Question

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 role="math" localid="1647784076635" ${20}^{th}$card to appear, what is the conditional probability that the card following it is the

1. ace of spades?
2. two of the clubs?

Short answer

1. The conditional probability card ${21}^{st}$is Ace of spade $=\frac{3}{128}.$
2. The conditional probability that the card is the two of clubs is$\frac{29}{1536}.$
   

Step-by-step solution

Given Information (Part-a)

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.

Computation of the Probability (Part-a)

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 $=\frac{3}{4}\times \frac{1}{32}$

$=\frac{3}{128}$

Final Answer (Part-a)

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

Given Information (Part-b)

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.

Find the Conditional Probability (Part-b)

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)

$=\frac{1}{48}\times \frac{29}{32}$

$\frac{29}{1536}$.

Final Answer (Part-b)

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