Probability | ICSE Class 10 Maths Notes

This note covers probability, or the numerical measure of chance, through random experiments, possible outcomes, sample spaces, events, equally likely outcomes, theoretical probability, complementary events, and calculations involving coins, dice, balls, cards and random selections.

What is a random experiment?

An experiment is an action or process that produces a result. A random experiment has known possible results, but its particular result cannot be predicted with certainty beforehand. An outcome is one possible result of that experiment.

Tossing a coin and throwing a die are examples. Before a coin lands, we know the possible results to consider, but we do not know which will occur. Knowing the possible outcomes is different from predicting the outcome of a particular toss.

What assumptions do we make about a coin?

A fair, or unbiased, coin has no reason to fall more often on one side than the other. A random toss allows it to fall freely, without interference. We consider head up and tail up, dismissing the possibility of the coin landing on its edge.

Use H to mean head and T to mean tail. These letters label outcomes. We can reasonably assume that head and tail are equally likely, meaning that each has the same chance of occurring under these assumptions.

What does a die experiment record?

A die is a cube whose faces, in the usual numbered example, bear 1, 2, 3, 4, 5 and 6. Throwing it once records the number on its upper face. For a fair die, each of these six outcomes has the same chance.

A trial means one performance of an experiment. Identifying the trial prevents confusion about what is being counted. One throw of a die gives a single face number; one draw from a bag gives a single selected object.

How do sample spaces and events describe outcomes?

Definition: The sample space is the set of all possible outcomes of an experiment. A set is a collection of distinct objects or results. We use the letter S for the sample space.

Braces, the symbols { and }, enclose the members of a set. The sign = means “equals”. For one coin toss, write S = {H, T}. For one throw of the numbered die, write S = {1, 2, 3, 4, 5, 6}.

An event is a subset of the sample space. A subset is a selection of members from a set, possibly including all or none of them. We use E as a label for an event and define its meaning in each problem.

When does an event occur?

An event occurs when the actual outcome belongs to it. For a die, let E mean getting a number greater than 4. Then E = {5, 6}. Either 5 or 6 makes this event occur; the event need not specify just one outcome.

An elementary event contains exactly one outcome. Getting a head in one coin toss is elementary. Getting a number greater than 4 on the numbered die is not elementary because both 5 and 6 satisfy the condition.

How should two distinguishable results be recorded?

For two different coins tossed together, an ordered pair records the first coin's result followed by the second coin's result. The sample space is {(H, H), (H, T), (T, H), (T, T)}. The position of each letter identifies its coin.

The pairs (H, T) and (T, H) are different outcomes, although both contain one head and one tail. Listing just “two heads, two tails, one of each” groups outcomes unevenly. A complete list should retain the distinction between the two coins.

Why must the outcomes being counted be equally likely?

Equally likely outcomes have the same chance of occurring. This condition is essential when probability is calculated by counting favourable outcomes and dividing by the total number of possible outcomes. Merely listing different descriptions does not establish equal chances.

For a fair coin, head and tail are equally likely. For a fair numbered die, the individual face results are equally likely. These assumptions justify counting each coin result or die face once when calculating a probability.

Why are colours not necessarily equally likely?

Consider a bag containing 4 red balls and 1 blue ball, with a ball drawn without looking. Red and blue are the two possible colours, but red is more likely because four individual balls are red and only one is blue.

The equally likely possibilities are the individual balls under random selection. Colour groups can contain different numbers of those possibilities. Therefore, the number of colour names is not the appropriate total for the counting formula in this experiment.

Note: Random selection means that each individual object is equally likely to be selected. It does not mean that every category of objects has the same probability.

How can a selection procedure support equal chances?

Identical name cards mixed thoroughly in a bag provide a way to select a student at random. A well-shuffled pack provides equally likely individual cards. In a ball problem, drawing without looking avoids deliberately choosing a particular colour.

Always identify what the procedure selects: a ball, a card, or a name card. Then count those objects. Only after identifying the individual possibilities should you group together the outcomes that satisfy the event being investigated.

How is theoretical probability calculated?

Theoretical probability, also called classical probability, calculates chance using the possible outcomes and an assumption of equal likelihood. An outcome is favourable to an event when it satisfies that event's stated condition. “Favourable” does not necessarily mean desirable.

The notation P(E) means the probability of event E. A slash, /, writes a fraction, and + means addition.

In the counting formula, the numerator is the number of favourable outcomes; the denominator is the number of all possible outcomes. A numerator is the upper part of a fraction, and a denominator is its lower part.

Definition: P(E) = number of outcomes favourable to E ÷ number of all possible outcomes, provided the outcomes being counted are equally likely. The symbol ÷ means division.

What steps give a complete calculation?

1. State the experiment and the event whose probability is required.
2. List or count all possible outcomes, checking that they are equally likely.
3. List or count the outcomes satisfying the event's exact wording.
4. Divide the favourable count by the total count and simplify the fraction.

Worked example 1. A fair coin is tossed once, with head and tail as the only outcomes considered. Find the probability of each result.

Answer: There are 2 equally likely outcomes. Head has 1 favourable outcome, so P(head) = 1/2. Tail also has 1 favourable outcome, so P(tail) = 1/2.

Worked example 2. A bag contains one red, one blue and one yellow ball, all of the same size. One ball is drawn without looking. Find the probability of each colour.

Answer: The 3 individual balls are equally likely to be drawn. Each colour has 1 favourable outcome. Therefore P(red) = 1/3, P(blue) = 1/3 and P(yellow) = 1/3.

In these expressions, words inside P( ) name the event directly. Equal probabilities follow from equal favourable counts and equal likelihood of the individual outcomes, rather than from the number of event names written down.

What results must every probability calculation satisfy?

Result: A probability lies between zero and one

0 ≤ P(E) ≤ 1, where ≤ means “less than or equal to”. The favourable count cannot be negative and cannot exceed the total count. Dividing it by the positive total therefore gives a value from zero to one, including both endpoints.

A proposed probability outside this range indicates an error. Check whether the denominator includes all possible outcomes and whether the numerator counts only outcomes satisfying the event. A probability describes a part of the possible outcomes, not a ratio of favourable outcomes to unfavourable outcomes.

Result: Impossible and certain events have probabilities zero and one

An impossible event has no possible outcome that makes it occur. A certain event, also called a sure event, occurs whichever possible outcome is obtained. These descriptions concern the stated experiment and its allowed outcomes.

Result: Elementary-event probabilities sum to one

The probabilities of all the elementary events of an experiment add to 1. Each possible outcome has been included once, so the complete collection accounts for every possible result. The word “all” matters in applying this result.

For the fair coin, the elementary events are head and tail, and 1/2 + 1/2 = 1. For the bag with one ball of each of three colours, the three elementary-event probabilities give 1/3 + 1/3 + 1/3 = 1.

Do not apply this check to an arbitrary list of events. First establish that the list covers the possible outcomes completely without counting an outcome more than once. Otherwise, adding the listed probabilities does not perform the intended completeness check.

How do we translate a die question into favourable outcomes?

In a single throw of a fair die numbered 1 to 6, the total number of equally likely outcomes is 6. The event's wording determines which of those numbers enter the favourable count. Write the qualifying numbers before writing the probability.

How do strict and inclusive comparisons differ?

At most means less than or equal to the stated limit.

“Greater than 4” excludes 4 itself. “Less than or equal to 4” includes 4. The phrase not greater than 4 has the same meaning as less than or equal to 4. Keeping the boundary value correct prevents a counting error.

Worked example 3. A fair die with faces numbered 1, 2, 3, 4, 5 and 6 is thrown once. Find the probabilities of getting a number greater than 4 and of getting a number less than or equal to 4.

Answer: Numbers greater than 4 are 5 and 6, giving 2 favourable outcomes. Thus P(number greater than 4) = 2/6 = 1/3. Numbers less than or equal to 4 are 1, 2, 3 and 4, giving P(number at most 4) = 4/6 = 2/3.

In the example, the two favourable lists together contain all six outcomes, with none repeated. Their probabilities add to 1 because one of those two conditions must hold.

How do number properties define events?

A prime number is an integer greater than 1 with exactly two positive factors, 1 and itself. An integer is a positive whole number, zero, or the negative of a positive whole number. A factor divides a number exactly. Among the die outcomes, the prime numbers are 2, 3 and 5.

An odd number is an integer not divisible by 2. The odd outcomes are 1, 3 and 5. These two lists each contain three outcomes, but they describe different events. Equal probabilities do not require identical favourable outcomes.

How do we solve problems involving balls and name cards?

For a random draw, count the individual objects that could be selected. A colour or a group describes an event, while the objects in that group supply the favourable outcomes. The total includes every object in the container, regardless of its group.

How do unequal colour groups affect the answer?

Worked example 4. A box contains 3 blue, 2 white and 4 red marbles. One marble is drawn at random, meaning each marble has the same chance. Find the probability of each colour.

Answer: The total number of marbles is 3 + 2 + 4 = 9. White has 2 favourable outcomes, blue has 3 and red has 4. Therefore P(white) = 2/9, P(blue) = 3/9 = 1/3 and P(red) = 4/9.

The denominator stays 9 throughout because the experiment remains one draw from the same box. The numerator changes with the colour requested. There are three colour categories, but their unequal numbers of marbles mean that the categories are not equally likely.

The three probabilities add to 1 because every marble has one of the listed colours. This provides a check on the counts. It does not turn the colour events into elementary events: each colour here includes more than one individual marble.

How does the same method apply to students?

Worked example 5. A class has 40 students: 25 girls and 15 boys. Each student's name is written on one identical card. The cards are thoroughly mixed and one is drawn. Find the probability of selecting a girl and of selecting a boy.

Answer: There are 40 equally likely name cards. The 25 girls' cards give P(girl) = 25/40 = 5/8. The 15 boys' cards give P(boy) = 15/40 = 3/8. Their sum is 1.

One card per student connects the number of cards directly with the number of students. The identical cards and thorough mixing explain the equal-likelihood assumption. The two group names alone would not justify treating the groups as equally probable.

How are probabilities calculated for playing cards?

A deck, or pack, contains 52 playing cards divided into four suits, which are named groups of cards: spades, hearts, diamonds and clubs. Each suit contains 13 cards. Spades and clubs are black; hearts and diamonds are red.

The cards in each suit are ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Face cards are kings, queens and jacks. An ace is a named card in each suit and is not a face card.

What does shuffling allow us to assume?

Drawing one card from a well-shuffled deck gives 52 equally likely individual outcomes. A suit name, colour or card name identifies which cards are favourable. The total is still 52 when the draw is from the complete deck.

The symbol − means subtraction.

Worked example 6. One card is drawn from a well-shuffled deck of 52 cards containing 4 aces. Find the probability that it is an ace and that it is not an ace.

Answer: There are 4 favourable cards for an ace, so P(ace) = 4/52 = 1/13. The number of cards that are not aces is 52 − 4 = 48. Thus P(not an ace) = 48/52 = 12/13.

Removing the ace count from the total counts the remaining possible cards; it does not describe physically removing a card before making this single draw.

How do we count face cards?

Worked example 7. A well-shuffled deck has 52 cards in four suits. Each suit contains one king, one queen and one jack, its three face cards. One card is drawn. Find the probability of a face card.

Answer: Each of the 4 suits contributes 3 face cards. The symbol × means multiplication, so the total favourable count is 4 × 3 = 12. Hence P(face card) = 12/52 = 3/13.

Count both the card type and its possible suits. Counting only the names king, queen and jack would overlook that each name appears once in every suit. The favourable count must count individual cards, just as the denominator does.

How does the complement help us find a probability?

The complement of an event E means the event “not E”. It contains all the outcomes in the sample space that are not favourable to E. We write it in words here to keep the condition visible.

Result: Complementary probabilities add to one

P(E) + P(not E) = 1, so P(not E) = 1 − P(E). The two events cover every possible outcome, with none counted in both. Either E occurs or it does not occur in the trial.

In the die example, “number greater than 4” and “number less than or equal to 4” are complementary events. In the card example, “ace” and “not an ace” are complementary events. The second event must include every outcome excluded by the first.

When does subtraction reduce the work?

For the complete pack with four aces, finding P(ace) = 1/13 immediately gives P(not an ace) = 1 − 1/13 = 12/13. This agrees with counting the 48 cards that are not aces directly.

Worked example 8. Sangeeta and Reshma play a tennis match in which one of them wins. The probability that Sangeeta wins is 0.62. Find the probability that Reshma wins.

Answer: The two winning events are complementary. Therefore P(Reshma wins) = 1 − P(Sangeeta wins) = 1 − 0.62 = 0.38.

The condition that one of the two players wins makes the complement clear. There is no need to assume equal winning chances or count possible scores. The supplied probability and the complementary relationship are sufficient for the calculation.

What must “not” include?

Check the full sample space before translating a negative condition. In a bag containing 3 red balls and 5 black balls, not red means black. The two listed colours account for every ball, so excluding the red balls leaves the black balls.

Do not replace “not an ace” with “a face card”. Numbered cards also satisfy “not an ace”. A complement excludes the named event and keeps all remaining outcomes, rather than choosing just one alternative category.

How do we interpret word problems and check our reasoning?

A word problem can describe an event through a decision rule rather than a colour or number. Translate that rule into the exact objects that qualify. The favourable count follows the wording, while the total count comes from the entire collection available for selection.

How do acceptance rules change the favourable count?

Worked example 9. A carton has 100 shirts: 88 good, 8 with minor defects and 4 with major defects. Minor and major defects are the two stated fault categories. Jimmy accepts only good shirts. Sujatha rejects only shirts with major defects. One shirt is drawn at random. Find each acceptance probability.

Answer: Jimmy accepts 88 shirts, giving P(acceptable to Jimmy) = 88/100 = 0.88. Sujatha accepts the good shirts and those with minor defects, totalling 88 + 8 = 96. Therefore P(acceptable to Sujatha) = 96/100 = 0.96.

Only is decisive in both conditions. Jimmy's rule excludes both defect categories. Sujatha's rule excludes just the major-defect category. The two probabilities use the same denominator because they refer to the same random draw from the carton.

How does theoretical probability differ from experimental probability?

Experimental probability, also called empirical probability, uses observed trials: the number of trials in which an event happened divided by the total number of trials. Theoretical probability instead uses assumptions about the possible outcomes to calculate a probability.

As the number of trials increases, we may expect the experimental and theoretical probabilities to be nearly the same. This does not promise exact agreement for a particular group of trials. A theoretical value is not a fixed schedule of future results.

What final checks should we make?

1. Check that the event matches the wording, including “only”, “not” and boundary values.
2. Check that both counts refer to individual outcomes from the same experiment.
3. Check that the equally likely assumption applies before using the counting formula.
4. Check the arithmetic and ensure the probability lies between 0 and 1.

Glossary

• Random experiment — A process with known possible outcomes whose particular result cannot be predicted with certainty beforehand.
• Trial — One performance of an experiment, such as tossing a coin or drawing one ball.
• Outcome — One possible result of an experiment, recorded according to what the experiment observes.
• Sample space — The set of all possible outcomes of the experiment being considered.
• Event — A subset of the sample space, specifying the outcomes that satisfy a stated condition.
• Equally likely outcomes — Outcomes that have the same chance of occurring under the experiment's stated assumptions.
• Favourable outcome — An outcome that satisfies the condition defining the event whose probability is required.
• Theoretical probability — The favourable-outcome count divided by the total possible-outcome count, when those outcomes are equally likely.
• Elementary event — An event containing exactly one of the possible outcomes of an experiment.
• Impossible event — An event with no possible outcome that makes it occur, giving probability zero.
• Certain event — An event that occurs for every possible outcome of the experiment, giving probability one.
• Complementary events — An event and its non-occurrence, together covering every possible outcome without any overlap.
• Empirical probability — The observed number of trials in which an event occurred divided by the total trials.

Common errors and misconceptions

• Misconception: Two named possibilities must have equal chances. Correct: Equal likelihood needs justification; red and blue are unequal colour groups in a bag with 4 red balls and 1 blue ball.
• Misconception: The denominator is the number of colours. Correct: When individual balls are equally likely, count all individual balls for the denominator.
• Misconception: Every event contains just one outcome. Correct: Only an elementary event has exactly one outcome; getting more than 4 on a numbered die includes 5 and 6.
• Misconception: “Not greater than 4” excludes 4. Correct: It means less than or equal to 4, so 4 is included.
• Misconception: An ace is a face card. Correct: Kings, queens and jacks are face cards; the ace is a separate card in each suit.
• Misconception: “Not an ace” means “a face card”. Correct: The complement includes every card except the aces, including the numbered cards.
• Misconception: Increasing the number of trials guarantees exact agreement with theory. Correct: We may expect experimental and theoretical probabilities to be nearly the same as trials increase.

Exam-style questions with model answers

Q1. A fair coin is tossed once, considering head and tail as its only possible outcomes. State the sample space and find the probability of a head. [2 marks]

1. Let H mean head and T mean tail. The sample space, denoted by S, is S = {H, T}.
2. There are two equally likely outcomes and one favourable outcome for head. Therefore the probability of a head is 1/2.

Q2. A fair die has faces numbered 1, 2, 3, 4, 5 and 6. It is thrown once. List the outcomes greater than 4, calculate their probability, and find the probability of a number not greater than 4. [3 marks]

1. The outcomes greater than 4 are 5 and 6. There are two favourable outcomes among the six equally likely die results.
2. The probability of a number greater than 4 is the favourable count divided by the total count: 2/6 = 1/3.
3. Not greater than 4 means 1, 2, 3 or 4. Its probability is 4/6 = 2/3, also equal to 1 − 1/3.

Q3. A box contains 3 blue, 2 white and 4 red marbles. One marble is selected, with every marble equally likely. State the total and calculate the probabilities of white, blue and red. [4 marks]

1. The total number of possible individual marbles is 3 + 2 + 4 = 9. This is the denominator for all three probabilities.
2. There are 2 white marbles, giving the probability of white as 2/9.
3. There are 3 blue marbles, giving the probability of blue as 3/9 = 1/3.
4. There are 4 red marbles, giving the probability of red as 4/9. The three colour probabilities together add to 1.

Q4. A class contains 25 girls and 15 boys. Each student has one identical name card; all cards are mixed thoroughly and one is drawn at random. Find the total number of outcomes, the probability of a girl, and the probability of a boy. [3 marks]

1. There are 25 + 15 = 40 students and therefore 40 name cards. Mixing identical cards makes the individual name cards equally likely outcomes.
2. The 25 cards naming girls are favourable to selecting a girl. The probability is 25/40 = 5/8.
3. The 15 cards naming boys are favourable to selecting a boy. The probability is 15/40 = 3/8, which also equals 1 − 5/8.

Q5. One card is drawn from a well-shuffled deck of 52 cards. The deck has four suits, each containing one ace and three face cards: a king, a queen and a jack. Find the probabilities of an ace, not an ace, and a face card, showing the relevant counts. [5 marks]

1. The random draw has 52 equally likely individual card outcomes. Each probability therefore uses 52 as the total count, since the whole deck is available.
2. Each of the four suits contains one ace, giving four aces altogether. The probability of drawing an ace is 4/52 = 1/13.
3. There are 52 − 4 = 48 cards that are not aces. The probability of not drawing an ace is 48/52 = 12/13.
4. Each suit has three face cards. Across four suits, the favourable face-card count is 4 × 3 = 12 cards.
5. The probability of drawing a face card is therefore 12/52 = 3/13. Aces are excluded because the stated face cards are kings, queens and jacks.

Q6. A carton contains 100 shirts: 88 good, 8 with minor defects and 4 with major defects. Jimmy accepts only good shirts; Sujatha rejects only shirts with major defects. One shirt is selected at random. Explain the total and each trader's favourable count, then calculate both acceptance probabilities. [5 marks]

1. There are 100 individual shirts, and random selection makes each equally likely to be drawn. The total possible-outcome count for both acceptance questions is 100.
2. Jimmy accepts only the 88 good shirts. Neither minor-defect nor major-defect shirts meet his condition, so his favourable count is 88.
3. Jimmy's acceptance probability is the favourable count divided by the total count: 88/100 = 0.88.
4. Sujatha rejects only the 4 major-defect shirts. She therefore accepts both the good shirts and those with minor defects: 88 + 8 = 96 shirts.
5. Sujatha's acceptance probability is 96/100 = 0.96. The larger favourable count follows from her different acceptance rule, while the total available shirts remains unchanged.

Q7. Sangeeta and Reshma play a tennis match in which one of them wins. Sangeeta's probability of winning is 0.62. Explain the relationship between their winning events and calculate Reshma's probability of winning. [2 marks]

1. Exactly one of the two players wins, so their winning events are complementary and their probabilities add to 1.
2. Reshma's probability of winning is 1 − 0.62 = 0.38.

Q8. A fair die numbered 1 to 6 is thrown once. Find the probabilities of getting 8 and of getting a number less than 7. Name each type of event and state the permitted range of a probability. [3 marks]

1. No face is marked 8, so its favourable count is zero and its probability is 0/6 = 0. Getting 8 is an impossible event.
2. All six faces are numbered below 7, so the probability of a number less than 7 is 6/6 = 1. This is a certain event.
3. Every probability lies between 0 and 1, including both endpoints, because a favourable count is non-negative and cannot exceed the total possible-outcome count.

Key takeaways

• A random experiment has identifiable possible outcomes, while its particular result cannot be predicted with certainty beforehand.
• The sample space lists all possible outcomes; an event selects those satisfying the condition being investigated.
• Use favourable outcomes divided by total outcomes only when the individual outcomes being counted are equally likely.
• Count individual balls, cards or name cards before grouping them by colour, type or student category.
• A probability lies between zero and one; impossible and certain events have these endpoint probabilities respectively.
• An event and its complement cover every possible outcome without overlap, so their probabilities add to one.
• Read restrictions carefully: “not”, “only” and “less than or equal to” determine which outcomes are favourable.
• As trials increase, we may expect experimental and theoretical probabilities to be nearly the same, without guaranteeing exact agreement.

Test yourself

What distinguishes an outcome from an event?

An outcome is one possible result; an event is a set of outcomes satisfying a condition.

A bag has 4 red balls and 1 blue ball. Under equally likely individual selection, why are the colours not equally likely?

Four individual balls produce red, while only one produces blue. The colour groups therefore have different favourable counts.

A fair die is numbered 1 to 6. Which outcomes satisfy “not greater than 4”?

The outcomes are 1, 2, 3 and 4, because the condition includes equality with 4.

Why does a certain event have probability 1?

Every possible outcome is favourable, so dividing the favourable count by the identical total count gives 1.

What is an elementary event, and what do all elementary-event probabilities sum to?

An elementary event contains exactly one outcome. The probabilities of all elementary events in the experiment sum to 1.

If E is an event with probability P(E) = 0.05, what is the probability of not E?

The complement rule gives P(not E) = 1 − 0.05 = 0.95, because the two probabilities sum to 1.

Which named cards are face cards, and is an ace included?

Kings, queens and jacks are face cards. An ace is a separate card and is not included.

What relationship may we expect between experimental and theoretical probability as trials increase?

We may expect them to be nearly the same. Exact agreement is not guaranteed for a particular group of trials.