Question

There are 52 cards in a pack. All jacks are removed from the pack. Two cards are chosen at a random without replacement. Find the probability that both cards are spade.

• A. 22/119
• B. 45/118
• C. 21/112
• D. 11/188

Hint:

When dealing with card probability questions, always write out the initial state and the state change.
A common trap is forgetting that removing "all jacks" also removes the Jack of Spades, which decreases the count of spades from 13 to 12. Always account for suit-specific cards when modifying the deck!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A standard deck contains 52 playing cards.
First, all "jacks" are removed from this deck.
We then draw two cards consecutively without replacement.
We need to determine the probability that both of the drawn cards are spades.

Step 2: Key Formula or Approach:
For dependent events (without replacement), the joint probability of drawing two cards of the same suit is:
\[ P(S_1 \cap S_2) = P(S_1) \times P(S_2 | S_1) \]
Where:
\(P(S_1)\) is the probability that the first card drawn is a spade.
\(P(S_2 | S_1)\) is the conditional probability that the second card drawn is a spade, given that the first card drawn was also a spade.

Step 3: Detailed Explanation:
$\bullet$

Step 1: Calculate the remaining cards in the deck after removing jacks:
A standard deck has 4 jacks (one for each of the four suits: Spades, Hearts, Diamonds, Clubs).
Total initial cards = 52.
If we remove all 4 jacks:
Total remaining cards = \(52 - 4 = 48\) cards.
$\bullet$

Step 2: Calculate the remaining spades in the deck:
A standard deck contains 13 spade cards.
One of these spade cards is the Jack of Spades, which has been removed.
Therefore, the number of spade cards left in the deck is:
Total spade cards remaining = \(13 - 1 = 12\) cards.
$\bullet$

Step 3: Calculate the probability of the first card being a spade (\(S_1\)):
We select 1 spade out of the 12 available spades, from a total of 48 cards:
\[ P(S_1) = \frac{12}{48} = \frac{1}{4} \]
$\bullet$

Step 4: Calculate the probability of the second card being a spade (\(S_2\)) given \(S_1\):
Since we do not replace the first card, we now have:
Remaining total cards = \(48 - 1 = 47\).
Remaining spade cards = \(12 - 1 = 11\).
\[ P(S_2 | S_1) = \frac{11}{47} \]
$\bullet$

Step 5: Calculate the combined probability:
\[ P(S_1 \cap S_2) = P(S_1) \times P(S_2 | S_1) \]
\[ P(S_1 \cap S_2) = \frac{1}{4} \times \frac{11}{47} \]
Multiply the denominators:
\[ 4 \times 47 = 188 \]
\[ P(S_1 \cap S_2) = \frac{11}{188} \]

Step 4: Final Answer:
The probability that both chosen cards are spades is 11/188.