MathsCBSE Class 10 · Probability

Probability Problems on Playing Cards for Class 10

Updated 14 October 2026 · Fact-checked

Playing card problems ask for the probability of drawing a certain type of card from a well-shuffled deck of 52 cards. You need the deck structure: 4 suits, 26 red, 26 black, 12 face cards, 4 aces. Some questions remove cards first, use 'or' and 'not', or hide an unknown number of removed cards.

Understand Probability Problems on Playing Cards

A standard deck has 52 cards in 4 suits of 13 cards each. Hearts and diamonds are red. Spades and clubs are black. Each suit has an ace, the number cards 2 to 10, and a jack, a queen and a king. So there are 26 red and 26 black cards, and 4 cards of each rank (4 aces, 4 kings, and so on).

Face cards are only the jack, queen and king. Each suit has 3, so the deck has $3 \times 4 = 12$ face cards, 6 red and 6 black. An ace is not a face card. Number cards (2 to 10) make $9 \times 4 = 36$ cards.

Why does the classical definition work here? 'Well-shuffled' and 'drawn at random' mean every card left in the deck has the same chance of coming out. So each card is one equally likely outcome, and $P (E) = \frac{favourable cards}{cards in the deck}$ . If the deck is not shuffled, or someone looks while drawing, the outcomes are not equally likely and this formula does not apply.

Two ideas from the rest of the chapter appear all the time. For 'A or B' (for example, a king or a red card), add the two counts and subtract the cards that are in both, so that no card is counted twice. For 'not' or 'neither ... nor', use $P (\overline{E}) = 1 - P (E)$ , or count directly what is left.

A common twist is removing cards. When some cards are taken out, the total falls, and the counts of every type may also fall. Before counting anything, write down the new total and work out exactly what each suit now contains. Sometimes the number removed is unknown: then you set the given probability equal to count over $(52 - x)$ and solve a linear equation, just as in the bags-and-balls problems.

Probability Problems on Playing Cards: formulas to remember

Deck structure: $52 = 4 suits \times 13 cards$; Hearts, diamonds are red; spades, clubs are black.
Colour counts: $26 red, 26 black$; Each colour has 2 suits.
Face cards: $3 \times 4 = 12 face cards (6 red, 6 black)$; Jack, queen, king only. Aces are not face cards.
Each rank: $4 aces, 4 kings, 4 queens, 4 jacks, \dots$; One of each rank per suit; 2 red and 2 black.
Probability of a card event: $P (E) = \frac{number of favourable cards}{number of cards in the deck}$; Use the total of cards actually left after any removal.
'Or' count without double counting: $n (A or B) = n (A) + n (B) - n (both)$; Example: king or red = 4 + 26 - 2 = 28.
Complement: $P (not E) = 1 - P (E)$; Use for 'not' and 'neither ... nor' questions.
Number cards and suits: $Number cards (2 to 10) = 9 \times 4 = 36 (18 red, 18 black)$; Each suit has 13 cards: an ace, 9 number cards and 3 face cards (J, Q, K).

How to solve Probability Problems on Playing Cards questions

Use this method for any card question, whether the deck is full or some cards are removed.

1Write the total: 52 for a full deck, or 52 minus the removed cards. Write it as a separate line.
2If cards are removed, note what each suit and each type (face, ace, colour) now contains.
3Read the event and mark key words: 'or', 'and', 'not', 'neither ... nor', 'face card', 'red'.
4Count favourable cards. For 'or', add and subtract the overlap. For 'not', count the rest or use the complement.
5Write $P (E)$ as favourable over total and reduce to lowest terms.
6If a probability is given and the number removed is unknown, form an equation with total $52 - x$ , solve, then check by substitution.
7Check that every answer lies between $0$ and $1$ .

Probability Problems on Playing Cards 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.

Probability Problems on Playing Cards

Imagine you are on a long train journey and your cousin brings a deck of 52 cards. For his game, he removes all four 2s and all four 3s and shuffles the rest well. You draw one card without looking.

You score a bonus if the card is a face card or an ace. Find the probability of a bonus, and the probability that the card is a red number card (2 to 10).

Try it first, then show the solutionSolution

Cards left $= 52 - 8 = 44$ .
Face cards are untouched: 12. Aces are untouched: 4.
Face cards and aces do not overlap, so favourable $= 12 + 4 = 16$ .
$P (bonus) = \frac{16}{44} = \frac{4}{11}$ .
Red number cards in a full deck $= 9 \times 2 = 18$ .
Red 2s and 3s removed $= 4$ , so red number cards left $= 14$ .
$P (red number card) = \frac{14}{44} = \frac{7}{22}$ .

Answer: $P (bonus) = \frac{4}{11}$ , $P (red number card) = \frac{7}{22}$

Memory hook: Cards taken out shrink the total and the group they belong to. Fix the new total first, then recount each group.

Tricks to remember Probability Problems on Playing Cards

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

4-13-26-52
Each rank 4, each suit 13, each colour 26, whole deck 52.
Use it when: Instant counts for kings, spades, red cards.
JQK only
Face cards have faces: Jack, Queen, King. The ace has no face.
Use it when: Stops you counting 16 face cards.
Add, then take away the both
For 'or', add the two groups and subtract the cards that belong to both.
Use it when: King or heart = 4 + 13 - 1 = 16.

Probability Problems on Playing Cards: solved examples

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

Example 1: MCQ after removing cards · 1 mark · MCQHarder board level

The four queens and the two red aces are removed from a well-shuffled deck of 52 cards. A card is drawn at random from the rest. The probability that it is neither a heart nor a king is

(a) $\frac{16}{23}$

(b) $\frac{9}{13}$

(d) $\frac{18}{23}$

Try it first, then show the solutionSolution

Cards left $= 52 - 4 - 2 = 46$ .
Hearts left $= 13 - 1$ (queen) $- 1$ (ace) $= 11$ . Kings are untouched: 4, and the king of hearts is in both groups. Hearts or kings $= 11 + 4 - 1 = 14$ .
Neither $= 46 - 14 = 32$ , so $P = \frac{32}{46} = \frac{16}{23}$ . ( $\frac{9}{13}$ is the full-deck answer, a trap.)

Answer: (a) $\frac{16}{23}$

Example 2: A 2-mark removal question · 2 marks · Very short answerBoard level

The king of hearts and the ace of spades are removed from a well-shuffled deck of 52 cards. A card is drawn at random from the rest. Find the probability that it is (i) a king, (ii) a black card.

Try it first, then show the solutionSolution

Cards left $= 52 - 2 = 50$ . (½ mark)
(i) Kings left $= 3$ , so $P = \frac{3}{50}$ . (½ mark)
(ii) Black cards left $= 26 - 1 = 25$ (the ace of spades has gone), so $P = \frac{25}{50} = \frac{1}{2}$ . (1 mark)

Answer: (i) $\frac{3}{50}$ (ii) $\frac{1}{2}$

Example 3: Assertion-reason on a changed deck · 1 mark · Assertion-reasonBoard level

Assertion (A): If all four aces are removed from a deck of 52 cards, the probability of drawing a face card from the rest is $\frac{3}{13}$ .
Reason (R): A deck of 52 cards has 12 face cards.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

R is true: J, Q and K in each of 4 suits give 12 face cards.
After removing the aces, 48 cards are left and all 12 face cards remain, so $P = \frac{12}{48} = \frac{1}{4}$ , not $\frac{3}{13}$ . ( $\frac{3}{13} = \frac{12}{52}$ uses the old total.) A is false.

Answer: (d) A is false but R is true.

Example 4: Kings and one ace removed · 3 marks · Short answerBoard level

All four kings and the ace of spades are removed from a well-shuffled deck of 52 cards. One card is drawn at random from the rest. Find the probability that it is (i) a face card, (ii) a black card, (iii) an ace.

Try it first, then show the solutionSolution

Cards left $= 52 - 5 = 47$ .
(i) Face cards left: 4 jacks + 4 queens $= 8$ , so $P = \frac{8}{47}$ . (1 mark)
(ii) Black cards left $= 26 - 2$ (black kings) $- 1$ (ace of spades) $= 23$ , so $P = \frac{23}{47}$ . (1 mark)
(iii) Aces left $= 3$ , so $P = \frac{3}{47}$ . (1 mark)

Answer: (i) $\frac{8}{47}$ (ii) $\frac{23}{47}$ (iii) $\frac{3}{47}$

Example 5: 'Or' events from a full deck · 3 marks · Short answerBoard level

One card is drawn from a well-shuffled deck of 52 cards. Find the probability that it is (i) a king or a red card, (ii) neither a red card nor a king, (iii) a red face card or an ace.

Try it first, then show the solutionSolution

(i) Red 26 + kings 4 − red kings 2 $= 28$ ; $P = \frac{28}{52} = \frac{7}{13}$ . (1 mark)
(ii) Complement of (i): $1 - \frac{7}{13} = \frac{6}{13}$ . (1 mark)
(iii) Red face cards 6, aces 4, no overlap: $10$ ; $P = \frac{10}{52} = \frac{5}{26}$ . (1 mark)

Answer: (i) $\frac{7}{13}$ (ii) $\frac{6}{13}$ (iii) $\frac{5}{26}$

Example 6: Red face cards and black aces removed · 3 marks · Short answerBoard level

All the red face cards and both black aces are removed from a well-shuffled deck of 52 cards. One card is drawn at random from the remaining cards. Find the probability that it is (i) a red card, (ii) a number card (2 to 10), (iii) neither a spade nor an ace.

Try it first, then show the solutionSolution

Cards removed $= 6 + 2 = 8$ , so cards left $= 52 - 8 = 44$ . (½ mark)
(i) Red cards left $= 26 - 6 = 20$ ; $P = \frac{20}{44} = \frac{5}{11}$ . (½ mark)
(ii) No number card was removed, so number cards $= 36$ ; $P = \frac{36}{44} = \frac{9}{11}$ . (½ mark)
(iii) Spades left $= 13 - 1 = 12$ (the ace of spades has gone). (½ mark)
Aces left $= 2$ (both red), and neither is a spade, so spade or ace $= 12 + 2 = 14$ . (½ mark)
Neither a spade nor an ace $= 44 - 14 = 30$ ; $P = \frac{30}{44} = \frac{15}{22}$ . (½ mark)

Answer: (i) $\frac{5}{11}$ (ii) $\frac{9}{11}$ (iii) $\frac{15}{22}$

Example 7: Unknown number of cards removed · 5 marks · Long answerHarder board level

From a well-shuffled deck of 52 cards, $x$ black cards, none of them face cards, are removed. A card is now drawn at random, and the probability that it is red is $\frac{13}{24}$ . Find $x$ . Then find the probability that the card drawn is (i) a face card, (ii) a black card.

Try it first, then show the solutionSolution

Cards left $= 52 - x$ ; red cards are untouched $= 26$ . (½ mark)
$\frac{26}{52 - x} = \frac{13}{24}$ . (1 mark)
$13 (52 - x) = 26 \times 24 \Rightarrow 52 - x = 48$ . (1 mark)
$x = 4$ ; check $\frac{26}{48} = \frac{13}{24}$ . (½ mark)
(i) Face cards untouched $= 12$ ; $P = \frac{12}{48} = \frac{1}{4}$ . (1 mark)
(ii) Black cards left $= 26 - 4 = 22$ ; $P = \frac{22}{48} = \frac{11}{24}$ . (1 mark)

Answer: $x = 4$ ; (i) $\frac{1}{4}$ (ii) $\frac{11}{24}$

Example 8: Equal unknown numbers removed from two suits · 5 marks · Long answerHarder board level

From a well-shuffled deck of 52 cards, $x$ number cards (2 to 10) of hearts and $x$ number cards of spades are removed. A card is drawn at random from the rest. The probability that it is neither a heart nor a spade is $\frac{13}{20}$ . Find $x$ . Hence find the probability that the card drawn is (i) a heart or a face card, (ii) a black card.

Try it first, then show the solutionSolution

Cards left $= 52 - 2 x$ . Diamonds and clubs are untouched, so cards that are neither a heart nor a spade $= 26$ . (½ mark)
$\frac{26}{52 - 2 x} = \frac{13}{20}$ . (1 mark)
$13 (52 - 2 x) = 26 \times 20 \Rightarrow 52 - 2 x = 40 \Rightarrow x = 6$ . (1 mark)
Check: $\frac{26}{40} = \frac{13}{20}$ . Cards left $= 40$ . (½ mark)
(i) Hearts left $= 13 - 6 = 7$ , including all 3 heart face cards. Face cards $= 12$ . Heart or face card $= 7 + 12 - 3 = 16$ ; $P = \frac{16}{40} = \frac{2}{5}$ . (1 mark)
(ii) Black cards left $= 26 - 6 = 20$ ; $P = \frac{20}{40} = \frac{1}{2}$ . (1 mark)

Answer: $x = 6$ ; (i) $\frac{2}{5}$ (ii) $\frac{1}{2}$

Example 9: Case study: card booth with lost cards · 4 marks · Case-basedHOTS / competency

Suppose your class runs a card game at the school fair using one deck of 52 cards. During the day, all six black face cards and some black number cards go missing. No red card is lost. At the end of the day, the organiser notes that the probability of drawing a red card from the cards left is $\frac{13}{20}$ .

(i) How many cards are left, and how many of them are black? (1 mark)

(ii) Find the probability that a card drawn at random from the cards left is a face card. (1 mark)

(iii) To make the game easier, the organiser adds $k$ red cards from another deck so that the probability of drawing a red card becomes $\frac{3}{4}$ . Find $k$ . (2 marks)

Try it first, then show the solutionSolution

(i) Red cards $= 26$ , so $\frac{26}{T} = \frac{13}{20}$ , giving $T = 40$ cards left. (½ mark)
Black cards left $= 40 - 26 = 14$ , so 12 black cards were lost: 6 face and 6 number cards. (½ mark)
(ii) Only the red face cards remain: $6$ ; $P = \frac{6}{40} = \frac{3}{20}$ . (1 mark)
(iii) After adding, red $= 26 + k$ and total $= 40 + k$ , so $\frac{26 + k}{40 + k} = \frac{3}{4}$ . (1 mark)
$4 (26 + k) = 3 (40 + k) \Rightarrow 104 + 4 k = 120 + 3 k \Rightarrow k = 16$ ; check $\frac{42}{56} = \frac{3}{4}$ . (1 mark)

Answer: (i) 40 cards, 14 of them black (ii) $\frac{3}{20}$ (iii) $k = 16$

Where students lose marks in Probability Problems on Playing Cards

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

Mistake: Counting aces as face cards
Students think every picture-like card is a face card.
Wrong working
Face cards $= 16$ , $P = \frac{16}{52} = \frac{4}{13}$
Right working
Face cards $= 12$ (J, Q, K), $P = \frac{12}{52} = \frac{3}{13}$
Do this: Say 'JQK' every time you read 'face card'.
Usually costs: The full mark for that part.
Mistake: Double counting in 'or' events
Adding two groups without checking their overlap.
Wrong working
King or red $= 4 + 26 = 30$ , $P = \frac{15}{26}$
Right working
King or red $= 4 + 26 - 2 = 28$ , $P = \frac{7}{13}$
Do this: Always ask: which cards are in both groups?
Usually costs: 1 mark, often the whole part.
Mistake: Keeping 52 as the total after removing cards
The removal is read, but the total is not updated.
Wrong working
After removing 4 kings, $P (queen) = \frac{4}{52}$
Right working
Total $= 48$ , $P (queen) = \frac{4}{48} = \frac{1}{12}$
Do this: Write 'cards left = ...' as your very first line.
Usually costs: Every part of the question.
Mistake: Removing cards from the wrong group
Not checking which colour or type the removed cards belong to.
Wrong working
Red face cards removed, black cards $= 26 - 6 = 20$
Right working
Red face cards removed, black cards still $= 26$ ; total $= 46$ , $P = \frac{13}{23}$
Do this: List, suit by suit, what is left before counting.
Usually costs: 1 to 2 marks.

Exam tips

Write the total number of cards as a separate line; it is often a step mark.
In 'or' parts, show the subtraction of the overlap so the examiner sees your method.
Reduce every fraction to lowest terms; MCQ options are usually given in reduced form.
For a 5-mark unknown-card question, check your value of $x$ by substituting back.
If a type of card is fully removed, its probability is $0$ ; write it, do not skip the part.

Practice questions on Probability Problems on Playing Cards

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

Q1 · 3 marks · Short answerHarder board level
From a well-shuffled deck of 52 cards, $x$ black number cards are removed. A card is then drawn at random, and the probability that it is a face card is $\frac{3}{11}$ . Find $x$ , and the probability that the card drawn is (i) a red number card, (ii) a black card.
Hint
Face cards are untouched, so only the total changes.
Answer
$\frac{12}{52 - x} = \frac{3}{11}$ , so $52 - x = 44$ and $x = 8$ . (i) Red number cards are untouched: 18, so $P = \frac{18}{44} = \frac{9}{22}$ . (ii) Black cards left $= 26 - 8 = 18$ , so $P = \frac{18}{44} = \frac{9}{22}$ .
Q2 · 3 marks · Short answerHarder board level
All the diamonds, the kings of the other three suits and the two black jacks are removed from a deck of 52 cards. A card is drawn at random from the rest. Find the probability that it is (i) a red face card or a black number card, (ii) neither a red card nor a face card, (iii) a king.
Hint
Write down what is left of each suit before counting.
Answer
Left $= 52 - 13 - 3 - 2 = 34$ : 12 hearts (no king), 11 spades and 11 clubs (no king, no jack). (i) Red face cards: $J ♡$ , $Q ♡$ $= 2$ ; black number cards $= 18$ ; total $20$ , $P = \frac{20}{34} = \frac{10}{17}$ . (ii) Black cards left $= 22$ , of which the face cards are the two black queens, so $20$ ; $P = \frac{10}{17}$ . (iii) No king is left, so $P = 0$ .
Q3 · 5 marks · Long answerHarder board level
One card is lost from a deck of 52 cards. A card is drawn at random from the remaining 51. The probability that it is an ace is $\frac{1}{17}$ and the probability that it is a heart is $\frac{4}{17}$ .
(i) Identify the lost card, giving reasons.
(ii) Find the probability that the card drawn is a red face card.
(iii) Find the probability that it is a black card or an ace.
Hint
Convert each probability into a count out of 51.
Answer
(i) Aces left $= \frac{1}{17} \times 51 = 3$ , so the lost card is an ace. Hearts left $= \frac{4}{17} \times 51 = 12$ , so the lost card is a heart. Hence the ace of hearts is lost. (2 marks)
(ii) All 6 red face cards remain: $P = \frac{6}{51} = \frac{2}{17}$ . (1 mark)
(iii) Black cards $= 26$ , and the aces left that are not black are the ace of diamonds only: $26 + 1 = 27$ . $P = \frac{27}{51} = \frac{9}{17}$ . (2 marks)
Q4 · 5 marks · Long answerHarder board level
From a deck of 52 cards, $x$ red number cards are removed. A card is then drawn at random, and the probability that it is black is $\frac{13}{22}$ . Find $x$ , and the probability that the card drawn is (i) a face card, (ii) a red card.
Hint
Black cards stay 26; the total is $52 - x$ .
Answer
$\frac{26}{52 - x} = \frac{13}{22} \Rightarrow 52 - x = 44$ , so $x = 8$ . (i) $\frac{12}{44} = \frac{3}{11}$ . (ii) Red left $18$ , $\frac{18}{44} = \frac{9}{22}$ .
Q5 · 3 marks · Short answerHarder board level
Meera writes: "P(red card or king or jack) $= \frac{26 + 4 + 4}{52} = \frac{17}{26}$ ." Find her error, give the correct probability, and find the probability that the card is neither red, nor a king, nor a jack.
Hint
Which kings and jacks are already counted among the red cards?
Answer
The red kings and red jacks (4 cards) are counted twice. Correct count $= 26 + 2$ (black kings) $+ 2$ (black jacks) $= 30$ , so $P = \frac{30}{52} = \frac{15}{26}$ . Neither: $52 - 30 = 22$ , so $P = \frac{22}{52} = \frac{11}{26}$ .
Q6 · 4 marks · Short answerHOTS / competency
Riya removes some cards from a full deck of 52 cards. Every card she removes is red. Afterwards, for a card drawn at random from the rest, P(red card) $= \frac{10}{23}$ and P(face card) $= \frac{3}{23}$ . Work out exactly which cards Riya removed, and find the probability that a card drawn now is a black card or a jack.
Hint
Let $r$ red cards be removed, so the total is $52 - r$ . Find $r$ first, then use the face-card probability.
Answer
$\frac{26 - r}{52 - r} = \frac{10}{23} \Rightarrow 598 - 23 r = 520 - 10 r \Rightarrow r = 6$ , so 46 cards are left. Face cards left $= \frac{3}{23} \times 46 = 6$ , so 6 face cards were removed. All 6 removed cards are red face cards, so Riya removed exactly the 6 red face cards (J, Q, K of hearts and diamonds). Black or jack: 26 black cards, and the only jacks left are black, so $26$ ; $P = \frac{26}{46} = \frac{13}{23}$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a quiz club's deck has lost all the 2s, 3s, 4s and 5s of every suit, and nothing else.
(i) How many cards are left, and how many of them are red? (1 mark)
(ii) Find the probability of drawing a face card. (1 mark)
(iii) A player wins a prize for a red card or an ace. Find the probability of winning. The club then adds $k$ red number cards from another deck so that P(red card) becomes $\frac{3}{5}$ . Find $k$ . (2 marks)
Hint
16 cards are lost, 8 of them red.
Answer
(i) $52 - 16 = 36$ cards; red $= 26 - 8 = 18$ . (1)
(ii) All 12 face cards remain: $P = \frac{12}{36} = \frac{1}{3}$ . (1)
(iii) Red 18, aces 4, red aces 2: $18 + 4 - 2 = 20$ , so $P = \frac{20}{36} = \frac{5}{9}$ . Then $\frac{18 + k}{36 + k} = \frac{3}{5}$ gives $90 + 5 k = 108 + 3 k$ , so $k = 9$ . Check: $\frac{27}{45} = \frac{3}{5}$ . (2)

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.How many face cards are in a deck?Show answer
12: jack, queen, king of each suit.
2.Which suits are red?Show answer
Hearts and diamonds.
3.How many black kings are there?Show answer
2
4.Is an ace a face card?Show answer
No.
5.How many number cards (2 to 10) are in a deck?Show answer
36
6.P(red queen) from a full deck?Show answer
$\frac{2}{52} = \frac{1}{26}$

Probability Problems on Playing Cards: frequently asked questions

What are face cards in a deck of 52 cards?

Face cards are the jack, queen and king of each suit. There are 12 of them, 6 red and 6 black. The ace is not a face card.

How do I find the probability of drawing a king or a red card?

Add the 4 kings and the 26 red cards, then subtract the 2 red kings counted twice. That gives 28 cards, so the probability is 28 out of 52, which is 7 out of 13.

What changes when some cards are removed?

The total becomes 52 minus the cards removed, and the count of each type may drop too. Work out exactly which cards are left before counting favourable ones.

Do I need to know card games for the exam?

No. You only need the structure of the deck: suits, colours, ranks and face cards. Every question can be solved from these counts.