91Ó°ÊÓ

Given in the question that the first ace is the ${20}^{th}$card to appear and the conditional probability that the card following it is the ace of spades.

The probability that ${20}^{\text{th}}$card is not Ace of spade $=\frac{3}{4}$

From 21 to$52=(32cards)$

In 32 cards there is one Ace of spade $\mathrm{p}=\frac{1}{32}$

The probability that role="math" localid="1647784603931" ${21}^{\mathrm{st}}$card is Ace of spade $=\frac{3}{4}×\frac{1}{32}$

$=\frac{3}{128}$

The probability card ${21}^{\text{st}}$is Ace of spade$\frac{3}{128}.$

Given in the question that the first ace is the ${20}^{th}$card to appear and the conditional probability that the card following it is the two of clubs.

The first $19$cards can be any of $48$the cards (excluding

From $21$to $52$($32$cards)

There are $3$Ace's and $2$$\mathrm{P}\left({21}^{\text{st}}\right.$card is two of clubs)

$\frac{29}{1536}$.

The conditional probability that the card is the two of clubs is $\frac{29}{1536}.$