Q 3.41 - A First Course in Probability

Question

A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace. Hint: Condition on whether or not the interchanged card is selected 

Step-by-Step Solution

Answer

The probability of getting an ace card is

$P\left(G\mid F^c\cap E\right)=\frac{3}{51}$

Step 1: Given information

A deck of cards is well-shuffled and then divided into two halves of 26 cards each.

A card is drawn from one of the halves; it turns out to be an ace and it placed in the second half

The half is then shuffled, and a card is drawn from it. 

We have to find the probability that this drawn card is an ace. 

Step 2: The conditional probability of getting an ace given that interchanged card

Let $E=\{$the first card drawn is an Ace$\}$

$F=\{$ The second card drawn is the original Ace drawn in the first round$\}$$G=\{$the Second card drawn is an Ace $\}$

The number of cards in the second half after an ace is placed is 27.

The conditional probability of getting an ace given that interchanged card is selected is, $P(F\mid E)=\frac{1}{27}$

Step 3: The conditional probability of getting an ace given that interchanged ace card is not selected

The conditional probability of getting an ace given that interchanged ace card is not selected is$P\left(F^c\mid E\right)=1-P(F\mid E)$

$=1-\frac{1}{27}$

$=\frac{26}{27}$

Step 4: Final answer

The probability of getting an ace out of the remaining 51 cards pack is,
$P\left(G\mid F^{*}\cap E\right)$$=\frac{3}{51}$