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 $20th$ card to appear, what is the conditional probability that the card following it is the

(a) ace of spades?

(b) two of clubs?

Short 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 $1.89\%$

Step-by-step solution

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 $21st$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 $|AB|$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 - $|AC|$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.

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.