Counting the Ways Something Can Happen Is the Whole of Measuring Chance
List the sample space of an experiment, compute the probability of an event by counting favourable outcomes, use the complement rule to avoid long lists, recognise sure and impossible events, and handle marbles, defective items, spinners and two dice.
How can you put a number on something that has not happened yet?
Throw a die. One of six faces will land uppermost, and there is no reason for any one of them to be preferred. So if you want the chance of getting a four, you count: one face out of six gives a four, so the probability is .
That is the whole method, and it rests on one assumption that must be stated: the outcomes are equally likely. A fair die, a fair coin, a well-shuffled deck, a ball drawn at random — each phrase in a question is there to tell you that the counting is legitimate.
So probability at this level is a counting problem, not a guessing problem. Two lists settle every question:
- The sample space — every possible outcome of the experiment, listed once each
- The favourable outcomes — those belonging to the event you are asked about
And the probability is the size of the second list divided by the size of the first.
The reason this is worth care rather than intuition is that people's intuitions about chance are unreliable. Ask how likely it is to get two heads when tossing two coins and many will say , reasoning that the outcomes are two heads, two tails, or one of each. **But "one of each" happens in two ways — head then tail, and tail then head — so the sample space has four members and the answer is . Writing the list out is what protects you.
Two ideas then make the counting manageable.
- The complement rule, which lets you count what you do not want when that list is shorter
- The bounds**: every probability lies between and , with for an impossible event and for a sure one
That second one is the fastest check in the whole chapter. A probability greater than or less than is not an unusual answer — it is a wrong one.
This page covers the CBSE Class 10 Maths chapter on probability: the classical definition, sample spaces, coins, dice and cards, the complement rule, and applied problems.
That is the whole method, and it rests on one assumption that must be stated: the outcomes are equally likely. A fair die, a fair coin, a well-shuffled deck, a ball drawn at random — each phrase in a question is there to tell you that the counting is legitimate.
So probability at this level is a counting problem, not a guessing problem. Two lists settle every question:
- The sample space — every possible outcome of the experiment, listed once each
- The favourable outcomes — those belonging to the event you are asked about
And the probability is the size of the second list divided by the size of the first.
The reason this is worth care rather than intuition is that people's intuitions about chance are unreliable. Ask how likely it is to get two heads when tossing two coins and many will say , reasoning that the outcomes are two heads, two tails, or one of each. **But "one of each" happens in two ways — head then tail, and tail then head — so the sample space has four members and the answer is . Writing the list out is what protects you.
Two ideas then make the counting manageable.
- The complement rule, which lets you count what you do not want when that list is shorter
- The bounds**: every probability lies between and , with for an impossible event and for a sure one
That second one is the fastest check in the whole chapter. A probability greater than or less than is not an unusual answer — it is a wrong one.
This page covers the CBSE Class 10 Maths chapter on probability: the classical definition, sample spaces, coins, dice and cards, the complement rule, and applied problems.
Formula
What is the classical definition of probability?
The number of outcomes favourable to the event, divided by the total number of equally likely outcomes.
And the two consequences that follow immediately:
Why the probability cannot leave that range. The favourable outcomes are a subset of all the outcomes, so their count is at least and at most the total. **Dividing gives a number between and , and nothing else is possible.
- — the event is impossible
- — the event is sure or certain
- Any value between — the event may or may not happen
Why the complement rule holds.** Every outcome either belongs to or does not, so the two counts add to the total:
**The event "not " is called the complement** of , and it is written or .
Worked example 1 — the sample space of a single die. List the sample space and find the probability of an even number, a prime number, and a number greater than .
The sample space is , with six equally likely outcomes.
- Even numbers: , , — three of them, so
- Prime numbers: , , — three of them, so
- **Greater than **: , — two of them, so
**Note that is not prime, which is why the prime list has three members and not four. That single fact decides the answer, and it comes from the real-numbers chapter rather than from this one.
Worked example 2 — a sure event and an impossible one.** For the same die, find the probability of getting a number less than , and of getting an .
- **Less than **: all six outcomes qualify, so . A sure event
- **An **: no outcome qualifies, so . An impossible event
Worked example 3 — the complement rule in use. If the probability that a student passes an examination is , find the probability that the student fails.
Check: . Correct, and that addition is worth writing because it is the whole content of the rule.
One thing to be careful about in the definition. The outcomes must be equally likely for the counting to be valid. **If a die were weighted so that a six came up more often, the sample space would still have six members but the probability of a six would not be ** — and no amount of counting would give it. That is why every question says "fair", "unbiased", "well-shuffled" or "at random", and those words are part of the data rather than decoration.
And the two consequences that follow immediately:
Why the probability cannot leave that range. The favourable outcomes are a subset of all the outcomes, so their count is at least and at most the total. **Dividing gives a number between and , and nothing else is possible.
- — the event is impossible
- — the event is sure or certain
- Any value between — the event may or may not happen
Why the complement rule holds.** Every outcome either belongs to or does not, so the two counts add to the total:
**The event "not " is called the complement** of , and it is written or .
Worked example 1 — the sample space of a single die. List the sample space and find the probability of an even number, a prime number, and a number greater than .
The sample space is , with six equally likely outcomes.
- Even numbers: , , — three of them, so
- Prime numbers: , , — three of them, so
- **Greater than **: , — two of them, so
**Note that is not prime, which is why the prime list has three members and not four. That single fact decides the answer, and it comes from the real-numbers chapter rather than from this one.
Worked example 2 — a sure event and an impossible one.** For the same die, find the probability of getting a number less than , and of getting an .
- **Less than **: all six outcomes qualify, so . A sure event
- **An **: no outcome qualifies, so . An impossible event
Worked example 3 — the complement rule in use. If the probability that a student passes an examination is , find the probability that the student fails.
Check: . Correct, and that addition is worth writing because it is the whole content of the rule.
One thing to be careful about in the definition. The outcomes must be equally likely for the counting to be valid. **If a die were weighted so that a six came up more often, the sample space would still have six members but the probability of a six would not be ** — and no amount of counting would give it. That is why every question says "fair", "unbiased", "well-shuffled" or "at random", and those words are part of the data rather than decoration.
How do you handle coins, dice and a deck of cards?
Write the sample space out for coins and dice; memorise the composition of a deck for cards. Both are counting exercises once the list is in front of you.
Two coins tossed together. The sample space is
four outcomes, because each coin independently shows a head or a tail. Then:
- Exactly one head: and , so
- At least one head: , , , so
- No head: only , so
- Two heads: only , so
**Notice that and are different outcomes.** They describe the same number of heads but not the same result of the experiment, and counting them once instead of twice is the error that produces the popular wrong answer of for two heads.
Check with the complement: . Agrees with the count.
Three coins tossed together. Now there are outcomes:
- Exactly two heads: , , , so
- At most one head: , , , , so
- At least two heads: the complement of the previous one, so
"At most" and "at least" are where marks are lost. At most one head means zero or one; at least two heads means two or three. Write out which cases the phrase covers before counting, because the two phrases partition the eight outcomes and each covers four.
**A deck of playing cards. The composition has to be known, and it is entirely factual:
- Four suits** — spades, clubs, hearts and diamonds — of cards each
- Spades and clubs are black; hearts and diamonds are red, so of each colour
- **Each suit has an ace, the numbers to , a jack, a queen and a king
- The jacks, queens and kings are the face cards**, so there are face cards in all, three per suit
- Aces are not face cards, which gives aces and face cards as separate counts
Worked example — one card drawn from a well-shuffled deck. Find the probability that it is:
- A king: kings, so
- A red card: red, so
- A spade: spades, so
- A face card: of them, so
- A red king: the king of hearts and the king of diamonds, so
- Neither a king nor a queen: there are such cards, leaving , so
The last one used the complement, and it is much quicker than listing the cards. Check it: , and . Correct.
One count worth double-checking every time. A question about a "black face card" wants the jacks, queens and kings of spades and clubs — six cards, giving . Halving the twelve face cards is right here, but it would be wrong for a question about aces and face cards together, where the two sets must be added and not scaled. Count the cards, do not scale a previous answer.
Two coins tossed together. The sample space is
four outcomes, because each coin independently shows a head or a tail. Then:
- Exactly one head: and , so
- At least one head: , , , so
- No head: only , so
- Two heads: only , so
**Notice that and are different outcomes.** They describe the same number of heads but not the same result of the experiment, and counting them once instead of twice is the error that produces the popular wrong answer of for two heads.
Check with the complement: . Agrees with the count.
Three coins tossed together. Now there are outcomes:
- Exactly two heads: , , , so
- At most one head: , , , , so
- At least two heads: the complement of the previous one, so
"At most" and "at least" are where marks are lost. At most one head means zero or one; at least two heads means two or three. Write out which cases the phrase covers before counting, because the two phrases partition the eight outcomes and each covers four.
**A deck of playing cards. The composition has to be known, and it is entirely factual:
- Four suits** — spades, clubs, hearts and diamonds — of cards each
- Spades and clubs are black; hearts and diamonds are red, so of each colour
- **Each suit has an ace, the numbers to , a jack, a queen and a king
- The jacks, queens and kings are the face cards**, so there are face cards in all, three per suit
- Aces are not face cards, which gives aces and face cards as separate counts
Worked example — one card drawn from a well-shuffled deck. Find the probability that it is:
- A king: kings, so
- A red card: red, so
- A spade: spades, so
- A face card: of them, so
- A red king: the king of hearts and the king of diamonds, so
- Neither a king nor a queen: there are such cards, leaving , so
The last one used the complement, and it is much quicker than listing the cards. Check it: , and . Correct.
One count worth double-checking every time. A question about a "black face card" wants the jacks, queens and kings of spades and clubs — six cards, giving . Halving the twelve face cards is right here, but it would be wrong for a question about aces and face cards together, where the two sets must be added and not scaled. Count the cards, do not scale a previous answer.
When should you use the complement rule instead of counting directly?
Whenever the outcomes you do not want are easier to count than the ones you do. That is almost always the case with the phrases "at least one" and "not".
Worked example 1 — two dice, product even. Two dice are thrown together. Find the probability that the product of the numbers shown is even.
Counting directly is unpleasant, because there are many ways to get an even product. Counting the complement is easy: the product is odd only when both numbers are odd.
There are odd numbers on each die, so the odd-product outcomes number
out of total. So
Nine outcomes counted instead of twenty-seven. That is the whole case for the rule.
Worked example 2 — the sample space of two dice. It has outcomes, written as ordered pairs from to . The order matters: and are different outcomes, just as and were for coins.
Now find the probability that the sum is:
- ****: the pairs , , , , — five of them, so
- ****: , , , , , — six of them, so
- **At most **: , , , , , — six of them, so
- ****: impossible, since the largest sum is , so
- **At most **: every outcome qualifies, so
**A sum of is the most likely of all**, with six ways, and the counts fall away on either side to one way each for and for . **That asymmetry between and — six ways against five — is the kind of thing intuition gets wrong and a list gets right.
Worked example 3 — defective items.** A lot of bulbs contains defective ones. One bulb is drawn at random.
Now the second part, which changes the sample space. Suppose the bulb drawn is not defective and is not replaced. A second bulb is drawn. Find the probability that it is not defective.
**After the first draw there are bulbs left, of which are good** — one good bulb has been removed. So
The total changed and so did the favourable count. That is the whole idea: without replacement, the sample space shrinks, and both numbers must be updated. **Answering again is the error this question is built to catch.
Worked example 4 — marbles.** A bag contains red balls and black balls. One ball is drawn at random. Find the probability that it is red, and that it is not red.
There are balls in all, so
Check: . Correct — and note that "not red" here means black, because those are the only two colours. If the bag also held white balls, "not red" would cover two colours, and that is worth checking against the question's wording.
Worked example 1 — two dice, product even. Two dice are thrown together. Find the probability that the product of the numbers shown is even.
Counting directly is unpleasant, because there are many ways to get an even product. Counting the complement is easy: the product is odd only when both numbers are odd.
There are odd numbers on each die, so the odd-product outcomes number
out of total. So
Nine outcomes counted instead of twenty-seven. That is the whole case for the rule.
Worked example 2 — the sample space of two dice. It has outcomes, written as ordered pairs from to . The order matters: and are different outcomes, just as and were for coins.
Now find the probability that the sum is:
- ****: the pairs , , , , — five of them, so
- ****: , , , , , — six of them, so
- **At most **: , , , , , — six of them, so
- ****: impossible, since the largest sum is , so
- **At most **: every outcome qualifies, so
**A sum of is the most likely of all**, with six ways, and the counts fall away on either side to one way each for and for . **That asymmetry between and — six ways against five — is the kind of thing intuition gets wrong and a list gets right.
Worked example 3 — defective items.** A lot of bulbs contains defective ones. One bulb is drawn at random.
Now the second part, which changes the sample space. Suppose the bulb drawn is not defective and is not replaced. A second bulb is drawn. Find the probability that it is not defective.
**After the first draw there are bulbs left, of which are good** — one good bulb has been removed. So
The total changed and so did the favourable count. That is the whole idea: without replacement, the sample space shrinks, and both numbers must be updated. **Answering again is the error this question is built to catch.
Worked example 4 — marbles.** A bag contains red balls and black balls. One ball is drawn at random. Find the probability that it is red, and that it is not red.
There are balls in all, so
Check: . Correct — and note that "not red" here means black, because those are the only two colours. If the bag also held white balls, "not red" would cover two colours, and that is worth checking against the question's wording.
How do you handle numbered discs, spinners and bags in real situations?
Identify the total number of equally likely outcomes, then count the favourable ones carefully — usually by listing or by a divisibility argument.
Worked example 1 — numbered discs. A box contains discs numbered from to . One disc is drawn at random. Find the probability that it bears:
A two-digit number. The one-digit numbers are to , nine of them, so the two-digit numbers number :
A perfect square. The perfect squares up to are , , , , , , , , — nine of them, since is inside the range and is not:
**A number divisible by .** These are , , , up to , and there are of them:
The perfect-square count is where care is needed. **Listing them is safer than computing and rounding, because the rounding rule is easy to get backwards — and the list takes ten seconds.
Worked example 2 — a spinning wheel.** A wheel is divided into eight equal sectors numbered to and is spun once. Find the probability of getting:
- An odd number: , , , — four of them, so
- **A number greater than **: to , six of them, so
- **A number less than **: all eight, so , a sure event
- **A multiple of **: and , so
The sectors must be equal for this counting to work. A wheel with unequal sectors would need the areas rather than the count, and a question always says "equal sectors" when it means them.
Worked example 3 — a bag with three colours. A bag contains red, white and green marbles. One marble is drawn at random. Find the probability that it is:
The total is .
- Red:
- White:
- Not green: the green ones number , so
- Red or white: , so
The last two agree, and they must, because "not green" and "red or white" describe the same set of marbles. That coincidence is a genuine check — if the two methods had disagreed, a colour had been miscounted.
Worked example 4 — removing items first. A bag contains balls, of which are black. If one ball is drawn at random, the probability of it being black is . If more black balls are put in, the probability of drawing a black ball becomes double what it was. Find .
**After adding, there are balls of which are black:**
Check: originally ; after adding, , which is double. Correct.
Notice that adding balls changed both the numerator and the denominator. That is the same principle as the without-replacement bulb question: whenever the contents change, recount the total as well as the favourable cases. Updating only one of the two is the single most common error in applied probability questions.
Worked example 5 — a game with two dice. Two dice are thrown. Find the probability that the two numbers are the same, and that they differ by .
The same: , , , , , — six outcomes, so .
**Differing by **: , , , , , , , , , — ten outcomes, so .
The ordered pairs had to be counted both ways round. There are five unordered pairs that differ by , and each gives two outcomes. Forgetting the reversal halves the answer, and it is the same mistake as counting and as one.
Worked example 1 — numbered discs. A box contains discs numbered from to . One disc is drawn at random. Find the probability that it bears:
A two-digit number. The one-digit numbers are to , nine of them, so the two-digit numbers number :
A perfect square. The perfect squares up to are , , , , , , , , — nine of them, since is inside the range and is not:
**A number divisible by .** These are , , , up to , and there are of them:
The perfect-square count is where care is needed. **Listing them is safer than computing and rounding, because the rounding rule is easy to get backwards — and the list takes ten seconds.
Worked example 2 — a spinning wheel.** A wheel is divided into eight equal sectors numbered to and is spun once. Find the probability of getting:
- An odd number: , , , — four of them, so
- **A number greater than **: to , six of them, so
- **A number less than **: all eight, so , a sure event
- **A multiple of **: and , so
The sectors must be equal for this counting to work. A wheel with unequal sectors would need the areas rather than the count, and a question always says "equal sectors" when it means them.
Worked example 3 — a bag with three colours. A bag contains red, white and green marbles. One marble is drawn at random. Find the probability that it is:
The total is .
- Red:
- White:
- Not green: the green ones number , so
- Red or white: , so
The last two agree, and they must, because "not green" and "red or white" describe the same set of marbles. That coincidence is a genuine check — if the two methods had disagreed, a colour had been miscounted.
Worked example 4 — removing items first. A bag contains balls, of which are black. If one ball is drawn at random, the probability of it being black is . If more black balls are put in, the probability of drawing a black ball becomes double what it was. Find .
**After adding, there are balls of which are black:**
Check: originally ; after adding, , which is double. Correct.
Notice that adding balls changed both the numerator and the denominator. That is the same principle as the without-replacement bulb question: whenever the contents change, recount the total as well as the favourable cases. Updating only one of the two is the single most common error in applied probability questions.
Worked example 5 — a game with two dice. Two dice are thrown. Find the probability that the two numbers are the same, and that they differ by .
The same: , , , , , — six outcomes, so .
**Differing by **: , , , , , , , , , — ten outcomes, so .
The ordered pairs had to be counted both ways round. There are five unordered pairs that differ by , and each gives two outcomes. Forgetting the reversal halves the answer, and it is the same mistake as counting and as one.
Exam tip
Which habits protect the marks in a probability answer?
Write the total number of outcomes first, then list or count the favourable ones, then divide and simplify. Showing the two counts separately is what earns the method marks.
- State the total number of outcomes explicitly: for a die, for two coins, for three coins, for two dice, for a deck
- List the sample space for coins and small dice problems rather than counting in your head
- Treat ordered outcomes as distinct: and , and and
- Unpack "at least" and "at most" into the cases they cover before counting
- Use the complement whenever the unwanted outcomes are fewer — "at least one", "not", "neither"
- Know the deck: cards, per suit, of each colour, face cards, aces, and aces are not face cards
- Recount the total when the contents change — without replacement, or after adding items
- Simplify the fraction, and give the answer as a fraction unless a decimal is asked for
- **Check that **, and that complementary probabilities add to
- Answer in a sentence naming the event
The misconception to name. "Equally likely" is an assumption, not an automatic fact. It is supplied by the words "fair", "unbiased", "well-shuffled", "at random" and "equal sectors" — and without one of them the counting method does not apply. A question describing a loaded die or unequal sectors is not asking for a count.
A second trap. Changing only one of the two numbers when the situation changes. **After a good bulb is removed without replacement, the probability is , not and not ** — both the good count and the total fell by one. The same applies when balls are added: , with both parts updated.
- State the total number of outcomes explicitly: for a die, for two coins, for three coins, for two dice, for a deck
- List the sample space for coins and small dice problems rather than counting in your head
- Treat ordered outcomes as distinct: and , and and
- Unpack "at least" and "at most" into the cases they cover before counting
- Use the complement whenever the unwanted outcomes are fewer — "at least one", "not", "neither"
- Know the deck: cards, per suit, of each colour, face cards, aces, and aces are not face cards
- Recount the total when the contents change — without replacement, or after adding items
- Simplify the fraction, and give the answer as a fraction unless a decimal is asked for
- **Check that **, and that complementary probabilities add to
- Answer in a sentence naming the event
The misconception to name. "Equally likely" is an assumption, not an automatic fact. It is supplied by the words "fair", "unbiased", "well-shuffled", "at random" and "equal sectors" — and without one of them the counting method does not apply. A question describing a loaded die or unequal sectors is not asking for a count.
A second trap. Changing only one of the two numbers when the situation changes. **After a good bulb is removed without replacement, the probability is , not and not ** — both the good count and the total fell by one. The same applies when balls are added: , with both parts updated.
Did you know
Why do so many people get the two-coin question wrong?
Ask a group how likely it is to get one head and one tail when two coins are tossed, and a good number will say . Their reasoning is clean: the results are two heads, two tails, or one of each — three possibilities, so each has probability one third.
The three descriptions are real, but they are not equally likely, and that is the whole difficulty. Writing the outcomes out shows why:
"One of each" occupies two of the four slots, so its probability is , while two heads and two tails get each. You are twice as likely to get a mixed result as a matching one.
And you can test it. Toss two coins forty times and record the results; mixed results should come up roughly twice as often as either matching result. The experiment will not give exactly twenty, ten and ten — but it will not give thirteen, thirteen and fourteen either, and that is the point.
The general lesson is worth stating plainly: a description is not an outcome. "One head" describes a set of outcomes, and a set's probability depends on how many outcomes it contains. Grouping outcomes by description and then assuming the groups are equally likely is the single most reliable way to get a probability wrong.
The same trap appears with two dice. "The sum is " and "the sum is " look like two descriptions of the same kind, but the first covers six outcomes and the second covers one. **A sum of is six times as likely as a sum of **, which is why is the sum every dice game is built around.
And the pattern of the sums is worth seeing once. Counting the ways to make each total from to gives — **rising to a peak at and falling away symmetrically.** Those eleven numbers add to , as they must, and the shape explains why totals near the middle are common and totals at the ends are rare.
There is a second kind of probability that this chapter mentions but does not compute. Experimental or empirical probability is found by actually performing the experiment many times and dividing the number of successes by the number of trials. Theoretical probability, which is what this chapter computes, is found by counting. For a fair coin the theoretical probability of a head is exactly ; the experimental probability from ten tosses might be . As the number of trials grows, the experimental value settles toward the theoretical one — which is why casinos and insurers can rely on probabilities even though any single event is unpredictable.
One last thing the bounds tell you. Because , **no probability can ever be expressed as a number like or , and no probability of an event can exceed the probability of a larger event containing it. Those are not conventions but consequences of dividing a part by a whole** — and they make the sanity check at the end of every answer genuinely free.
The three descriptions are real, but they are not equally likely, and that is the whole difficulty. Writing the outcomes out shows why:
"One of each" occupies two of the four slots, so its probability is , while two heads and two tails get each. You are twice as likely to get a mixed result as a matching one.
And you can test it. Toss two coins forty times and record the results; mixed results should come up roughly twice as often as either matching result. The experiment will not give exactly twenty, ten and ten — but it will not give thirteen, thirteen and fourteen either, and that is the point.
The general lesson is worth stating plainly: a description is not an outcome. "One head" describes a set of outcomes, and a set's probability depends on how many outcomes it contains. Grouping outcomes by description and then assuming the groups are equally likely is the single most reliable way to get a probability wrong.
The same trap appears with two dice. "The sum is " and "the sum is " look like two descriptions of the same kind, but the first covers six outcomes and the second covers one. **A sum of is six times as likely as a sum of **, which is why is the sum every dice game is built around.
And the pattern of the sums is worth seeing once. Counting the ways to make each total from to gives — **rising to a peak at and falling away symmetrically.** Those eleven numbers add to , as they must, and the shape explains why totals near the middle are common and totals at the ends are rare.
There is a second kind of probability that this chapter mentions but does not compute. Experimental or empirical probability is found by actually performing the experiment many times and dividing the number of successes by the number of trials. Theoretical probability, which is what this chapter computes, is found by counting. For a fair coin the theoretical probability of a head is exactly ; the experimental probability from ten tosses might be . As the number of trials grows, the experimental value settles toward the theoretical one — which is why casinos and insurers can rely on probabilities even though any single event is unpredictable.
One last thing the bounds tell you. Because , **no probability can ever be expressed as a number like or , and no probability of an event can exceed the probability of a larger event containing it. Those are not conventions but consequences of dividing a part by a whole** — and they make the sanity check at the end of every answer genuinely free.
Exam relevance
How does probability at this level prepare you for JEE and NEET?
This is foundation work for Class 11 Probability and Permutations and Combinations, and for Class 12 Probability, all of which are examined in JEE Main and JEE Advanced, with probability also appearing in NEET Biology through genetics.
Where the classical definition leads. Class 11 keeps unchanged and adds the addition rule for the union of two events, . The card questions you answer here by counting — "a king or a queen", "a red face card" — are exactly the union and intersection problems that rule formalises, and recognising that saves learning it as a new idea.
Where the counting leads. Class 11 Permutations and Combinations supplies the tools for sample spaces too large to list: the number of ways to draw three cards from , or to arrange objects. The habit of asking "how many outcomes in total, and how many favourable?" is unchanged — only the method of counting them grows. JEE Main sets probability questions whose whole difficulty is the counting.
Where the complement rule leads. It remains the standard route for "at least one" problems at every level, and in Class 12 it combines with independence: the probability of at least one success in independent trials is . The rule you use here on nine odd-product outcomes is the same rule, and JEE Advanced relies on it heavily.
Where the without-replacement idea leads. It becomes conditional probability in Class 12, written , and the bulb question — the second draw after a good bulb has been removed — is a conditional probability computed without the notation. Recognising that now makes the Class 12 chapter feel like naming something you already do, and it leads on to Bayes' theorem, a reliable JEE Main topic.
Where probability leads in NEET. Mendelian genetics is applied probability: the chance of a particular genotype from a cross is counted from a sample space of gametes, exactly as coin outcomes are counted here. A Punnett square is a sample space drawn as a grid, and the two-coin trap has a genetics twin — the heterozygous combination arises in two ways, which is why a monohybrid cross gives a genotype ratio rather than .
Question types to expect. At this level: sample spaces, coins, dice, cards, complements, and applied bag-and-item problems. In competitive papers: addition and multiplication rules, conditional probability, Bayes' theorem, binomial probabilities, and counting-heavy sample spaces.
The single trap that costs marks. Treating descriptions as equally likely outcomes. "One head" is not one outcome but two, and the identical error in genetics gives a ratio for a monohybrid cross instead of . List the outcomes, not the descriptions.
A second trap. Failing to update both counts when the situation changes. Without replacement the total falls as well as the favourable count, and in Class 12 the same oversight breaks a conditional probability at its first step.
Board versus competitive emphasis. The CBSE paper marks the total, the favourable count, the fraction and the simplification; a competitive paper marks a single value reached through a rule or a large count. The transferable habit is writing down the sample space's size before anything else — because at every level the denominator is where probability questions are won or lost.
Where the classical definition leads. Class 11 keeps unchanged and adds the addition rule for the union of two events, . The card questions you answer here by counting — "a king or a queen", "a red face card" — are exactly the union and intersection problems that rule formalises, and recognising that saves learning it as a new idea.
Where the counting leads. Class 11 Permutations and Combinations supplies the tools for sample spaces too large to list: the number of ways to draw three cards from , or to arrange objects. The habit of asking "how many outcomes in total, and how many favourable?" is unchanged — only the method of counting them grows. JEE Main sets probability questions whose whole difficulty is the counting.
Where the complement rule leads. It remains the standard route for "at least one" problems at every level, and in Class 12 it combines with independence: the probability of at least one success in independent trials is . The rule you use here on nine odd-product outcomes is the same rule, and JEE Advanced relies on it heavily.
Where the without-replacement idea leads. It becomes conditional probability in Class 12, written , and the bulb question — the second draw after a good bulb has been removed — is a conditional probability computed without the notation. Recognising that now makes the Class 12 chapter feel like naming something you already do, and it leads on to Bayes' theorem, a reliable JEE Main topic.
Where probability leads in NEET. Mendelian genetics is applied probability: the chance of a particular genotype from a cross is counted from a sample space of gametes, exactly as coin outcomes are counted here. A Punnett square is a sample space drawn as a grid, and the two-coin trap has a genetics twin — the heterozygous combination arises in two ways, which is why a monohybrid cross gives a genotype ratio rather than .
Question types to expect. At this level: sample spaces, coins, dice, cards, complements, and applied bag-and-item problems. In competitive papers: addition and multiplication rules, conditional probability, Bayes' theorem, binomial probabilities, and counting-heavy sample spaces.
The single trap that costs marks. Treating descriptions as equally likely outcomes. "One head" is not one outcome but two, and the identical error in genetics gives a ratio for a monohybrid cross instead of . List the outcomes, not the descriptions.
A second trap. Failing to update both counts when the situation changes. Without replacement the total falls as well as the favourable count, and in Class 12 the same oversight breaks a conditional probability at its first step.
Board versus competitive emphasis. The CBSE paper marks the total, the favourable count, the fraction and the simplification; a competitive paper marks a single value reached through a rule or a large count. The transferable habit is writing down the sample space's size before anything else — because at every level the denominator is where probability questions are won or lost.
Key takeaways
What must you be able to do from this chapter?
One definition, one complement rule and two bounds.
- **, valid only when the outcomes are equally likely
- **, with for an impossible event and for a sure event
- ****, and the two add to
- A die has sample space : even gives , prime gives since is not prime, and greater than gives
- Two coins give , , , : exactly one head is , at least one head is , two heads is
- ** and are different outcomes** — this is why two heads is and not
- Three coins give outcomes: exactly two heads is , at most one head is
- A deck has cards, per suit, of each colour, face cards and aces, and aces are not face cards
- From a deck: a king is , a red card , a spade , a face card , a red king , neither a king nor a queen , a black face card
- Two dice give ordered outcomes: sum is , sum is , at most is , sum is , same numbers is , differing by is
- **An even product is **, found by counting the odd-odd outcomes and complementing
- **The ways of making each sum from to ** are , adding to
- ** defectives in bulbs** give defective and not; after a good bulb is removed, the next is good with probability
- Recount both numbers when the contents change, whether items are removed or added
- ** discs numbered to **: two-digit is , a perfect square is with nine squares up to , divisible by is
- **A bag of red, white and green**: not green is , which equals red or white
- ** balls with black, doubling after more black are added**, gives
- A description is not an outcome — group sizes differ, so groups are not equally likely
- Experimental probability settles toward the theoretical value as the number of trials grows
The most convincing self-test is forty tosses. Throw two coins forty times, tally how often you get two heads, two tails and one of each, and see whether the mixed result really does turn up about twice as often as either of the others — then ask yourself why the counts are not exactly , and .
- **, valid only when the outcomes are equally likely
- **, with for an impossible event and for a sure event
- ****, and the two add to
- A die has sample space : even gives , prime gives since is not prime, and greater than gives
- Two coins give , , , : exactly one head is , at least one head is , two heads is
- ** and are different outcomes** — this is why two heads is and not
- Three coins give outcomes: exactly two heads is , at most one head is
- A deck has cards, per suit, of each colour, face cards and aces, and aces are not face cards
- From a deck: a king is , a red card , a spade , a face card , a red king , neither a king nor a queen , a black face card
- Two dice give ordered outcomes: sum is , sum is , at most is , sum is , same numbers is , differing by is
- **An even product is **, found by counting the odd-odd outcomes and complementing
- **The ways of making each sum from to ** are , adding to
- ** defectives in bulbs** give defective and not; after a good bulb is removed, the next is good with probability
- Recount both numbers when the contents change, whether items are removed or added
- ** discs numbered to **: two-digit is , a perfect square is with nine squares up to , divisible by is
- **A bag of red, white and green**: not green is , which equals red or white
- ** balls with black, doubling after more black are added**, gives
- A description is not an outcome — group sizes differ, so groups are not equally likely
- Experimental probability settles toward the theoretical value as the number of trials grows
The most convincing self-test is forty tosses. Throw two coins forty times, tally how often you get two heads, two tails and one of each, and see whether the mixed result really does turn up about twice as often as either of the others — then ask yourself why the counts are not exactly , and .