MathsCBSE Class 11 · Probability

Classical Probability: Dice, Coins, Cards and Selections (Class 11)

Updated 17 October 2026 · Fact-checked

When all outcomes of an experiment are equally likely, P(E) = n(E)/n(S), the number of favourable outcomes over the total number of outcomes. List the sample space for dice and coins; use combinations to count favourable and total outcomes for cards, balls and committees. A probability always lies between 0 and 1.

Count all 36 equally likely ordered pairs, then the 6 that sum to 7, to get P = n(E)/n(S) = 1/6.

Illustration drawn for QuestionBuddy.

Understand Classical Probability: Dice, Coins, Cards and Selections

The sample space $S$ is the set of all possible outcomes; an event $E$ is a subset of $S$ . If the $n (S)$ outcomes are equally likely,
$ $P (E) = \frac{n ( E )}{n ( S )}, 0 \leq P (E) \leq 1, P (S) = 1, P (\emptyset) = 0.$ $
The skill is counting both numbers correctly and being sure the outcomes are equally likely.

Dice and coins. Two dice have $36$ equally likely ordered pairs; $(2, 5)$ and $(5, 2)$ are different. Three coins have $8$ outcomes; $n$ coins have $2^{n}$ . For repeated tosses, the number of outcomes with exactly $k$ heads in $n$ tosses is $^{n} C_{k}$ .

Cards. A pack has $52$ cards: $4$ suits of $13$ , $26$ red and $26$ black, $12$ face cards (J, Q, K), $4$ aces. For several cards drawn together, use combinations: total $^{52} C_{k}$ , favourable by choosing from the right groups. For example two aces: $\frac{^{4} C _{2}}{^{52} C _{2}} = \frac{6}{1326} = \frac{1}{221}$ .

Balls and committees (selection without replacement). If $a$ red and $b$ blue balls are in a bag and $k$ are drawn together, the number of outcomes is $^{a + b} C_{k}$ and "exactly $j$ red" has $^{a} C_{j}^{b} C_{k - j}$ favourable outcomes. The same method applies to committees and quality-control samples. "At least one" is easiest as $1 - P (none)$ .

Number problems. For numbers formed from digits, count the total of arrangements and the favourable arrangements; a divisibility by $3$ test uses the digit sum. Select the digit sets first, then arrange.

Calendar problems. A leap year has $52$ weeks and $2$ extra consecutive days; the pair can be (Sun,Mon), (Mon,Tue), ..., (Sat,Sun), seven equally likely cases. Exactly $2$ contain a Sunday, so $P (53 Sundays) = \frac{2}{7}$ .

Check. An answer above $1$ or below $0$ signals a counting error. The probability of an event and of its complement add up to $1$ .

Classical Probability: Dice, Coins, Cards and Selections: formulas to remember

Classical probability: $P (E) = n (E) / n (S)$; Outcomes must be equally likely.
Complement: $P (n o tE) = 1 - P (E)$; Use for "at least one".
Two dice: $36 o u t co m es; s u m 7 occ u r s 6 t im es$; Outcomes are ordered pairs.
n coins: $2^{n} o u t co m es; e x a c tl y k h e a d s : n C k$; Count by combinations.
Cards or balls drawn together: $P = (f a v o u r ab l ese l ec t i o n s) / (t o t a l se l ec t i o n s), b o t h w i t hn C r$; No order, no replacement.

How to solve Classical Probability: Dice, Coins, Cards and Selections questions

Use these steps for classical probability.

1Describe the experiment and write the size of the sample space (list it for dice and coins; use $^{n} C_{r}$ for selections).
2Check that the outcomes are equally likely.
3Define the event in words and count the favourable outcomes carefully (list them or use combinations by group).
4Write $P (E) = n (E) / n (S)$ and simplify the fraction.
5For "at least one", count the complement instead.
6Check that $0 \leq P \leq 1$ .

Classical Probability: Dice, Coins, Cards and Selections 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 with two dice

Suppose a board game lets a player move out of the start square only if the sum of two fair dice is 10 or more.

What is the probability of getting out on a throw?

Try it first, then show the solutionSolution

Favourable outcomes: sum 10: $(4, 6), (5, 5), (6, 4)$ ; sum 11: $(5, 6), (6, 5)$ ; sum 12: $(6, 6)$ . Total $6$ .
Sample space size $36$ .
$P = \frac{6}{36} = \frac{1}{6}$ .

Answer: $\frac{1}{6}$

Memory hook: List the ordered pairs: (4,6) and (6,4) are different outcomes.

Quality inspection

Suppose a box has 10 bulbs, 3 of them defective, and an inspector picks 2 bulbs together at random.

Find the probability that neither bulb is defective.

Try it first, then show the solutionSolution

Total selections: $^{10} C_{2} = 45$ .
Both good: $^{7} C_{2} = 21$ .
$P = \frac{21}{45} = \frac{7}{15}$ .

Answer: $\frac{7}{15}$

Memory hook: Choose from the good group only for "none defective".

Tricks to remember Classical Probability: Dice, Coins, Cards and Selections

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

Ordered pairs for dice
(1,2) and (2,1) are two outcomes, so 36 in all.
Use it when: Dice counts.
Same style top and bottom
If you use combinations for the total, use combinations for the favourable.
Use it when: Cards and balls.
Complement for at least one
1 − P(none) is almost always shorter.
Use it when: Committees, tosses, balls.
Digit sum for 3
A number is divisible by 3 when its digit sum is; pick the digit sets first.
Use it when: Number problems.

Classical Probability: Dice, Coins, Cards and Selections: solved examples

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

Example 1: Two dice · 2 marks · Very short answerBoard level

Two fair dice are thrown. Find the probability that the sum is greater than 9.

Try it first, then show the solutionSolution

Sums 10, 11, 12 occur in $3 + 2 + 1 = 6$ ways. (1 mark)
$P = \frac{6}{36} = \frac{1}{6}$ . (1 mark)

Answer: $\frac{1}{6}$

Example 2: Two cards · 3 marks · Short answerBoard level

Two cards are drawn together from a well-shuffled pack of 52. Find the probability that (i) both are aces, (ii) one is a king and the other a queen.

Try it first, then show the solutionSolution

Total ways: $^{52} C_{2} = 1326$ . (½ mark)
(i) $^{4} C_{2} = 6$ , so $P = \frac{6}{1326} = \frac{1}{221}$ . (1 mark)
(ii) Choose 1 king and 1 queen: $4 \times 4 = 16$ , so $P = \frac{16}{1326} = \frac{8}{663}$ . (1½ marks)

Answer: (i) $\frac{1}{221}$ (ii) $\frac{8}{663}$

Example 3: Five coin tosses · 3 marks · Short answerHarder board level

A fair coin is tossed 5 times. Find the probability of (i) exactly 3 heads, (ii) at most 1 head, (iii) at least one tail.

Try it first, then show the solutionSolution

Sample space size $2^{5} = 32$ . (½ mark)
(i) Exactly 3 heads: $^{5} C_{3} = 10$ , so $\frac{10}{32} = \frac{5}{16}$ . (1 mark)
(ii) At most 1 head: $^{5} C_{0} +^{5} C_{1} = 6$ , so $\frac{6}{32} = \frac{3}{16}$ . (1 mark)
(iii) At least one tail $= 1 - P (all heads) = 1 - \frac{1}{32} = \frac{31}{32}$ . (½ mark)

Answer: (i) $\frac{5}{16}$ (ii) $\frac{3}{16}$ (iii) $\frac{31}{32}$

Example 4: Divisibility by 3 · 4 marks · Short answerHarder board level

A three-digit number is formed from the digits 1, 2, 3, 4, 5 without repetition. Find the probability that it is divisible by 3.

Try it first, then show the solutionSolution

Total numbers: $^{5} P_{3} = 60$ . (1 mark)
A number is divisible by 3 if its digit sum is. Triples with sum a multiple of 3: ${1, 2, 3}$ (6), ${1, 3, 5}$ (9), ${2, 3, 4}$ (9), ${3, 4, 5}$ (12): four sets. (1½ marks)
Each set gives $3! = 6$ numbers: $4 \times 6 = 24$ favourable. (1 mark)
$P = \frac{24}{60} = \frac{2}{5}$ . (½ mark)

Answer: $\frac{2}{5}$

Example 5: Balls from a bag · 5 marks · Long answerHOTS / competency

A bag contains 4 red, 5 blue and 6 green balls. Three balls are drawn together. Find the probability that (i) one ball of each colour is drawn, (ii) all three are of the same colour.

Try it first, then show the solutionSolution

Total selections: $^{15} C_{3} = 455$ . (1 mark)
(i) One of each colour: $4 \times 5 \times 6 = 120$ . (1 mark)
$P = \frac{120}{455} = \frac{24}{91}$ . (1 mark)
(ii) All red $^{4} C_{3} = 4$ , all blue $^{5} C_{3} = 10$ , all green $^{6} C_{3} = 20$ : total $34$ . (1 mark)
$P = \frac{34}{455}$ (does not simplify, since $455 = 5 \times 7 \times 13$ ). (1 mark)

Answer: (i) $\frac{24}{91}$ (ii) $\frac{34}{455}$

Where students lose marks in Classical Probability: Dice, Coins, Cards and Selections

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

Mistake: Counting dice outcomes as unordered
(2,5) and (5,2) are different outcomes.
Wrong working
Sum 7: 3 ways out of 21.
Right working
6 ways out of 36.
Do this: Always use 36 for two dice.
Usually costs: The answer.
Mistake: Mixing permutations and combinations
The numerator and the denominator use different counting styles.
Wrong working
Two aces: $\frac{^{4} C _{2}}{^{52} P _{2}}$ .
Right working
$\frac{^{4} C _{2}}{^{52} C _{2}}$ .
Do this: Use the same style on both.
Usually costs: The answer.
Mistake: Leaving the fraction unsimplified or above 1
Counting errors are not caught.
Wrong working
$P = \frac{6}{36}$ left, or $P = \frac{40}{36}$ accepted.
Right working
Simplify to $\frac{1}{6}$ ; an answer above 1 means recount.
Do this: Check that $0 \leq P \leq 1$ .
Usually costs: 1 mark.
Mistake: Treating non-equally likely outcomes as equally likely
Sum of two dice values are not equally likely.
Wrong working
P(sum = 7) = 1/11 (11 possible sums).
Right working
$\frac{6}{36} = \frac{1}{6}$ .
Do this: Use the 36 equally likely pairs, not the 11 sums.
Usually costs: The answer.

Exam tips

List the sample space for one or two dice or coins; it makes counting visible and earns method marks.
State the total number of outcomes first, then the favourable ones.
For cards and balls drawn together use combinations on both top and bottom.
For "at least one", write "1 − P(none)" as the first line.
Simplify fractions and check the answer lies between 0 and 1.

Practice questions on Classical Probability: Dice, Coins, Cards and Selections

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
A fair die is thrown. The probability of getting a prime number is:
(a) $\frac{1}{3}$
(b) $\frac{1}{2}$
(c) $\frac{2}{3}$
(d) $\frac{1}{6}$
Hint
Primes on a die: 2, 3, 5.
Answer
(b). $\frac{3}{6} = \frac{1}{2}$ .
Q2 · 1 mark · MCQHarder board level
Two fair dice are thrown. The probability that the product of the numbers is even is:
(a) $\frac{1}{4}$
(b) $\frac{1}{2}$
(c) $\frac{3}{4}$
(d) $\frac{5}{6}$
Hint
Use the complement: both numbers odd.
Answer
(c). Both odd: $3 \times 3 = 9$ outcomes; $1 - \frac{9}{36} = \frac{3}{4}$ .
Q3 · 3 marks · Short answerHarder board level
Five cards are drawn together from a pack of 52. Find the probability that all five are hearts.
Hint
Use $^{13} C_{5}$ and $^{52} C_{5}$ .
Answer
$^{13} C_{5} = 1287$ and $^{52} C_{5} = 2598960$ . $P = \frac{1287}{2598960} = \frac{33}{66640}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The probability that a leap year has 53 Sundays is $\frac{2}{7}$ .
Reason (R): A leap year has 52 full weeks and 2 extra consecutive days, and 2 of the 7 equally likely pairs of consecutive days contain a Sunday.
(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.
Hint
List the seven pairs of consecutive days.
Answer
(a). Pairs with a Sunday: (Sat, Sun) and (Sun, Mon), so $\frac{2}{7}$ .
Q5 · 4 marks · Short answerHOTS / competency
Three cards are drawn together from a pack of 52. Find the probability that they are of three different suits.
Hint
Choose 3 suits, then one card from each.
Answer
Choose the 3 suits: $^{4} C_{3} = 4$ ; one card from each: $1 3^{3} = 2197$ . Favourable $= 4 \times 2197 = 8788$ . Total $^{52} C_{3} = 22100$ . $P = \frac{8788}{22100} = \frac{169}{425}$ .

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.What is P(E) when outcomes are equally likely?Show answer
n(E)/n(S).
2.How many outcomes for two dice?Show answer
36.
3.How many outcomes for n coins?Show answer
2^n.
4.How do you find P(at least one)?Show answer
1 − P(none).
5.How many ways to choose 2 cards from 52?Show answer
1326.

Classical Probability: Dice, Coins, Cards and Selections: frequently asked questions

Why do we not use 11 outcomes for the sum of two dice?

The 11 sums are not equally likely. The 36 ordered pairs are, so the sample space uses pairs.

When do I use combinations in probability?

When several objects are chosen together without order, as with cards, balls and committees.

Is a sample space needed for every question?

You should at least state its size. Describing it helps you check that the outcomes are equally likely.

Is the introductory section on random experiments examined?

The chapter page says it is assessed only formatively in 2026-27, but sample spaces are used for every classical probability problem.