Chapter 14 · Probability

14.2 Probability with Coins, Dice and Cards

Two coins, three coins, two dice, a pack of 52 cards: list the equally likely outcomes without missing or repeating any, count the favourable ones, and the probability follows. The 36-cell dice grid and the card board do the counting with you.

In Section 14.1 you found probabilities from short lists: six faces of a die, a handful of balls in a bag. Real games are richer. A ludo player throws two dice and moves by their sum; a gambler's favourite is the pair of coins; a card player cares about hearts, kings and face cards. The formula is the same, $P(E)=\dfrac mn$, but now both $m$ and $n$ are larger and it is easy to lose count. The craft of this section is listing a sample space in an orderly way, so that every outcome is counted exactly once.

We build the sample spaces step by step: the $4$ outcomes of two coins, the $8$ of three coins, the $36$ of two dice, and the $52$ cards of a standard pack. Along the way you meet a counting rule that tells you the size of such a list in advance, and the complement trick of Theorem 14.2, which turns "at least one" problems into one-line calculations.

Tossing coins

One coin has the sample space $\{H,T\}$. Toss it twice (or toss two coins together; the two ways of doing it are the same experiment). The first toss can be $H$ or $T$, and for each of these the second toss can be $H$ or $T$:

first coinsecond coinoutcome
HHHH
HTHT
THTH
TTTT

The four outcomes $HH,HT,TH,TT$ are equally likely. Notice that $HT$ and $TH$ are two different outcomes: in $HT$ the first coin shows heads and the second tails; in $TH$ it is the other way round. (If you toss two identical coins together, imagine one of them painted red to keep them apart.) The same table-building idea gives the size of any such list.

📐 Theorem 14.3 — Counting the outcomes of two experiments done one after the other
Suppose a first experiment has $p$ equally likely outcomes, and that, whichever outcome occurs, a second experiment then has $q$ equally likely outcomes. Then the pair of experiments has $p\times q$ equally likely outcomes (the ordered pairs). In particular $n$ tosses of a fair coin have $2^n$ equally likely outcomes, and a throw of two fair dice has $6\times6=36$.

Reason. Write the outcomes of the first experiment as $p$ rows. In each row, list the $q$ outcomes of the second experiment. Every pair appears once and only once, and there are $p$ rows of $q$ pairs: $p\times q$ in all. Each pair is equally likely because every step is. For three coins, apply the rule twice: $2\times2\times2=8$.

✏️ Example 1 — Two coins
Two fair coins are tossed together. Find the probability of getting (i) two heads, (ii) exactly one head, (iii) at least one head.
Sample space $\{HH,HT,TH,TT\}$, so $n=4$.
(i) Two heads Only $HH$: $P=\dfrac14$.
(ii) Exactly one head $HT$ and $TH$: $P=\dfrac24=\dfrac12$.
(iii) At least one head "At least one head" means one head or two: $HH,HT,TH$, so $P=\dfrac34$.
Check The complement of "at least one head" is "no head", i.e. $TT$, with probability $\dfrac14$; and $1-\dfrac14=\dfrac34$ ✓.
✏️ Example 2 — Three coins
Three fair coins are tossed together. Find the probability of getting (i) exactly two heads, (ii) at least one tail, (iii) all three coins showing the same face.
Sample space $2\times2\times2=8$ outcomes: $HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$.
(i) Two heads $HHT, HTH, THH$: $P=\dfrac38$.
(ii) At least one tail The only outcome with no tail is $HHH$, so by the complement $P=1-\dfrac18=\dfrac78$.
(iii) Same face $HHH$ and $TTT$: $P=\dfrac28=\dfrac14$.
Check Listing for (ii) gives $7$ outcomes with at least one tail: $\dfrac78$ ✓.

Throwing two dice

Throw a red die and a blue die together. The red die shows one of $6$ numbers and, for each, the blue die shows one of $6$, so there are $6\times6=36$ equally likely outcomes (Theorem 14.3). It is best to write them as ordered pairs $(a,b)$ where $a$ is the red number and $b$ the blue number, arranged in a $6\times6$ grid: row $a$, column $b$. The grid in Interactive 1 below is exactly this list.

📘 Definition 14.6 — Doublet
When two dice are thrown, a doublet is an outcome in which both dice show the same number: $(1,1),(2,2),\dots,(6,6)$.

The sums of the two numbers run from $2$ to $12$, but the eleven sums are not equally likely: a sum of $7$ can be made in six ways, a sum of $2$ in only one. Counting along the grid gives this table, which is worth remembering:

sum23456789101112
ways12345654321

The ways add up to $36$. For instance, a sum of $5$ happens in $4$ ways, $(1,4),(2,3),(3,2),(4,1)$, so $P(\text{sum }5)=\dfrac4{36}=\dfrac19$.

⚠️ Things to be careful about

1. $(2,5)$ and $(5,2)$ are different outcomes. Do not count them as one. Only the doublets $(a,a)$ are single outcomes. The total is $36$, not $21$.

2. The eleven sums are not equally likely. The answer to "probability of a sum of $7$" is $\dfrac6{36}$, not $\dfrac1{11}$. Counting the sums $2,\dots,12$ as equally likely outcomes is the most common error in dice problems — use the $36$ pairs instead.

3. "At least", "at most", "more than". "At least $10$" means $10$, $11$ or $12$. "More than $10$" means $11$ or $12$ only. When "at least one" is hard to list, count the complement: no 6 at all, no head at all.

✏️ Example 3 — Sums and doublets
Two fair dice are thrown together. Find the probability that (i) the sum is $9$, (ii) the dice show a doublet, (iii) the sum is less than $4$.
Total $n=36$ equally likely ordered pairs.
(i) Sum 9 $(3,6),(4,5),(5,4),(6,3)$: $m=4$, $P=\dfrac4{36}=\dfrac19$.
(ii) Doublet $(1,1),\dots,(6,6)$: $m=6$, $P=\dfrac6{36}=\dfrac16$.
(iii) Sum less than 4 The sum is $2$ or $3$: $(1,1)$, $(1,2)$, $(2,1)$. So $m=3$ and $P=\dfrac3{36}=\dfrac1{12}$.
✏️ Example 4 — Using the complement with two dice
Two fair dice are thrown. Find the probability that (i) the sum is not $7$, (ii) at least one die shows $6$, (iii) the sum is a prime number.
(i) Not 7 A sum of $7$ has $6$ ways, so $P(\text{sum }7)=\dfrac6{36}=\dfrac16$ and $P(\text{not }7)=1-\dfrac16=\dfrac56$ (that is $30$ pairs out of $36$).
(ii) At least one 6 The complement is "no die shows $6$": each die has $5$ other faces, so $5\times5=25$ pairs. $P=1-\dfrac{25}{36}=\dfrac{11}{36}$. (Listing: the row $a=6$ has $6$ pairs, the column $b=6$ has $6$, and $(6,6)$ is in both, so $6+6-1=11$ ✓.)
(iii) Prime sum The prime sums are $2,3,5,7,11$ with $1,2,4,6,2$ ways: $1+2+4+6+2=15$ pairs. $P=\dfrac{15}{36}=\dfrac5{12}$.
Interactive 1 — The 36 Outcomes of Two Dice

Each cell is one of the $36$ equally likely outcomes: the red die's number is the row, the blue die's number is the column. Choose an event, and the favourable cells light up. Count them and divide by $36$ — the readouts show the working. Press Play to tour the events. Try "sum is $7$" and then "sum is not $7$": the lit cells and the dim cells swap.

favourable outcomes $m$—
all outcomes $n$36
$P(E)=m/n$—
$P(\overline{E})$—

A pack of 52 playing cards

📘 Definition 14.7 — A standard pack of playing cards
A standard pack has $52$ cards in $4$ suits of $13$ cards each: spades (♠) and clubs (♣), which are black, and hearts (♥) and diamonds (♦), which are red. Each suit has the cards ace, $2,3,\dots,10$, jack, queen, king. The jacks, queens and kings are the face cards: $3$ per suit, $12$ in the pack. When a card is drawn from a well-shuffled pack, each of the $52$ cards is equally likely.

So there are $26$ red cards and $26$ black cards, $4$ aces, $4$ kings, $4$ queens, $4$ jacks, $12$ face cards, and $36$ cards numbered $2$ to $10$. Interactive 2 shows the pack as a board of $4$ rows and $13$ columns so that you can see these counts.

🧭 Problem-solving strategy — cards, dice and coins
  1. Fix the total: $2^n$ for $n$ coins, $36$ for two dice, $52$ for a full pack (fewer if cards have been removed — recount!).
  2. Describe the event and list or count the favourable outcomes in an organised way: by rows of the grid, by suits, by rank.
  3. If the event contains "at least one" or "not", try the complement: count the opposite and subtract from $1$.
  4. Reduce the fraction. Check that $0\le P\le1$.
✏️ Example 5 — One card from a full pack
A card is drawn at random from a well-shuffled pack of $52$ cards. Find the probability that it is (i) a king, (ii) a red card, (iii) a face card, (iv) the ace of spades, (v) a red king, (vi) not a heart.
Total $n=52$.
(i) King $4$ kings: $P=\dfrac4{52}=\dfrac1{13}$.
(ii) Red card $26$ red cards: $P=\dfrac{26}{52}=\dfrac12$.
(iii) Face card $12$ face cards: $P=\dfrac{12}{52}=\dfrac3{13}$.
(iv) Ace of spades Exactly one such card: $P=\dfrac1{52}$.
(v) Red king The kings of hearts and diamonds: $P=\dfrac2{52}=\dfrac1{26}$.
(vi) Not a heart $52-13=39$ cards: $P=\dfrac{39}{52}=\dfrac34$ (or $1-\dfrac14$).
✏️ Example 6 — Cards removed first
All four kings are removed from a well-shuffled pack, and a card is drawn from the remaining cards. Find the probability that it is (i) a heart, (ii) a face card, (iii) an ace, (iv) a king.
New total $52-4=48$ cards remain, so $n=48$ (not $52$).
(i) Heart The king of hearts has gone: $13-1=12$ hearts. $P=\dfrac{12}{48}=\dfrac14$.
(ii) Face card Only jacks and queens remain: $4+4=8$. $P=\dfrac8{48}=\dfrac16$.
(iii) Ace All $4$ aces remain: $P=\dfrac4{48}=\dfrac1{12}$.
(iv) King There are none left, so $P=0$ — an impossible event.
Interactive 2 — The Pack of 52: Suits, Colours and Face Cards

The board shows the whole pack, one row per suit and one column per rank ($A,2,\dots,10,J,Q,K$). Choose an event and the cards favourable to it are lit. The readouts give $m$, $n=52$, the probability in lowest terms and the probability of the complement. Press Play for a tour.

favourable cards $m$—
all cards $n$52
$P(E)=m/n$—
$P(\overline{E})$—

Exercises

I
Coins and dice.
  • 1 Two coins are tossed. Find the probability of getting two tails.
  • 2 Two coins are tossed. Find the probability of getting at least one tail.
  • 3 Three coins are tossed. Find the probability of getting three heads.
  • 4 Three coins are tossed. Find the probability of getting exactly one head.
  • 5 One die is thrown. Find the probability of getting a perfect square.
  • 6 Two dice are thrown. Find the probability that the sum is $5$.
  • 7 Two dice are thrown. Find the probability that the sum is $12$.
  • 8 Two dice are thrown. Find the probability that the sum is greater than $9$.
1 Only $TT$ out of $4$: $P=\dfrac14$.
2 Complement of "no tail" ($HH$): $1-\dfrac14=\dfrac34$.
3 Only $HHH$ out of $8$: $P=\dfrac18$.
4 $HTT, THT, TTH$: $P=\dfrac38$.
5 Perfect squares on a die: $1$ and $4$. $P=\dfrac26=\dfrac13$.
6 $(1,4),(2,3),(3,2),(4,1)$: $P=\dfrac4{36}=\dfrac19$.
7 Only $(6,6)$: $P=\dfrac1{36}$.
8 Sums $10,11,12$ have $3+2+1=6$ pairs. $P=\dfrac6{36}=\dfrac16$.
II
Two dice and one card.
  • 1 Two dice are thrown. Find the probability that the sum is even.
  • 2 Two dice are thrown. Find the probability that the product of the two numbers is $6$.
  • 3 Two dice are thrown. Find the probability that at least one die shows $5$.
  • 4 Two dice are thrown. Find the probability that the two numbers differ by exactly $2$.
  • 5 A card is drawn from a pack of $52$. Find the probability that it is a queen.
  • 6 Find the probability that the card is a diamond.
  • 7 Find the probability that the card is a red face card.
  • 8 Find the probability that the card is numbered from $2$ to $10$.
1 Even sums $2,4,6,8,10,12$ have $1+3+5+5+3+1=18$ pairs. $P=\dfrac{18}{36}=\dfrac12$.
2 $(1,6),(2,3),(3,2),(6,1)$: $P=\dfrac4{36}=\dfrac19$.
3 Complement: no $5$ at all, $5\times5=25$ pairs. $P=1-\dfrac{25}{36}=\dfrac{11}{36}$.
4 $(1,3),(2,4),(3,5),(4,6)$ and their reverses $(3,1),(4,2),(5,3),(6,4)$: $8$ pairs. $P=\dfrac8{36}=\dfrac29$.
5 $4$ queens: $P=\dfrac4{52}=\dfrac1{13}$.
6 $13$ diamonds: $P=\dfrac{13}{52}=\dfrac14$.
7 Red face cards: $J,Q,K$ of hearts and diamonds, $6$ cards. $P=\dfrac6{52}=\dfrac3{26}$.
8 $9$ ranks $\times$ $4$ suits $=36$ cards. $P=\dfrac{36}{52}=\dfrac9{13}$.
III
Complements and harder counting.
  • 1 Two dice are thrown. Find the probability that the sum is less than $12$.
  • 2 Two dice are thrown. Find the probability that neither die shows $4$.
  • 3 Two dice are thrown. Find the probability that the product of the numbers is even.
  • 4 Two dice are thrown. Find the probability that the sum is a multiple of $3$.
  • 5 All four aces are removed from a pack and a card is drawn from the other $48$. Find the probability that it is a spade, and the probability that it is a king.
  • 6 A card is drawn from a full pack. Find the probability that it is neither a heart nor a king.
  • 7 Three coins are tossed. Find the probability of getting at least two heads.
  • 8 Asha says: "Tossing two coins, I can get two heads, two tails or one of each, so each of these has probability $\tfrac13$." Find her mistake and the correct probability of one head and one tail.
1 Only the sum $12$ is excluded, which has one pair: $P=1-\dfrac1{36}=\dfrac{35}{36}$.
2 Each die has $5$ faces other than $4$: $5\times5=25$ pairs, $P=\dfrac{25}{36}$.
3 The product is odd only if both numbers are odd: $3\times3=9$ pairs. So $P(\text{even})=1-\dfrac9{36}=\dfrac{27}{36}=\dfrac34$.
4 Sums $3,6,9,12$ have $2+5+4+1=12$ pairs. $P=\dfrac{12}{36}=\dfrac13$.
5 No spade is removed except the ace of spades: $13-1=12$ spades. $P(\text{spade})=\dfrac{12}{48}=\dfrac14$. All $4$ kings remain: $P(\text{king})=\dfrac4{48}=\dfrac1{12}$.
6 Cards that are hearts or kings: the $13$ hearts and the $3$ other kings, $16$ cards. Neither: $52-16=36$. $P=\dfrac{36}{52}=\dfrac9{13}$.
7 $HHH,HHT,HTH,THH$: $4$ out of $8$. $P=\dfrac12$.
8 The three results are not equally likely, because "one of each" can happen in two ways ($HT$ and $TH$) while the others can happen in one way. The equally likely outcomes are $HH,HT,TH,TT$, so $P(\text{one of each})=\dfrac24=\dfrac12$, $P(\text{two heads})=P(\text{two tails})=\dfrac14$.
IV
Applications.
  • 1 A spinner has $8$ equal sectors: $3$ red, $2$ blue, $2$ yellow and $1$ green. Find the probability that it stops on blue, and on a colour other than green.
  • 2 In ludo you need a six. Find the probability of a six on one throw of a die, of not a six, and of getting at least one six if you throw two dice together.
  • 3 In a game, two coins are tossed. You win $\text{₹}\,10$ for two heads and $\text{₹}\,5$ for exactly one head. Find the probability of winning $\text{₹}\,10$ and the probability of winning some money.
  • 4 In a board game two dice are thrown and the player moves forward by their sum. Find the probability of moving at least $9$ squares.
  • 5 The queen of hearts and the jack of clubs are lost from a pack. A card is drawn from the remaining $50$. Find the probability that it is a heart and the probability that it is a face card.
  • 6 A raffle drum holds $100$ tickets numbered $1$ to $100$. One ticket is drawn. Find the probability that the number is a perfect square, and that it is a multiple of $7$.
1 $P(\text{blue})=\dfrac28=\dfrac14$. Not green: $8-1=7$ sectors, $P=\dfrac78$.
2 One die: $P(6)=\dfrac16$, $P(\text{not }6)=\dfrac56$. Two dice: no six at all has $5\times5=25$ pairs, so $P(\text{at least one 6})=1-\dfrac{25}{36}=\dfrac{11}{36}$.
3 $P(\text{₹}\,10)=P(HH)=\dfrac14$. Winning something means at least one head: $\dfrac34$.
4 Sums $9,10,11,12$ have $4+3+2+1=10$ pairs. $P=\dfrac{10}{36}=\dfrac5{18}$.
5 Hearts left: $13-1=12$, so $P=\dfrac{12}{50}=\dfrac6{25}$. Face cards left: $12-2=10$, so $P=\dfrac{10}{50}=\dfrac15$.
6 Perfect squares $1,4,9,\dots,100$: ten of them, $P=\dfrac{10}{100}=\dfrac1{10}$. Multiples of $7$: $7,14,\dots,98$, fourteen of them, $P=\dfrac{14}{100}=\dfrac7{50}$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE Main

How this is examined: a board paper almost always has a two-dice or two-coin question (2–3 marks, often with a complement), a pack-of-cards question (usually with some cards removed, 3 marks) and one probability MCQ or Assertion–Reason item. JEE Main-style questions ask for the same counting, with trickier conditions. Show the sample space size and list the favourable outcomes.

E1. Board · 1 mark Two fair coins are tossed together. The probability of getting at least one head is  (A) $\tfrac14$  (B) $\tfrac12$  (C) $\tfrac34$  (D) $1$

Idea The outcomes are $HH,HT,TH,TT$. At least one head: $HH,HT,TH$, so $\dfrac34$. (Or $1-P(TT)=1-\dfrac14$.)
Trap (B) $\tfrac12$ is the probability of exactly one head; (A) is the probability of two heads.
Answer (C).

E2. Board · 1 mark Assertion (A): When two coins are tossed, the probability of getting one head and one tail is $\tfrac13$. Reason (R): When two coins are tossed, the four outcomes $HH$, $HT$, $TH$, $TT$ are equally likely.
(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.

Check R True: the four ordered outcomes are equally likely.
Check A "One head and one tail" is the two outcomes $HT$ and $TH$, so the probability is $\tfrac24=\tfrac12$, not $\tfrac13$. A is false (the "$\tfrac13$" comes from treating "one of each" as a single outcome).
Answer (D).

E3. Board · 2 marks Two dice are thrown together. Find the probability that the sum of the numbers is a perfect square.

Total (½ mark) $n=6\times6=36$ outcomes.
Favourable (1 mark) The sum lies between $2$ and $12$, so the perfect squares possible are $4$ and $9$. Sum $4$: $(1,3),(2,2),(3,1)$ — $3$ ways. Sum $9$: $(3,6),(4,5),(5,4),(6,3)$ — $4$ ways. So $m=3+4=7$.
Answer (½ mark) $P=\dfrac7{36}$.

E4. Board · 3 marks All the red face cards are removed from a well-shuffled pack of $52$ cards. A card is drawn at random from the remaining cards. Find the probability that it is (i) a king, (ii) a heart, (iii) a face card.

Total (1 mark) The red face cards are the jack, queen and king of hearts and of diamonds: $6$ cards. $52-6=46$ cards remain, so $n=46$.
(i) King (½ mark) Both red kings are gone; the $2$ black kings remain. $P=\dfrac2{46}=\dfrac1{23}$.
(ii) Heart (1 mark) $13-3=10$ hearts remain. $P=\dfrac{10}{46}=\dfrac5{23}$.
(iii) Face card (½ mark) Only the black jacks, queens and kings are left: $6$ cards. $P=\dfrac6{46}=\dfrac3{23}$.

E5. JEE Main Two fair dice are thrown. The probability that the product of the two numbers is a perfect square is  (A) $\tfrac16$  (B) $\tfrac29$  (C) $\tfrac14$  (D) $\tfrac5{18}$

Idea Go through the first number $a=1,\dots,6$ and ask which second numbers $b$ make $ab$ a perfect square.
Count $a=1$: $b=1,4$. $a=2$: $b=2$ (the other candidate $b=8$ is not on a die). $a=3$: $b=3$. $a=4$: $b=1,4$. $a=5$: $b=5$. $a=6$: $b=6$. That is $2+1+1+2+1+1=8$ pairs: the six doublets, plus $(1,4)$ and $(4,1)$.
Probability $P=\dfrac8{36}=\dfrac29$. Option (A) $\tfrac16=\dfrac6{36}$ counts only the doublets; (D) is $\dfrac{10}{36}$.
Answer (B).

E6. Board · Case study A game of two dice. Priya and Rahul play a game with two fair dice, one red and one blue. Each throws both dice together on their turn.
(i) How many equally likely outcomes are there, and what is the probability that the sum is $7$? (2 marks)
(ii) Rahul scores if both numbers are odd. Find the probability that he scores. (1 mark)
(iii) Priya scores unless the sum is $7$. Find the probability that she scores. (1 mark)

(i) There are $6\times6=36$ outcomes. A sum of $7$: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — $6$ pairs. $P=\dfrac6{36}=\dfrac16$. (2 marks)
(ii) Odd numbers are $1,3,5$ on each die: $3\times3=9$ pairs. $P=\dfrac9{36}=\dfrac14$.
(iii) "Sum is not $7$" is the complement of (i): $P=1-\dfrac16=\dfrac56$.
🔬 Research–Teaching NexusMini project · optional, about 30 minutes

Research question. When two dice are thrown, is the sum $7$ really the most common, and are the eleven possible sums really not equally likely?

Why it matters. Board-game designers choose rules using exactly this counting: in many games a roll of $7$ has a special meaning because it is the likeliest. Casinos, lotteries and insurance companies all work out what to expect by counting equally likely outcomes, and then check it against records.

What to do.

  1. Draw the $6\times6$ grid of outcomes and write in each cell the sum of its row and column numbers. Count how many cells show each sum $2,\dots,12$ and turn these into probabilities out of $36$.
  2. With two dice (or two dice-rolling apps), throw them $180$ times; with friends, share the work, say $3$ people with $60$ throws each. Tally the sum each time.
  3. For each sum, multiply its probability by $180$ to get the expected count, and compare with your tally in a table.
  4. Draw a bar chart of your tally. Mark the expected counts as a line on the same chart.
  5. Write a short note: which sum was most common? Were the eleven sums close to equally likely?

Tools: Two dice (or a dice app), a tally sheet, squared paper, a calculator

Hand in: The grid with sums, the table of expected and observed counts, the bar chart and your note.

Going further. Repeat with the product of the two numbers. Which products are likeliest? How many cells give a product of $12$?

Teacher/parent note: with $180$ throws the expected counts for the sums $2,3,\dots,12$ are $5,10,15,20,25,30,25,20,15,10,5$. Observed counts commonly differ by $4$–$6$ from these, so the shape (a peak at $7$, symmetric) matters more than exact numbers. The sums are far from equally likely: $7$ should come up about six times as often as $2$ or $12$.

🧪 Self-check quiz

0/5 correct

Q1. Two fair coins are tossed together. The probability of getting exactly one head is

Q2. The number of equally likely outcomes when two dice are thrown together is

Q3. Two dice are thrown. The probability that the sum is $8$ is

Q4. A card is drawn from a well-shuffled pack of $52$ cards. The probability that it is a red king is

Q5. A card is drawn from a well-shuffled pack of $52$ cards. The probability that it is not a face card is

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