91Ó°ÊÓ

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:

$\text{a)}P(Bâˆ\pounds A)\mathbf{b}\left)P\right(Câˆ\pounds A)$

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 $48·47·…·30=\frac{48!}{29!}$different ways.

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

In event $Bâˆ\copyright A$, the number of different permutations $|AB|$is:

The first $19$cards can be any non ace cards, and order matters, so same as before $48·47·…·30=\frac{48\text{!}}{29!}$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 $Câˆ\copyright A$, 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.

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

$$\begin{gathered} P(Bâˆ\pounds A)=\frac{\left|AB\right|}{\left|A\right|} \\ =\frac{\frac{48!}{29!}â‹\dots 3â‹\dots 1â‹\dots 31!}{\frac{48!}{29!}â‹\dots 4â‹\dots 32{!}_{32}} \\ =\frac{3}{4â‹\dots 32}=\frac{3}{128} \\ ≈2.34\mathrm{\%} \end{gathered}$$

$$\begin{gathered} P(Câˆ\pounds A)=\frac{\left|AB\right|}{\left|A\right|} \\ =\frac{\frac{47y}{28!}â‹\dots 4â‹\dots 1â‹\dots 31!}{\frac{48\mathrm{\%}}{29!}\frac{48}{29}â‹\dots 4â‹\dots 32{!}_{32}} \\ =\frac{1}{\frac{48}{29}â‹\dots 32} \\ =\frac{29}{1536} \\ ≈1.89\mathrm{\%} \end{gathered}$$

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.