Q. 3.28 - A First Course in Probability

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

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

Step-by-Step Solution

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 1: 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.

Step 2: Computation of the Probability (Part-a)

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

From 21 to $52=(32 cards)$

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

The probability that ${21}^{\mathrm{st}}$card is Ace of spade $=\frac{3}{4}\times \frac{1}{32}$

$=\frac{3}{128}$

Step 3: Final Answer (Part-a)

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

Step 4: 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. 

Step 5: 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)

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

$\frac{29}{1536}$.

Step 6: Final Answer (Part-b)

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