Two Coins Have Four Outcomes, Not Three, and the Difference Decides the Answer
Learn to list the sample space for a random experiment, write an event as a set of outcomes and find its probability, draw a tree diagram for a two-stage experiment, and handle at-least-one questions.
How many outcomes does tossing two coins really have?
Toss two coins. A natural way to list what can happen is: two heads, one head, no heads — three possibilities.
From that list, the probability of exactly one head looks like .
It is not. Toss two coins a hundred times and exactly one head turns up around half the time, not a third.
The reason is that one head can happen two different ways. If the coins are a rupee and a two-rupee coin, then head-on-the-rupee with tail-on-the-two is a different outcome from tail-on-the-rupee with head-on-the-two. Written properly:
Four outcomes, each equally likely, and exactly one head covers two of them:
The counting formula from the previous part of this chapter is only as good as the list it is applied to, and listing the outcomes properly is the whole skill of this page. It covers the second part of the CBSE Class 9 Mathematics chapter on probability — the sample space, events as sets, tree diagrams, and compound events.
From that list, the probability of exactly one head looks like .
It is not. Toss two coins a hundred times and exactly one head turns up around half the time, not a third.
The reason is that one head can happen two different ways. If the coins are a rupee and a two-rupee coin, then head-on-the-rupee with tail-on-the-two is a different outcome from tail-on-the-rupee with head-on-the-two. Written properly:
Four outcomes, each equally likely, and exactly one head covers two of them:
The counting formula from the previous part of this chapter is only as good as the list it is applied to, and listing the outcomes properly is the whole skill of this page. It covers the second part of the CBSE Class 9 Mathematics chapter on probability — the sample space, events as sets, tree diagrams, and compound events.
How do you list the sample space for an experiment?
Write down every distinct result the experiment can produce, in a fixed order so nothing is missed or repeated. That complete list is the sample space, written , and the number of outcomes is .
Worked example 1 — one die.
Worked example 2 — one coin. , so .
Worked example 3 — two coins.
Worked example 4 — three coins. Each coin doubles the list:
Worked example 5 — two dice. Record the result as an ordered pair, first die then second:
The pairs and are different outcomes, exactly as and are.
Worked example 6 — a card from a pack. , one outcome per card.
Worked example 7 — a coin and a die together. Each of the coin results pairs with each of the die results:
and the list runs .
The multiplication pattern. If the first stage has outcomes and the second has , the pair has . Two coins give ; three give ; two dice give ; a coin and a die give .
Order matters when the stages are distinguishable. For two dice thrown together it may feel as though and are the same event, and for the sum they give the same value. But they are two separate outcomes of the experiment, and the equally-likely count needs both. **Collapsing them would give outcomes instead of , and those are not equally likely — which is the two-coin error from the opening, one size larger.
Write the list in a system, not at random. For three coins, fixing the first letter and then working through the rest produces all eight without hesitation. A haphazard list is where a missing outcome hides**, and a missing outcome changes the denominator of every answer that follows.
Worked example 1 — one die.
Worked example 2 — one coin. , so .
Worked example 3 — two coins.
Worked example 4 — three coins. Each coin doubles the list:
Worked example 5 — two dice. Record the result as an ordered pair, first die then second:
The pairs and are different outcomes, exactly as and are.
Worked example 6 — a card from a pack. , one outcome per card.
Worked example 7 — a coin and a die together. Each of the coin results pairs with each of the die results:
and the list runs .
The multiplication pattern. If the first stage has outcomes and the second has , the pair has . Two coins give ; three give ; two dice give ; a coin and a die give .
Order matters when the stages are distinguishable. For two dice thrown together it may feel as though and are the same event, and for the sum they give the same value. But they are two separate outcomes of the experiment, and the equally-likely count needs both. **Collapsing them would give outcomes instead of , and those are not equally likely — which is the two-coin error from the opening, one size larger.
Write the list in a system, not at random. For three coins, fixing the first letter and then working through the rest produces all eight without hesitation. A haphazard list is where a missing outcome hides**, and a missing outcome changes the denominator of every answer that follows.
Formula
How do you write an event as a set and find its probability?
List the outcomes that make the event happen, count them, and divide by the size of the sample space:
**Worked example 1 — two coins, .**
- At least one head: , so and
- Exactly one head: , so
- No head: , so
- Two heads: , so
Check: , covering two heads, one head and none.
**Worked example 2 — three coins, .**
- Exactly two heads: , so
- At least one head: everything except , so
- All three the same: , so
**Worked example 3 — two dice, .**
- *Sum is *: , so and
- *Sum is *: only , so
- A doublet: , so
- *Sum more than *: sum has pairs, sum has , sum has , giving in all, so
- *Sum is *: impossible,
**Worked example 4 — a card, .**
- A face card: of them, so
- Not a heart: cards, so
- A red king: cards, so
**Sum is the most likely total on two dice**, with ways, while and have one each. That asymmetry is invisible until the sample space is written out, and it is the clearest illustration of why the list comes before the division.
An impossible event is the empty set and a certain event is the whole sample space. For two dice, *sum * has , and *sum between and * has . **So and are not special rules** — they are what the same formula returns at the two extremes.
**Worked example 1 — two coins, .**
- At least one head: , so and
- Exactly one head: , so
- No head: , so
- Two heads: , so
Check: , covering two heads, one head and none.
**Worked example 2 — three coins, .**
- Exactly two heads: , so
- At least one head: everything except , so
- All three the same: , so
**Worked example 3 — two dice, .**
- *Sum is *: , so and
- *Sum is *: only , so
- A doublet: , so
- *Sum more than *: sum has pairs, sum has , sum has , giving in all, so
- *Sum is *: impossible,
**Worked example 4 — a card, .**
- A face card: of them, so
- Not a heart: cards, so
- A red king: cards, so
**Sum is the most likely total on two dice**, with ways, while and have one each. That asymmetry is invisible until the sample space is written out, and it is the clearest illustration of why the list comes before the division.
An impossible event is the empty set and a certain event is the whole sample space. For two dice, *sum * has , and *sum between and * has . **So and are not special rules** — they are what the same formula returns at the two extremes.
How do you draw a tree diagram for a two-stage experiment?
Draw one branch per outcome of the first stage, then repeat the whole set of second-stage branches at the end of each. Every complete path from left to right is one outcome.
Worked example 1 — two coins. The first coin gives two branches, and . At the tip of each, the second coin gives two more:
- then — the path
- then — the path
- then — the path
- then — the path
Four paths, matching . **The tree makes and visibly different, because they are different routes through the diagram — which is precisely the point the opening section had to argue for in words.
Worked example 2 — a coin then a die.** Two first-stage branches, each splitting into six:
So , since exactly one path is . And , from , , .
Worked example 3 — two draws from a bag, with replacement. A bag holds red and blue balls. One is drawn, its colour noted, and it is put back before the second draw.
Each branch carries its probability. First stage: with , with . Because the ball went back, the second stage has the same two probabilities on every branch.
Multiply along each path:
- :
- :
- :
- :
Check: . Correct.
So and .
Multiply along a path, add between paths. That is the whole arithmetic of a tree: and along the branches, or down the list of paths.
Worked example 4 — three coins on a tree. Three stages give paths, each with probability , and the three paths with exactly two heads give as before.
**The branches at every fork must sum to .** In worked example 3 each fork is , and if they do not sum to an outcome has been left out. That check is worth making at every fork before multiplying anything, because a missing branch produces a full set of answers that are all quietly wrong.
Replacement matters. Without putting the ball back, the second fork would read and after a red, and and after a blue — different numbers on different branches. The tree handles that without difficulty, which is exactly why it is worth drawing rather than reaching for a formula.
Worked example 1 — two coins. The first coin gives two branches, and . At the tip of each, the second coin gives two more:
- then — the path
- then — the path
- then — the path
- then — the path
Four paths, matching . **The tree makes and visibly different, because they are different routes through the diagram — which is precisely the point the opening section had to argue for in words.
Worked example 2 — a coin then a die.** Two first-stage branches, each splitting into six:
So , since exactly one path is . And , from , , .
Worked example 3 — two draws from a bag, with replacement. A bag holds red and blue balls. One is drawn, its colour noted, and it is put back before the second draw.
Each branch carries its probability. First stage: with , with . Because the ball went back, the second stage has the same two probabilities on every branch.
Multiply along each path:
- :
- :
- :
- :
Check: . Correct.
So and .
Multiply along a path, add between paths. That is the whole arithmetic of a tree: and along the branches, or down the list of paths.
Worked example 4 — three coins on a tree. Three stages give paths, each with probability , and the three paths with exactly two heads give as before.
**The branches at every fork must sum to .** In worked example 3 each fork is , and if they do not sum to an outcome has been left out. That check is worth making at every fork before multiplying anything, because a missing branch produces a full set of answers that are all quietly wrong.
Replacement matters. Without putting the ball back, the second fork would read and after a red, and and after a blue — different numbers on different branches. The tree handles that without difficulty, which is exactly why it is worth drawing rather than reaching for a formula.
What is the fastest way to find the probability of at least one head?
**Find the probability of NONE and subtract from .** At least one is the complement of none, and none is almost always a single outcome.
Worked example 1 — two coins. No head is the single outcome , so and
Check by listing: gives . The same answer, reached by counting three outcomes instead of one.
Worked example 2 — three coins. No head is only :
Listing would have meant writing out seven outcomes. The saving grows with the number of stages, and that is why the complement is the standard first move.
Worked example 3 — four coins. Now and no head is one outcome:
Worked example 4 — two draws with replacement. From the bag of red and blue, at least one blue is the complement of both red:
Check by adding paths: . Correct, and slower.
Worked example 5 — two dice, at least one six. No six means both dice show one of the other five faces, giving outcomes:
Check by listing: the first die shows six in pairs, the second die shows six in pairs, and was counted in both — so . The same answer, and the subtraction of the double-counted case is the step the complement method avoids entirely.
"At least one" and "exactly one" are different events. For two coins, at least one head is and exactly one head is , because at least one includes and exactly one excludes it. Both phrases appear in the same question paper, and reading the wrong one produces a perfectly correct answer to the other question.
The complement is not a trick, it is the same counting. Every outcome is either in or not, so . Dividing through by gives — which means the shortcut is guaranteed to agree with the long method, and the two can always be used to check each other.
Worked example 1 — two coins. No head is the single outcome , so and
Check by listing: gives . The same answer, reached by counting three outcomes instead of one.
Worked example 2 — three coins. No head is only :
Listing would have meant writing out seven outcomes. The saving grows with the number of stages, and that is why the complement is the standard first move.
Worked example 3 — four coins. Now and no head is one outcome:
Worked example 4 — two draws with replacement. From the bag of red and blue, at least one blue is the complement of both red:
Check by adding paths: . Correct, and slower.
Worked example 5 — two dice, at least one six. No six means both dice show one of the other five faces, giving outcomes:
Check by listing: the first die shows six in pairs, the second die shows six in pairs, and was counted in both — so . The same answer, and the subtraction of the double-counted case is the step the complement method avoids entirely.
"At least one" and "exactly one" are different events. For two coins, at least one head is and exactly one head is , because at least one includes and exactly one excludes it. Both phrases appear in the same question paper, and reading the wrong one produces a perfectly correct answer to the other question.
The complement is not a trick, it is the same counting. Every outcome is either in or not, so . Dividing through by gives — which means the shortcut is guaranteed to agree with the long method, and the two can always be used to check each other.
Exam tip
Exam tip: write the sample space in full before you divide
**Write out and state before finding any probability. A wrong denominator spoils every part of a multi-part question.
Two coins have FOUR outcomes** — , , , — and three coins have eight. Never collapse and .
**Two dice have outcomes**, and differs from . Never reduce to .
List in a system: fix the first letter or the first die, then work through the rest. Random listing is where a missing outcome hides.
Write the event as a set, not just a count: , so . Marks are given for the set.
On a tree, multiply along a path and add between paths — and along, or down.
**Check each fork's branches sum to before multiplying. A missing branch makes every answer wrong at once.
Check the path probabilities sum to ** at the end: .
**For "at least one", use .** Three coins: . Two dice, at least one six: .
"At least one" is not "exactly one" — against for two coins. Underline which the question says.
And note whether there is replacement — without it, the second fork's probabilities differ on each branch.
Two coins have FOUR outcomes** — , , , — and three coins have eight. Never collapse and .
**Two dice have outcomes**, and differs from . Never reduce to .
List in a system: fix the first letter or the first die, then work through the rest. Random listing is where a missing outcome hides.
Write the event as a set, not just a count: , so . Marks are given for the set.
On a tree, multiply along a path and add between paths — and along, or down.
**Check each fork's branches sum to before multiplying. A missing branch makes every answer wrong at once.
Check the path probabilities sum to ** at the end: .
**For "at least one", use .** Three coins: . Two dice, at least one six: .
"At least one" is not "exactly one" — against for two coins. Underline which the question says.
And note whether there is replacement — without it, the second fork's probabilities differ on each branch.
Did you know
Why a sum of 7 comes up more often than a sum of 2
Roll two dice and add. The totals run from to , which is eleven possibilities — but they are nowhere near equally likely.
Count the ways for each total out of the outcomes:
- and : one way each
- and : two ways each
- and : three ways each
- and : four ways each
- and : five ways each
- : six ways
Adding those up: . Every outcome accounted for.
So is the most likely total at , and is the least at — six times rarer. The counts climb to a peak in the middle and fall away symmetrically, which is why board games that need a spread of results use two dice rather than one. A single die gives every number equally often; two dice give a middle that turns up far more.
The reason wins is that it can be made in the most ways: , , and each of those reversed. A total of needs , and there is no reversing a pair of ones.
The effect strengthens with more dice. Three dice give totals from to , and and dominate while and need all three dice to agree. Keep adding dice and the shape of the distribution becomes the familiar bell.
All of which is hiding inside the plain instruction to write out the sample space. **The pairs look tediously uniform, and the eleven sums they produce are dramatically not** — and nothing but the full list reveals the difference.
Count the ways for each total out of the outcomes:
- and : one way each
- and : two ways each
- and : three ways each
- and : four ways each
- and : five ways each
- : six ways
Adding those up: . Every outcome accounted for.
So is the most likely total at , and is the least at — six times rarer. The counts climb to a peak in the middle and fall away symmetrically, which is why board games that need a spread of results use two dice rather than one. A single die gives every number equally often; two dice give a middle that turns up far more.
The reason wins is that it can be made in the most ways: , , and each of those reversed. A total of needs , and there is no reversing a pair of ones.
The effect strengthens with more dice. Three dice give totals from to , and and dominate while and need all three dice to agree. Keep adding dice and the shape of the distribution becomes the familiar bell.
All of which is hiding inside the plain instruction to write out the sample space. **The pairs look tediously uniform, and the eleven sums they produce are dramatically not** — and nothing but the full list reveals the difference.
Exam relevance
Why does JEE Main keep returning to the sample space?
Because almost every probability question at a higher level is a counting question, and a mis-listed sample space is the failure that no later step can repair.
This is the foundation for Class 11 Mathematics Permutations and Combinations and Probability, and Class 12 Probability, all examined in JEE Main. The definition is unchanged; what changes is that becomes far too large to list. The multiplication pattern seen here — two coins give , a coin and a die give — is the fundamental principle of counting, stated formally in Class 11 and used to find without enumeration.
The tree diagram becomes conditional probability. When the branches carry different probabilities — the without-replacement case flagged on this page — the numbers on the second fork are exactly , and multiply along a path is the multiplication theorem
of Class 12. Bayes' theorem is the same tree read backwards, and students who drew trees in Class 9 meet it as a new notation rather than a new idea.
The complement is the single most reused move. Any question containing at least one is a candidate for , and in Class 11 and 12 the alternative is a sum of many terms. The two-dice example here, where direct counting needed the inclusion-and-exclusion correction , is why: the complement route avoided double counting altogether, and inclusion-exclusion is itself a Class 11 topic for the cases where it cannot be avoided.
Where the dice sums lead. The shape found in the previous section — counts rising to a peak at — is a binomial distribution in miniature, treated in Class 12 Probability. Questions on the number of successes in repeated trials use it directly.
For NEET Biology, the sample-space idea is what makes a genetic cross calculable: the four boxes of a Punnett square are the sample space of two independent gametes, exactly as the four outcomes of two coins are, and the ratio is out of .
What the questions look like. For board work, expect **list the sample space and state , write an event as a set and find its probability, draw a tree diagram for two stages, and an at-least-one compound event. Full marks need the set written out, not only the fraction. For JEE Main, the counting is done with combinations, and the questions turn on choosing the right sample space and using the complement.
How board and competitive emphasis differ. A board paper rewards the written list** and the labelled tree. A competitive paper assumes both and tests whether you can count and for a situation far too large to write down.
The single trap that costs the most marks. Confusing at least one with exactly one. For two coins they are and , both perfectly plausible answers, and only the wording distinguishes them. Underline the phrase in the question before starting, and for at least reach straight for the complement.
This is the foundation for Class 11 Mathematics Permutations and Combinations and Probability, and Class 12 Probability, all examined in JEE Main. The definition is unchanged; what changes is that becomes far too large to list. The multiplication pattern seen here — two coins give , a coin and a die give — is the fundamental principle of counting, stated formally in Class 11 and used to find without enumeration.
The tree diagram becomes conditional probability. When the branches carry different probabilities — the without-replacement case flagged on this page — the numbers on the second fork are exactly , and multiply along a path is the multiplication theorem
of Class 12. Bayes' theorem is the same tree read backwards, and students who drew trees in Class 9 meet it as a new notation rather than a new idea.
The complement is the single most reused move. Any question containing at least one is a candidate for , and in Class 11 and 12 the alternative is a sum of many terms. The two-dice example here, where direct counting needed the inclusion-and-exclusion correction , is why: the complement route avoided double counting altogether, and inclusion-exclusion is itself a Class 11 topic for the cases where it cannot be avoided.
Where the dice sums lead. The shape found in the previous section — counts rising to a peak at — is a binomial distribution in miniature, treated in Class 12 Probability. Questions on the number of successes in repeated trials use it directly.
For NEET Biology, the sample-space idea is what makes a genetic cross calculable: the four boxes of a Punnett square are the sample space of two independent gametes, exactly as the four outcomes of two coins are, and the ratio is out of .
What the questions look like. For board work, expect **list the sample space and state , write an event as a set and find its probability, draw a tree diagram for two stages, and an at-least-one compound event. Full marks need the set written out, not only the fraction. For JEE Main, the counting is done with combinations, and the questions turn on choosing the right sample space and using the complement.
How board and competitive emphasis differ. A board paper rewards the written list** and the labelled tree. A competitive paper assumes both and tests whether you can count and for a situation far too large to write down.
The single trap that costs the most marks. Confusing at least one with exactly one. For two coins they are and , both perfectly plausible answers, and only the wording distinguishes them. Underline the phrase in the question before starting, and for at least reach straight for the complement.
Key takeaways
Sample space, tree diagrams and compound events: quick revision
- The sample space is the complete list of distinct outcomes, and is how many there are.
- One die: . One coin: . **Two coins: ** — . Three coins: . Two dice: . A card: . A coin and a die: .
- Stages multiply: outcomes then outcomes gives .
- ** and are different**, and so are and . Collapsing them gives outcomes that are not equally likely.
- List in a system — fix the first result, then work through the rest.
- **, and the event should be written as a set**.
- Two coins: at least one head , exactly one head , no head , two heads .
- Three coins: exactly two heads from ; at least one head ; all the same .
- Two dice: sum is ; sum is ; a doublet is ; sum over is ; sum is .
- A card: face card , not a heart , red king .
- ** and are the formula's extremes — the empty set and the whole sample space.
- On a tree, each complete path is one outcome; multiply along a path, add between paths**.
- A coin then a die gives paths, so .
- Bag of red and blue with replacement: , , , summing to .
- **Check every fork sums to ** before multiplying, and check the paths sum to at the end.
- Without replacement the second fork differs on each branch — and after a red, and after a blue.
- **For "at least one", use **: two coins , three coins , four coins .
- Two dice, at least one six: ; direct counting needs .
- At least one blue from the bag: .
- "At least one" is not "exactly one" — against for two coins.
- Dice sums are not equally likely: the counts run for totals to , adding to .
Toss two coins thirty times, tally , , and separately, then see whether one head really turned up near half the time rather than a third.
- One die: . One coin: . **Two coins: ** — . Three coins: . Two dice: . A card: . A coin and a die: .
- Stages multiply: outcomes then outcomes gives .
- ** and are different**, and so are and . Collapsing them gives outcomes that are not equally likely.
- List in a system — fix the first result, then work through the rest.
- **, and the event should be written as a set**.
- Two coins: at least one head , exactly one head , no head , two heads .
- Three coins: exactly two heads from ; at least one head ; all the same .
- Two dice: sum is ; sum is ; a doublet is ; sum over is ; sum is .
- A card: face card , not a heart , red king .
- ** and are the formula's extremes — the empty set and the whole sample space.
- On a tree, each complete path is one outcome; multiply along a path, add between paths**.
- A coin then a die gives paths, so .
- Bag of red and blue with replacement: , , , summing to .
- **Check every fork sums to ** before multiplying, and check the paths sum to at the end.
- Without replacement the second fork differs on each branch — and after a red, and after a blue.
- **For "at least one", use **: two coins , three coins , four coins .
- Two dice, at least one six: ; direct counting needs .
- At least one blue from the bag: .
- "At least one" is not "exactly one" — against for two coins.
- Dice sums are not equally likely: the counts run for totals to , adding to .
Toss two coins thirty times, tally , , and separately, then see whether one head really turned up near half the time rather than a third.