Q.3.28 - A First Course in Probability

Question

3.28. Suppose that an ordinary deck of $52$ cards is shuffled and the cards are then turned over one at a time until the first ace appears. Given that the first ace is the $20 th$ card to appear, what is the conditional probability that the card following it is the

(a) ace of spades?

(b) two of clubs?

Step-by-Step Solution

Answer

Apply the definition of conditional probability to define events that are equally likely..

Probability that $21$ . is ace of spades is $2.34\%$

Probability that $21$. is two of clubs is $1.89\%$

Step1:Find ace of spades(part a)

All five he cards are drawn at random from a deck of 52 cards.

Considered events:

A - The $20th$card is the first ace.

B - the $2lst$ card is the ace of spades

C - the $21 st$ card is the two of clubs

Calculate:

$\text{ a) }P(B\mid A)\mathbf{b}\left)P\right(C\mid A)$

All $52!$permutations of the cards are equally likely.

In event A, the number of different permutations -|A| is:

The first card can be any of the $48$ non ace cards, the second card can be any of the $47$non ace cards (and not the first card). the first $19$ can be drawn in $48·47·\dots ·30=\frac{48!}{29!}$different ways.

The remaining $32$ cards can be permuted in $32 !$ ways.

In event $B\cap A$, the number of different permutations $|A B|$ is:

The first $19$ cards can be any non ace cards, and order matters, so same as before $48·47·\dots ·30=\frac{48\text{ ! }}{29!}$different choices for the first $19$cards.

The twentieth card could be one of the three aces.

The $21$.st card is the ace of spades, and the remaining $31$cards can be permuted in $31 !$ ways.

 In event $C\cap A$, the number of different permutations - $|A C|$ is:

The first $19$ cards can be any non ace cards that are not two of clubs either, so there  are $47$options for the first card, $46$ for the second and so on.. $\frac{47!}{28!}$ different choices for the first$19$ cards.

One of the four aces could be the twentieth card.

The $21$.st card is the two of clubs, and the remaining $31$cards can be permuted in $31 !$ ways.

Step2: Find two of clubs(part b)

We have conditional probability and probability on an equally likely set of events by definition.

$$\begin{gathered} P(B\mid A)=\frac{\left|AB\right|}{\left|A\right|} \\ =\frac{\frac{48!}{29!}\cdot 3\cdot 1\cdot 31!}{\frac{48!}{29!}\cdot 4\cdot 32{!}_{32}} \\ =\frac{3}{4\cdot 32}=\frac{3}{128} \\ \approx 2.34\mathrm{\%} \end{gathered}$$

$$\begin{gathered} P(C\mid A)=\frac{\left|AB\right|}{\left|A\right|} \\ =\frac{\frac{47y}{28!}\cdot 4\cdot 1\cdot 31!}{\frac{48\mathrm{\%}}{29!}\frac{48}{29}\cdot 4\cdot 32{!}_{32}} \\ =\frac{1}{\frac{48}{29}\cdot 32} \\ =\frac{29}{1536} \\ \approx 1.89\mathrm{\%} \end{gathered}$$

The possibility is greater for the ace because the likelihood that it will be used before the ace is higher$21$. card is$1 / 4$, and the probability that two of clubs is used before is that it is one of the $19$ from $48$ non ace cards.