Probability with Coins, Dice and Cards

For outcomes that are all equally likely, the probability of an event is the number of outcomes in which it happens divided by the total number of outcomes:

P(event) = favourable outcomes ÷ total outcomes

Probability always lies between 0 (impossible) and 1 (certain), and the probability that an event does not happen is 1 minus the probability that it does. Exam questions on probability are therefore counting questions: count the total, count the favourable cases, and divide. The three common settings, coins, dice and cards, each have a fixed sample space worth knowing by heart.

Each coin has 2 outcomes, so n coins have 2ⁿ outcomes: 4 for two coins, 8 for three, 16 for four. List them for small n; for larger n, count with combinations. The number of outcomes with exactly k heads is nCk.

Worked example 1: exactly one head with two coins

The outcomes are HH, HT, TH and TT. Exactly one head occurs in HT and TH, so P = 2/4 = 1/2.

Worked example 2: at least one head with three coins

The only outcome with no head is TTT, 1 of 8. So P(at least one head) = 1 − 1/8 = 7/8.

Two dice give 6 × 6 = 36 outcomes. For sums, it helps to know how many ways each total can occur:

Sum	2	3	4	5	6	7	8	9	10	11	12
Ways	1	2	3	4	5	6	5	4	3	2	1

The pattern rises by one up to 7 and falls by one after it, so you can rebuild the table in seconds.

Worked example 3: a total of 7

There are 6 ways to make 7, so P = 6/36 = 1/6.

Worked example 4: a total of 8

There are 5 ways: (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2). P = 5/36.

Worked example 5: an even product

A product is odd only when both dice are odd, which happens in 3 × 3 = 9 outcomes. So P(even product) = 1 − 9/36 = 3/4.

Group	Number of cards
Suits: spades, clubs (black); hearts, diamonds (red)	13 each
Red cards; black cards	26 each
Each rank, such as aces or kings	4
Face cards (jack, queen, king)	12

Worked example 6: a face card

There are 12 face cards, so P = 12/52 = 3/13.

For "A or B", add the two counts and subtract the cards that are both, so nothing is counted twice.

Worked example 7: a king or a heart

4 kings plus 13 hearts counts the king of hearts twice, so the favourable cards are 4 + 13 − 1 = 16. P = 16/52 = 4/13.

When several balls are drawn together, the order does not matter, so count with combinations. The total is the number of ways to choose that many balls from the bag; the favourable count is the number of ways to choose them from the required colours.

Worked example 8: two red balls

A bag has 5 red and 3 blue balls. Two are drawn. Ways to draw 2 red: 5C2 = 10. Ways to draw any 2: 8C2 = 28. P = 10/28 = 5/14.

If the balls are drawn one at a time without replacement, the answer is the same: 5/8 × 4/7 = 20/56 = 5/14. Drawing with replacement is different, because the bag is restored after each draw.

Two events are independent when one has no effect on the other, such as tossing a coin and rolling a die. The probability that both happen is the product of their separate probabilities. This is often faster than listing the combined sample space.

Worked example 9: a head and a six

A coin is tossed and a die is rolled. P(head) = 1/2 and P(six) = 1/6, so P(head and six) = 1/2 × 1/6 = 1/12. Listing confirms it: the 12 combined outcomes include exactly one, (H, 6).

✘ Counting (3, 4) and (4, 3) as one outcome with two dice. They are different.
✘ Double counting cards that belong to both groups in an "or" question.
✘ Counting aces as face cards.
✘ Adding many cases for "at least one" instead of using 1 − P(none).

Counting methods are on permutations and combinations, and dice faces in reasoning papers are covered in dice problems. The formula list is in the Quant cheat code.

Q1. Three coins are tossed. What is the probability of getting exactly two heads? easy

(a) 1/2 (b) 3/8 (c) 1/4 (d) 2/3

✅ Answer: (b) 3/8

💡 Explanation: There are 2³ = 8 outcomes. Exactly two heads happens in 3 of them: HHT, HTH and THH. The probability is 3/8.

Q2. Two dice are thrown. What is the probability that the total is 9? easy

(a) 1/6 (b) 1/12 (c) 5/36 (d) 1/9

✅ Answer: (d) 1/9

💡 Explanation: Out of 36 outcomes, a total of 9 comes from (3, 6), (4, 5), (5, 4) and (6, 3), which is 4 outcomes. 4/36 = 1/9.

Q3. A card is drawn from a well-shuffled deck of 52. What is the probability that it is a red king? easy

(a) 1/26 (b) 1/13 (c) 1/52 (d) 1/2

✅ Answer: (a) 1/26

💡 Explanation: There are two red kings, of hearts and of diamonds. The probability is 2/52 = 1/26.

Q4. A bag holds 4 red and 6 green balls. Two balls are drawn at random. What is the probability that both are green? medium

(a) 9/25 (b) 3/5 (c) 1/3 (d) 2/15

✅ Answer: (c) 1/3

💡 Explanation: Ways to pick 2 green: 6C2 = 15. Ways to pick any 2: 10C2 = 45. The probability is 15/45 = 1/3.

Q5. Two dice are thrown. What is the probability of a doublet, the same number on both? easy

(a) 1/6 (b) 1/36 (c) 1/12 (d) 1/3

✅ Answer: (a) 1/6

💡 Explanation: The doublets are (1, 1) through (6, 6), which is 6 of the 36 outcomes. 6/36 = 1/6.

Q6. Four coins are tossed. What is the probability of at least one head? medium

(a) 1/16 (b) 3/4 (c) 1/2 (d) 15/16

✅ Answer: (d) 15/16

💡 Explanation: The only outcome with no head is TTTT, 1 of 2⁴ = 16. So P(at least one head) = 1 − 1/16 = 15/16.

Which cards count as face cards?

The jack, queen and king of each suit, which is 12 cards in all. Aces are not face cards. Some older books count aces as well, so read the question carefully, but the common modern convention is 12 face cards.

Why is P(at least one) usually found by subtraction?

'At least one' covers many cases, such as one, two or three heads, while its opposite, 'none', is a single case. Finding P(none) and subtracting it from 1 is faster and less likely to miss a case.

All problems and solutions on this page are original. No outside facts are relied on; every numeric answer is recomputed by software before publishing.

How this page was prepared. Drafted with AI assistance (Claude (Anthropic AI model)) from the sources listed above. 15 calculations on this page were recomputed by software and matched. Automated checks also covered structure, internal links and duplicate text. Software checks cannot confirm facts. Reviewed by praveenkumar on 2026-10-23 against the sources listed above.

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Quant Cheat Code: 20 Must-Know Shortcuts & Formulas