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 Concept:
This problem involves determining probability from a modified deck of playing cards.
There are 52 standard cards. Removing cards (the Jacks) changes both the total sample space and the number of favorable cards in specific suits.
Since the draw is "without replacement," the events are dependent.
The probability of the second card being a spade depends on the fact that a spade was already removed in the first draw.
Step 2: Key Formula or Approach:
For dependent events \( A \) and \( B \):
\[ P(A \cap B) = P(A) \times P(B|A) \]
Where \( P(B|A) \) is the probability of B happening given that A has already occurred.
Step 3: Detailed Explanation:
Initial State: Total 52 cards. Each suit (Spades, Hearts, Diamonds, Clubs) has 13 cards.
Removing all Jacks: There are 4 Jacks in total (one in each suit).
New Total Cards = \( 52 - 4 = 48 \).
Number of Spades remaining = \( 13 - 1 (\text{the Jack of Spades}) = 12 \).
Calculate the first draw (\( P_1 \)):
Probability that the first card is a spade:
\[ P_1 = \frac{12}{48} = \frac{1}{4} \]
Calculate the second draw (\( P_2 \)):
Since we did not replace the first spade, there are now \( 48 - 1 = 47 \) cards left in total.
There are now \( 12 - 1 = 11 \) spades left in total.
Probability that the second card is a spade:
\[ P_2 = \frac{11}{47} \]
Calculate the joint probability:
\[ P = P_1 \times P_2 = \frac{1}{4} \times \frac{11}{47} \]
Multiply the denominators: \( 4 \times 47 = 188 \).
\[ P = \frac{11}{188} \]
Step 4: Final Answer:
The probability that both chosen cards are spades is 11/188.