$

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.

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

$=1$

Explanation-5

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

so,

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

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

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

$=0.105498$

$=0.105$

Final Answer

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