MathsICSE Class 10 · Probability

Playing Cards Probability: The 52-Card Pack and Cards Removed (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

A standard pack has 52 cards in four suits of 13: spades and clubs are black, hearts and diamonds are red. Each suit has an ace, 2 to 10, jack, queen and king; the 12 jacks, queens and kings are face cards. Probability = favourable cards ÷ cards in the pack.

Each suit has 13 cards, so black and red each have 26; use P = favourable ÷ 52, and count the queen of diamonds only once in 'or' events.

Illustration drawn for QuestionBuddy.

Understand Playing Cards Probability

Drawing one card from a well-shuffled pack is a standard single-event problem under the CISCE 2027 syllabus ('simple problems on single events').

The pack.

	Spades	Clubs	Hearts	Diamonds
Colour	black	black	red	red
Cards	13	13	13	13

Each suit: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K.

Counts to know.

26 red, 26 black.
4 of each rank: 4 aces, 4 kings, 4 sevens, …
Face (picture) cards: J, Q, K of each suit, so 12 in all; 6 red, 6 black. Aces are not face cards.
Number cards 2 to 10: $9 \times 4 = 36$ .

'Or' events (one card). Count each card once. 'A diamond or a queen': 13 diamonds plus the 3 queens that are not diamonds $= 16$ . Do not add 13 + 4, which counts the queen of diamonds twice.

'Neither … nor' events. Find the cards that are one or the other, then subtract from the total: neither a club nor a face card $= 52 - (13 + 9) = 30$ .

Cards removed. When cards are taken out, work with the new total and recount the favourable cards that remain.

1. Write which cards are removed and how many: new $n (S)$ .
2. For each event, count only cards still in the pack.
3. Then divide and simplify.

A removal can leave some probabilities unchanged: removing all four aces leaves 12 of 48 cards as spades, still $\frac{1}{4}$ .

Playing Cards Probability: formulas to remember

One card: $P (E) = \frac{favourable cards}{52}$; Full, well-shuffled pack.
Face cards: $3 \times 4 = 12$; J, Q, K of each suit; aces are not face cards.
Either A or B (one card): $n (A or B) = n (A) + n (B) - n (both)$; Count the overlap once.
Cards removed: $P (E) = \frac{favourable cards left}{52 - cards removed}$; Recount both.

How to solve Playing Cards Probability questions

Picture the 4 × 13 grid of the pack: rows are suits, columns are ranks.

1Write $n (S)$ : 52, or 52 minus the cards removed.
2Translate the event: colour, suit, rank, face or number card.
3Count from the grid: by suit (13 each), by rank (4 each), or both.
4For 'or', subtract the overlap; for 'neither', subtract from the total.
5If cards were removed, check each counted card is still in the pack.
6Divide and simplify.

Playing Cards Probability in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A picture-card game

Suppose in a family card game you draw one card from a full shuffled pack, and you win a turn if it is a picture card (jack, queen or king).

What is your chance of winning the turn?

Try it first, then show the solutionSolution

Picture cards: 3 in each of 4 suits $= 12$ .
$P = \frac{12}{52} = \frac{3}{13}$ .

Answer: $\frac{3}{13}$

Memory hook: Twelve faces in the pack; the ace has no face.

A pack with cards missing

Imagine your younger brother has lost all four aces from the family pack, and you draw one card from the 48 left.

What is the probability of drawing a spade now? Has it changed?

Try it first, then show the solutionSolution

Spades left: $13 - 1 = 12$ ; total 48.
$P = \frac{12}{48} = \frac{1}{4}$ , the same as with a full pack, because one card was removed from every suit.

Answer: $\frac{1}{4}$ ; unchanged

Memory hook: Remove evenly from every suit, and suit chances stay the same.

Tricks to remember Playing Cards Probability

Short memory aids. Say each one aloud once, then use it on a question the same day.

4 × 13
Four suits down, thirteen ranks across: suit questions count rows, rank questions count columns.
Use it when: Any card count.
Black S and C, red H and D
Spades and Clubs are black; Hearts and Diamonds are red (like a red heart).
Use it when: Colour questions.
Don't count the queen of diamonds twice
In 'diamond or queen', the overlap card belongs to both groups.
Use it when: 'Or' events.

Playing Cards Probability: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: One rank · 1 mark · MCQBoard level

A card is drawn from a well-shuffled pack of 52 cards. The probability that it is a king is:
(a) $\frac{1}{52}$
(b) $\frac{1}{13}$
(c) $\frac{1}{4}$
(d) $\frac{3}{13}$

Try it first, then show the solutionSolution

4 kings: $\frac{4}{52} = \frac{1}{13}$ .

Answer: (b)

Example 2: Colour, rank and 'neither' · 3 marks · Short answerBoard level

One card is drawn from a well-shuffled pack of 52 cards. Find the probability that it is (a) a black card with an even number (2, 4, 6, 8 or 10), (b) a red ace, (c) neither a club nor a face card.

Try it first, then show the solutionSolution

(a) 5 even number cards in each of 2 black suits: 10. $\frac{10}{52} = \frac{5}{26}$ .
(b) Ace of hearts and ace of diamonds: $\frac{2}{52} = \frac{1}{26}$ .
(c) Clubs 13, plus face cards that are not clubs 9: 22. Neither: $52 - 22 = 30$ , so $\frac{30}{52} = \frac{15}{26}$ .

Answer: (a) $\frac{5}{26}$ (b) $\frac{1}{26}$ (c) $\frac{15}{26}$

Example 3: 'Or' and number cards · 3 marks · Short answerHarder board level

One card is drawn from a pack of 52. Find the probability that it is (a) a diamond or a queen, (b) a number card from 2 to 10, (c) not a spade.

Try it first, then show the solutionSolution

(a) 13 diamonds + 3 other queens $= 16$ : $\frac{16}{52} = \frac{4}{13}$ .
(b) $9 \times 4 = 36$ : $\frac{36}{52} = \frac{9}{13}$ .
(c) $1 - \frac{13}{52} = \frac{3}{4}$ .

Answer: (a) $\frac{4}{13}$ (b) $\frac{9}{13}$ (c) $\frac{3}{4}$

Example 4: Red face cards removed · 3 marks · Short answerHarder board level

All the red face cards are removed from a pack of 52 cards. A card is drawn from the remaining cards. Find the probability that it is (a) a face card, (b) a red card, (c) a heart.

Try it first, then show the solutionSolution

Removed: J, Q, K of hearts and of diamonds, 6 cards. $n (S) = 46$ .
(a) Face cards left: 6 black ones: $\frac{6}{46} = \frac{3}{23}$ .
(b) Red cards left: $26 - 6 = 20$ : $\frac{20}{46} = \frac{10}{23}$ .
(c) Hearts left: $13 - 3 = 10$ : $\frac{10}{46} = \frac{5}{23}$ .

Answer: (a) $\frac{3}{23}$ (b) $\frac{10}{23}$ (c) $\frac{5}{23}$

Example 5: Low cards removed · 3 marks · Short answerHOTS / competency

The 2, 3 and 4 of every suit are removed from a pack of 52 cards. One card is drawn from the rest. Find the probability that it is (a) a number card showing an even number, (b) a number card showing a prime number.

Try it first, then show the solutionSolution

Removed $3 \times 4 = 12$ cards; $n (S) = 40$ . Number cards left: 5 to 10.
(a) 6, 8, 10 in 4 suits: 12. $\frac{12}{40} = \frac{3}{10}$ .
(b) Primes left: 5 and 7 (2 and 3 were removed): 8 cards. $\frac{8}{40} = \frac{1}{5}$ .

Answer: (a) $\frac{3}{10}$ (b) $\frac{1}{5}$

Example 6: Case study: a magician's pack · 4 marks · Case-basedHOTS / competency

Before a trick, a magician removes all four jacks and the queen of spades from a pack of 52 cards, shuffles the rest and asks you to draw one card. Find the probability that your card is:
(a) a queen;
(b) a black face card;
(c) a spade;
(d) not a face card.

Try it first, then show the solutionSolution

Removed 5 cards: $n (S) = 47$ .
(a) Queens left: 3: $\frac{3}{47}$ .
(b) Black face cards were J♠, J♣, Q♠, Q♣, K♠, K♣; J♠, J♣ and Q♠ are gone, leaving 3: $\frac{3}{47}$ .
(c) Spades left: $13 - 2 = 11$ : $\frac{11}{47}$ .
(d) Face cards left: $12 - 5 = 7$ , so non-face cards $= 40$ : $\frac{40}{47}$ .

Answer: (a) $\frac{3}{47}$ (b) $\frac{3}{47}$ (c) $\frac{11}{47}$ (d) $\frac{40}{47}$

Where students lose marks in Playing Cards Probability

The slips examiners see most often, and what to write instead.

Mistake: Counting aces as face cards
Face cards are only jacks, queens and kings.
Wrong working
Face cards $= 16$
Right working
Face cards $= 12$
Do this: Face = has a face (a person) on it.
Usually costs: The answer mark.
Mistake: Counting the overlap twice in 'or'
One card can belong to both groups.
Wrong working
Diamond or queen: $13 + 4 = 17$
Right working
$13 + 3 = 16$
Do this: Ask: is there a card that is both?
Usually costs: 1 mark.
Mistake: Old total after cards are removed
The sample space has shrunk.
Wrong working
Red face cards removed; $P (face) = \frac{6}{52}$
Right working
$\frac{6}{46} = \frac{3}{23}$
Do this: Write the new $n (S)$ first.
Usually costs: Every part of the question.
Mistake: Unsimplified fractions
Answers should be in lowest terms.
Wrong working
$\frac{20}{46}$
Right working
$\frac{10}{23}$
Do this: Cancel at the end.
Usually costs: The CISCE examiners' comments on 2024 Q8(ii) and 2025 Q9(ii) report unsimplified answers in card questions of this kind.

Exam tips

Card questions in ICSE use a 'well-shuffled pack of 52 cards'; know the 4 × 13 structure without being told.
Write the favourable cards (for example 'J♠, J♣, Q♠') when the count is small; it earns the method mark.
When cards are removed, state the new total on its own line before any probability.
The CISCE examiners ask that answers be given in lowest terms (2024 Q8(ii), 2025 Q9(ii)).

Practice questions on Playing Cards Probability

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The number of face cards in a standard pack of 52 cards is:
(a) 4
(b) 12
(c) 16
(d) 26
Hint
J, Q, K in each suit.
Answer
(b)
Q2 · 2 marks · Short answerBoard level
A card is drawn from a pack of 52. Find the probability that it is (a) a red queen, (b) a club.
Hint
Two red suits.
Answer
(a) $\frac{2}{52} = \frac{1}{26}$ (b) $\frac{13}{52} = \frac{1}{4}$
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The probability of drawing a face card from a pack of 52 cards is $\frac{3}{13}$ .
Reason (R): There are 12 face cards in a pack of 52 cards.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
$\frac{12}{52}$ .
Answer
(c)
Q4 · 3 marks · Short answerHarder board level
A card is drawn from a pack of 52. Find the probability that it is (a) a black card that is not a face card, (b) the jack of hearts, (c) an ace or a spade.
Hint
(c) has one overlap card.
Answer
(a) $26 - 6 = 20$ : $\frac{5}{13}$ (b) $\frac{1}{52}$ (c) $13 + 3 = 16$ : $\frac{4}{13}$
Q5 · 3 marks · Short answerHarder board level
All the spades are removed from a pack of 52 cards. A card is drawn from the rest. Find the probability that it is (a) a heart, (b) a face card, (c) a black card.
Hint
39 cards remain.
Answer
(a) $\frac{13}{39} = \frac{1}{3}$ (b) $\frac{9}{39} = \frac{3}{13}$ (c) only clubs: $\frac{13}{39} = \frac{1}{3}$
Q6 · 3 marks · Short answerHarder board level
The face cards of diamonds and all four aces are removed from a pack. A card is drawn from the rest. Find the probability that it is (a) a diamond, (b) a red card.
Hint
7 cards removed; the ace of diamonds is one of them.
Answer
$n (S) = 45$ . (a) Diamonds left: $13 - 3 - 1 = 9$ : $\frac{1}{5}$ (b) Red left: $26 - 3 - 2 = 21$ : $\frac{7}{15}$
Q7 · 3 marks · Short answerHOTS / competency
Some red cards are lost from a pack of 52, and no other cards. A card is now drawn, and the probability that it is red is $\frac{1}{3}$ . How many red cards were lost? Give one way this could happen.
Hint
$\frac{26 - x}{52 - x} = \frac{1}{3}$ .
Answer
$78 - 3 x = 52 - x$ , so $x = 13$ : for example, all the hearts were lost (then $\frac{13}{39} = \frac{1}{3}$ ).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Cards in each suit?Show answer
13
2.Number of face cards?Show answer
12
3.Black suits?Show answer
Spades and clubs.
4. $P (ace)$ ?Show answer
$\frac{1}{13}$
5.Number cards 2 to 10 in a pack?Show answer
36

Playing Cards Probability: frequently asked questions

How many face cards are there in a pack of 52 cards?

There are 12 face cards: the jack, queen and king of each of the four suits. Aces are not face cards.

What is the probability of drawing a red card?

There are 26 red cards (hearts and diamonds) in 52, so the probability is 26/52 = 1/2.

How do I find a probability after some cards are removed?

Subtract the removed cards from 52 to get the new total, recount the favourable cards that are still in the pack, then divide and simplify.