Probability with Playing Cards
Probability with playing cards is one of the most common problem types in CBSE Class 10. A standard deck has 52 cards, and students must calculate the probability of drawing specific cards or combinations.
Playing card problems test the ability to count favourable outcomes correctly. The key challenge is knowing the deck structure — suits, face cards, number cards, and colour distribution.
The formula used is the same: P(E) = Number of favourable outcomes / Total number of outcomes. The difficulty lies in identifying what counts as a favourable outcome.
What is Probability with Playing Cards?
Definition: A standard deck of playing cards consists of 52 cards divided into 4 suits of 13 cards each.
Structure of a deck:
- 4 suits: Hearts (♥), Diamonds (♦), Clubs (♣), Spades (♠)
- Red suits: Hearts and Diamonds (26 cards total)
- Black suits: Clubs and Spades (26 cards total)
- Each suit has: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King
Card categories:
- Face cards (court cards): Jack, Queen, King — 12 total (3 per suit × 4 suits)
- Number cards: Ace through 10 — 40 total (10 per suit × 4 suits)
- Aces: 4 total (one per suit)
- Honours: Ace, King, Queen, Jack, 10 — 20 total (5 per suit × 4 suits)
Important:
- Ace is NOT a face card. Face cards are only Jack, Queen, and King.
- There are 2 red and 2 black cards of each denomination.
- Jokers are NOT included in standard probability problems (52-card deck).
Probability with Playing Cards Formula
Formula:
P(E) = Number of favourable outcomes / Total number of outcomes = n(E) / 52
Where:
- P(E) = probability of event E
- n(E) = number of cards that satisfy event E
- 52 = total cards in a standard deck
Complementary probability:
P(not E) = 1 − P(E)
Derivation and Proof
Quick-Reference Card Counts:
| Category | Count | Details |
|---|---|---|
| Total cards | 52 | 4 suits × 13 cards |
| Red cards | 26 | 13 Hearts + 13 Diamonds |
| Black cards | 26 | 13 Clubs + 13 Spades |
| Face cards | 12 | 4 Jacks + 4 Queens + 4 Kings |
| Red face cards | 6 | 3 from Hearts + 3 from Diamonds |
| Black face cards | 6 | 3 from Clubs + 3 from Spades |
| Aces | 4 | One per suit |
| Cards of one suit | 13 | A, 2, 3, ..., 10, J, Q, K |
| Cards of one denomination | 4 | One per suit |
| Number cards (2–10) | 36 | 9 per suit × 4 |
Types and Properties
Common types of playing card probability questions:
- Type 1: Probability of drawing a specific card (e.g., queen of hearts) → n(E) = 1
- Type 2: Probability of drawing a card of a specific suit (e.g., a spade) → n(E) = 13
- Type 3: Probability of drawing a face card → n(E) = 12
- Type 4: Probability of drawing a card that is NOT something (complementary) → use P(not E) = 1 − P(E)
- Type 5: Probability involving colour (red/black) → n(E) = 26
- Type 6: Probability involving "or" conditions (e.g., king or queen) → add individual counts if mutually exclusive
- Type 7: Probability involving "and" conditions (e.g., red king) → count cards satisfying BOTH conditions
Methods
Steps to solve playing card probability problems:
- Identify total outcomes: Total = 52 (unless stated otherwise).
- Identify the event: Read the question carefully — what card(s) are we looking for?
- Count favourable outcomes: Use the deck structure to count exactly how many cards satisfy the condition.
- Apply the formula: P(E) = favourable outcomes / 52.
- Simplify the fraction.
Common mistakes to avoid:
- Counting Ace as a face card (it is NOT).
- Forgetting that "red king" means only 2 cards (king of hearts and king of diamonds).
- Double-counting when the event involves overlapping categories (e.g., "a heart or a king" — the king of hearts is counted in both).
- Using 54 instead of 52 (jokers are not included).
Solved Examples
Example 1: Probability of Drawing a King
Problem: A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting a king.
Solution:
Given:
- Total cards = 52
- Number of kings = 4 (one per suit)
Using P(E) = n(E)/52:
- P(king) = 4/52 = 1/13
Answer: P(king) = 1/13.
Example 2: Probability of a Red Card
Problem: Find the probability of drawing a red card from a deck of 52 cards.
Solution:
Given:
- Total cards = 52
- Red cards = 26 (13 hearts + 13 diamonds)
Using P(E) = n(E)/52:
- P(red) = 26/52 = 1/2
Answer: P(red card) = 1/2.
Example 3: Probability of NOT Getting a Face Card
Problem: A card is drawn from a deck of 52 cards. Find the probability that it is NOT a face card.
Solution:
Given:
- Face cards = 12 (J, Q, K in each of 4 suits)
- Non-face cards = 52 − 12 = 40
Method 1 (Direct):
- P(not face card) = 40/52 = 10/13
Method 2 (Complement):
- P(face card) = 12/52 = 3/13
- P(not face card) = 1 − 3/13 = 10/13
Answer: P(not a face card) = 10/13.
Example 4: Probability of a Red Queen
Problem: Find the probability of drawing a red queen from a deck of 52 cards.
Solution:
Given:
- Queens = 4 (one per suit)
- Red queens = 2 (queen of hearts + queen of diamonds)
Using P(E) = n(E)/52:
- P(red queen) = 2/52 = 1/26
Answer: P(red queen) = 1/26.
Example 5: Probability of a Heart or a Spade
Problem: A card is drawn at random. Find the probability that it is a heart or a spade.
Solution:
Given:
- Hearts = 13, Spades = 13
- Heart and spade are mutually exclusive (a card cannot be both)
- Favourable outcomes = 13 + 13 = 26
Using P(E) = n(E)/52:
- P(heart or spade) = 26/52 = 1/2
Answer: P(heart or spade) = 1/2.
Example 6: Probability of a King or a Queen
Problem: Find the probability of drawing a king or a queen.
Solution:
Given:
- Kings = 4, Queens = 4
- These are mutually exclusive events
- Favourable outcomes = 4 + 4 = 8
Using P(E) = n(E)/52:
- P(king or queen) = 8/52 = 2/13
Answer: P(king or queen) = 2/13.
Example 7: Probability of a Heart or a King (Overlapping)
Problem: Find the probability that a card drawn is a heart or a king.
Solution:
Given:
- Hearts = 13, Kings = 4
- But king of hearts is in BOTH categories — counted once
- Favourable = 13 + 4 − 1 = 16
Why subtract 1: The king of hearts is both a heart and a king. If we add 13 + 4, we count it twice.
Using P(E) = n(E)/52:
- P(heart or king) = 16/52 = 4/13
Answer: P(heart or king) = 4/13.
Example 8: Probability of a Card Between 3 and 7
Problem: A card is drawn from a deck. Find the probability that its number is between 3 and 7 (inclusive).
Solution:
Given:
- Cards between 3 and 7 (inclusive): 3, 4, 5, 6, 7 → 5 denominations
- Each denomination has 4 cards (one per suit)
- Favourable outcomes = 5 × 4 = 20
Using P(E) = n(E)/52:
- P(3 to 7) = 20/52 = 5/13
Answer: P(card between 3 and 7) = 5/13.
Example 9: Neither a Spade nor a King
Problem: Find the probability that a card drawn is neither a spade nor a king.
Solution:
Step 1: Count "spade or king":
- Spades = 13, Kings = 4
- King of spades is in both → Spade or King = 13 + 4 − 1 = 16
Step 2: Neither spade nor king = 52 − 16 = 36
Using P(E) = n(E)/52:
- P(neither spade nor king) = 36/52 = 9/13
Answer: P(neither spade nor king) = 9/13.
Example 10: Black Card or a Face Card
Problem: A card is drawn at random. Find the probability that it is a black card or a face card.
Solution:
Given:
- Black cards = 26 (clubs + spades)
- Face cards = 12 (J, Q, K × 4 suits)
- Black face cards = 6 (counted in both categories)
Using inclusion-exclusion:
- Black or face = 26 + 12 − 6 = 32
Using P(E) = n(E)/52:
- P(black or face) = 32/52 = 8/13
Answer: P(black card or face card) = 8/13.
Real-World Applications
Uses of playing card probability:
- CBSE Board Exams: Playing card problems appear almost every year in the probability section — typically worth 2–3 marks.
- Competitive exams: Olympiads, NTSE, and entrance exams frequently use card-based probability.
- Game theory: Understanding card probabilities is the foundation of strategic card games.
- Combinatorics training: Counting favourable outcomes in a structured set builds foundational skills for higher combinatorics.
- Understanding conditional probability: In higher classes, card problems extend to drawing without replacement and Bayes' theorem.
Key Points to Remember
- A standard deck has 52 cards: 4 suits × 13 cards each.
- Face cards = Jack, Queen, King = 12 total. Ace is NOT a face card.
- Red cards = 26 (Hearts + Diamonds). Black cards = 26 (Clubs + Spades).
- Each denomination (e.g., King, 7, Ace) has exactly 4 cards — one per suit.
- For "or" events, check if they overlap. If overlapping, use: n(A or B) = n(A) + n(B) − n(A and B).
- For complementary events: P(not E) = 1 − P(E).
- The probability of any event lies between 0 and 1 (inclusive).
- Common exam trap: "Neither A nor B" = Total − n(A or B).
- Always verify that your favourable count does not exceed 52.
- Simplify all fractions to lowest terms in the final answer.
Practice Problems
- A card is drawn from a deck of 52 cards. Find the probability of getting (i) a jack of hearts (ii) a red face card.
- Find the probability that a card drawn is (i) a diamond (ii) not a diamond.
- What is the probability of drawing a card which is neither a king nor a queen?
- Find the probability of getting a black card or an ace.
- A card is drawn at random. What is the probability that it is a face card of a red suit?
- Find the probability that the card drawn is (i) a 10 of spades (ii) an even-numbered card.
- What is the probability of drawing a card which is a red card and a face card?
Frequently Asked Questions
Q1. How many face cards are there in a deck?
There are 12 face cards: 4 Jacks, 4 Queens, and 4 Kings. Ace is NOT a face card.
Q2. Is the Ace a face card?
No. In CBSE Class 10 probability, face cards are only Jack, Queen, and King. Each suit has 3 face cards, giving 12 face cards in total.
Q3. How many red cards are in a deck?
26 red cards — 13 Hearts and 13 Diamonds.
Q4. What is an honour card?
Honour cards are Ace, King, Queen, Jack, and 10 — five per suit, making 20 honour cards in a deck.
Q5. How do you handle 'or' in card problems?
If the events are mutually exclusive (e.g., king or queen), add the counts. If they overlap (e.g., heart or king), use n(A or B) = n(A) + n(B) − n(A and B) to avoid double-counting.
Q6. How many cards of each denomination are there?
Each denomination (Ace, 2, 3, ..., 10, Jack, Queen, King) has exactly 4 cards — one in each suit.
Q7. What is the probability of drawing the queen of clubs?
There is exactly 1 queen of clubs in the deck. P(queen of clubs) = 1/52.
Q8. Are jokers included in probability problems?
No. Standard CBSE problems use a 52-card deck without jokers, unless the question explicitly states otherwise.
Q9. What does 'neither A nor B' mean in card probability?
It means the card does not satisfy condition A and does not satisfy condition B. Count = 52 − n(A or B). Use inclusion-exclusion to find n(A or B) first.
Q10. How do you find the probability of drawing a number card?
Number cards are 2 through 10 (nine per suit). Total number cards = 9 × 4 = 36. P(number card) = 36/52 = 9/13. Note: If the question includes Ace as a number card, it becomes 10 × 4 = 40.
