Quantitative Aptitude · Probability

Counting Techniques and Calculating Simple Probability

Updated 1 October 2026 · Fact-checked

Probability of an event is the number of favourable outcomes divided by the total number of equally likely outcomes. To solve a question, find the size of the sample space, count favourable outcomes using multiplication or nCr, then divide. Use nCr when order does not matter, and simplify the fraction.

Understand Counting Techniques and Calculating Simple Probability

Probability measures how likely an event is. It is a number from 0 to 1. A probability of 0 means the event cannot happen. A probability of 1 means it is certain.

In simple problems, all outcomes are equally likely. A fair coin, a fair die, a well-shuffled pack of cards and a random draw from a bag all work this way. Then the classical definition applies: P(E) = favourable outcomes ÷ total outcomes.

The real work is counting. For one die or one coin you can list outcomes. For two dice there are 6 × 6 = 36 outcomes. For drawing 3 balls from 12, listing is impossible, so you use combinations. If the order of the items drawn does not matter, count with nCr. If the order matters (for example, arranging letters or forming numbers), count with nPr or the multiplication principle.

The key rule: count the favourable outcomes and the total outcomes in the same way. If the total uses nCr (unordered selection), the favourable count must also use nCr. Mixing ordered and unordered counts is the most common source of wrong answers.

Odds are another way to state the same chance. If m outcomes are favourable out of n, the odds in favour are m : (n − m) and the odds against are (n − m) : m. Odds compare favourable with unfavourable. Probability compares favourable with total.

Key formulas to remember

Classical probability: P(E) = m ÷ n; m = favourable outcomes, n = total outcomes. Valid only when all outcomes are equally likely.
Range of probability: 0 ≤ P(E) ≤ 1; A result above 1 or below 0 means you have made an error.
Complement: P(not E) = 1 − P(E); Use it for 'at least one' questions: P(at least one) = 1 − P(none).
Odds in favour: m : (n − m); Favourable : unfavourable.
Odds against: (n − m) : m; Unfavourable : favourable.
Probability from odds: If odds in favour are a : b, P(E) = a ÷ (a + b); If odds against are a : b, then P(E) = b ÷ (a + b).
Combinations: nCr = n! ÷ [r! × (n − r)!]; Use when order of selection does not matter. Also nCr = nC(n − r).
Permutations: nPr = n! ÷ (n − r)!; Use when order matters.
Size of sample space for coins and dice: n coins: 2ⁿ outcomes. n dice: 6ⁿ outcomes.; Two dice: 36. Three coins: 8. Three dice: 216.
Pack of cards facts: 52 cards = 4 suits × 13 cards. 26 red, 26 black. 12 face cards (J, Q, K). 4 aces.; Each suit has 13 cards: A, 2 to 10, J, Q, K.
Drawing r items from a mixed group: P = (aCx × bCy) ÷ (a+b)C(x+y); Choose x from group a and y from group b, without replacement, out of a + b items in total.

How to solve Counting Techniques and Calculating Simple Probability questions

This method works for dice, coins, cards and bag problems. Follow it in order and write each count before you divide.

1Identify the experiment: what is being tossed, rolled or drawn, and how many items. Check whether the draw is with or without replacement.
2Decide whether order matters. If the question only asks which items are drawn, use nCr. If it asks for a sequence or arrangement, use nPr or multiplication.
3Find the total number of outcomes n. For dice and coins use 6ⁿ or 2ⁿ. For draws use nCr from the full group.
4Find the favourable outcomes m with the same counting method. For several groups, multiply the nCr of each group.
5Compute P(E) = m ÷ n and reduce the fraction.
6Check the answer lies between 0 and 1. If the question says 'at least', consider 1 − P(none).
7If the question asks for odds, convert: odds in favour = m : (n − m), odds against = (n − m) : m. Then match the option.

Quickest way: Option elimination and the complement shortcut

When to use it: Use this when you have about a minute per question and four fractions in the options.

Estimate the answer first. If the event is rare, the answer should be small. Strike out options that are clearly too large.
For two dice, think of a 6 × 6 grid with 36 cells. Count the favourable cells directly. Sums: 2 and 12 have 1 way each, 7 has 6 ways, and the counts rise to 7 and then fall.
For 'at least one', compute 1 − P(none). It is usually one short calculation.
For balls in a bag, write only the nCr products. Cancel common factors before multiplying large numbers.
Check the denominator. After reduction, the denominator must be a divisor of the total count. For example, with a total of 12C3 = 220, the reduced denominator must divide 220 (1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110 or 220). Eliminate options that do not fit.
If the count needs more than about two minutes, skip it. Each wrong answer costs 0.25 marks, so do not guess blindly from a long calculation.

Common mistakes in Counting Techniques and Calculating Simple Probability

Counting the favourable outcomes with nCr but the total with a different method, such as nPr or by order.
Students rush the total and use whichever formula comes first.
Fix: Choose one method for the whole question. For unordered draws, use nCr for both the numerator and the denominator.
Treating (H, T) and (T, H) as one outcome for two coins, or treating a sum of 7 on two dice as one outcome.
Students count outcomes by type instead of by equally likely cases.
Fix: List ordered pairs. Two coins give 4 outcomes. Two dice give 36 ordered pairs, and a sum of 7 has 6 of them.
Confusing odds with probability, for example writing P = 3/5 when odds in favour are 3 : 5.
Both use the same two numbers, so students skip the conversion.
Fix: Odds in favour a : b means P = a ÷ (a + b). Add the two numbers to get the total.
Using the same denominator for each draw when the draws are without replacement.
Students forget that the bag shrinks after each draw.
Fix: Either use nCr for the whole draw at once, or reduce the numerator and denominator by 1 after every ball drawn.
Miscounting the cards in a pack, such as taking 13 face cards or 4 cards per suit.
The pack facts are half-remembered.
Fix: Memorise: 52 cards, 4 suits of 13, 12 face cards, 4 aces, 26 red and 26 black.
Computing 'at least one' by adding many cases and missing one.
Students avoid the complement rule.
Fix: Use P(at least one) = 1 − P(none) whenever 'none' is a single simple case.

Worked examples

Example 1

Two fair dice are rolled together. What is the probability that the sum of the numbers is 8?

(a) 1/6
(b) 5/36
(c) 1/9
(d) 7/36

Show the solution

Total outcomes = 6 × 6 = 36.
List pairs with sum 8: (2,6), (3,5), (4,4), (5,3), (6,2).
Favourable outcomes = 5.
P = 5 ÷ 36.

Answer: (b) 5/36

Example 2

A bag has 5 red, 4 blue and 3 green balls. Three balls are drawn at random without replacement. What is the probability that one ball of each colour is drawn?

(a) 3/11
(b) 2/11
(c) 5/22
(d) 1/4

Show the solution

Total balls = 5 + 4 + 3 = 12.
Total ways to draw 3 balls = 12C3 = (12 × 11 × 10) ÷ (3 × 2 × 1) = 220.
Favourable ways = 5C1 × 4C1 × 3C1 = 5 × 4 × 3 = 60.
P = 60 ÷ 220 = 3/11.

Answer: (a) 3/11

Example 3

Two cards are drawn at random without replacement from a well-shuffled pack of 52 cards. What is the probability that both are aces?

(a) 1/221
(b) 1/169
(c) 4/663
(d) 1/26

Show the solution

Total ways to choose 2 cards = 52C2 = (52 × 51) ÷ 2 = 1326.
Favourable ways = 4C2 = 6, since the pack has 4 aces.
P = 6 ÷ 1326 = 1/221.
Check with sequential draws: (4/52) × (3/51) = 12/2652 = 1/221. Both methods agree.

Answer: (a) 1/221

Exam tips

Questions in this topic are mostly direct counts. Practise the pack of cards facts and the 36-outcome dice grid until they are automatic.
Always check whether the draw is with or without replacement. The words are easy to miss, and they change the answer.
If options include odds such as 3 : 5 and probabilities such as 3/8, read the question for which one is asked. Examiners often include both as options.
Use 1 − P(none) for 'at least one' questions. It saves time and cuts errors.
If a question needs a long case-by-case count, mark it and return later. With 0.25 negative marking, take time on questions you can finish and skip the rest.

Practice questions from Probability

Counting Techniques and Calculating Simple Probability: frequently asked questions

How do I solve probability questions with balls from a bag?

Count the total balls and find the total ways to draw using nCr. Then find favourable ways by multiplying nCr for each colour you need. Divide favourable by total and simplify.

When should I use permutation and when combination in probability?

Use combination when you only care which items are selected, as in most draws of balls or cards. Use permutation or multiplication when the order or position matters, such as forming numbers or arranging letters. Use the same method for the total and the favourable count.

What is the formula for odds in favour and odds against?

If m outcomes are favourable out of n, odds in favour are m : (n − m) and odds against are (n − m) : m. To get probability from odds in favour a : b, use a ÷ (a + b).

How many outcomes are there when two dice are rolled?

There are 6 × 6 = 36 equally likely ordered outcomes. Treat (2,5) and (5,2) as different outcomes. This is why a sum of 7 has 6 ways and not 3.