Probability

The chapter has a short, tight spine. It opens by contrasting the empirical probability already met in Class IX with a new, calculation-based idea: theoretical (classical) probability, which needs no repeated trials, only the assumption that an experiment's outcomes are equally likely. From that single definition, $P(E) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$, the chapter draws out its consequences: the elementary event and the fact that all elementary-event probabilities sum to $1$; the range rule $0 \leq P(E) \leq 1$, with the boundary cases of an impossible event ($P=0$) and a sure event ($P=1$); and the complement rule $P(E) + P(\text{not }E) = 1$, used to shortcut "at least one" and "not" questions. The bulk of the chapter is then practice at identifying the sample space and counting favourable outcomes across a varied set of experiments — coins, dice (singly, then two together), a standard deck of playing cards, bags of coloured balls or candies, boxes of numbered discs, and word problems dressed as class elections or defective-item inspections. A short closing idea, marked by NCERT as beyond examination requirements, extends the favourable-to-total ratio from counting outcomes to measuring them — lengths and areas.

Random experiments and equally likely outcomes

An experiment whose outcome cannot be predicted with certainty, but whose set of possible outcomes is known in advance, is a random experiment. Tossing a coin is the standard first example: we assume the coin is fair or unbiased — symmetrical, with no reason to favour one face — and that the toss is random, meaning it falls freely with no interference. Under these assumptions the two outcomes, head (H) and tail (T), are equally likely. Throwing a fair (unbiased) die is the second standard example — the six outcomes $1,2,3,4,5,6$ are all equally likely. Not every experiment has equally likely outcomes. A bag with $4$ red balls and $1$ blue ball, drawn from without looking, does not make "getting a red ball" and "getting a blue ball" equally likely — you are four times more likely to draw red, simply because there are four times as many red balls. What is equally likely here is drawing any one particular ball from the five. From this chapter onward, unless stated otherwise, we work only with experiments where the equally likely assumption genuinely holds — and checking that it does, before applying any formula, is itself part of the mathematics.

Events, elementary events, and the sample space

The complete list of possible outcomes of an experiment is its sample space. For one coin toss, the sample space is $\{H, T\}$; for one die throw, it is $\{1,2,3,4,5,6\}$. An event is any outcome, or collection of outcomes, that we are interested in — "getting a head," "getting a number greater than 4," "drawing a red ball." An event with exactly one favourable outcome is called an elementary event. In the coin toss, "getting a head" and "getting a tail" are both elementary events; in a bag with one red, one blue, and one yellow ball, "drawing the red ball" is an elementary event, and so are the other two colour-events. A key structural fact follows immediately: since every outcome in the sample space belongs to exactly one elementary event, and every elementary event has probability (favourable $1$)/(total outcomes), adding up the probabilities of all the elementary events of an experiment must give exactly $1$ — together they account for the whole sample space, with nothing left over and nothing double-counted.

The classical (theoretical) definition of probability

With the sample space and equally likely outcomes in place, probability becomes a single ratio.

where: $P(E)$ is the theoretical (classical) probability of event $E$, the numerator counts only those outcomes in the sample space that make $E$ happen, and the denominator counts every outcome in the sample space, whether or not it favours $E$ — and this counting is valid only because the outcomes are assumed equally likely. This definition was given by Pierre-Simon Laplace in 1795, and unlike empirical probability it requires no repeated trials at all: for tossing a coin once, the number of favourable outcomes for "getting a head" is $1$ and the total is $2$, so $P(\text{head}) = \tfrac{1}{2}$ directly from counting, with no coin ever actually thrown. The empirical and theoretical values are not unrelated, though — as the number of trials in an experiment grows very large, the empirical probability tends to settle down close to the theoretical value, which is exactly why casinos, insurers, and quality inspectors trust the theoretical number in advance.

The range of a probability: impossible and sure events

Because the numerator of formula $(1)$ counts a subset of the outcomes counted in the denominator, the number of favourable outcomes can never exceed the total number of outcomes, and it can never be negative (a count of outcomes is never less than zero). This pins down every probability to a fixed range.

where: the lower bound $P(E) = 0$ occurs exactly when no outcome favours $E$ — getting an $8$ on a single throw of an ordinary die is impossible, since no face is marked $8$, so $P(\text{getting } 8) = \tfrac{0}{6} = 0$; such an event is called an impossible event. The upper bound $P(E) = 1$ occurs exactly when every outcome favours $E$ — getting a number less than $7$ on a die throw is certain, since all six faces qualify, so $P(E) = \tfrac{6}{6} = 1$; such an event is called a sure event or a certain event. Every other event's probability is a genuine fraction strictly between $0$ and $1$, reflecting a genuine mix of favourable and unfavourable outcomes.

Complementary events

Every event $E$ has an opposite: the event that $E$ does not happen, written $\overline{E}$ or "not $E$." If $E$ is "getting a head," then $\overline{E}$ is "getting a tail" — and for one coin toss, since $\overline{E}$ has exactly the outcomes the sample space has minus the outcomes for $E$, the two probabilities must add to $1$.

where: $\overline{E}$, the complement of $E$, is the event "not $E$," and $E$, $\overline{E}$ are called complementary events. The practical payoff of $(3)$ is enormous: whenever "not $E$" is easier to count than $E$ itself — as in "not an ace" out of a 52-card deck, where counting the $48$ non-aces directly is more work than counting the $4$ aces and subtracting — the complement rule turns a long count into a single subtraction, $P(\overline{E}) = 1 - P(E)$.

Compound experiments: two coins and two dice together

When an experiment involves more than one object — two coins tossed together, or two dice thrown together — the sample space is built from ordered outcomes, and getting this count right is the single most important step before any probability can be computed. Tossing two different coins together (say a $\textrm{₹}1$ coin and a $\textrm{₹}2$ coin) gives four equally likely outcomes, $(H,H), (H,T), (T,H), (T,T)$ — $(H,T)$ is a different outcome from $(T,H)$, because the two coins are distinguishable. The same idea, scaled up, governs two dice thrown together (equivalently, one die thrown twice). Writing the first number for a "blue" die and the second for a "grey" die, the sample space is every ordered pair $(i,j)$ with $i,j \in \{1,2,3,4,5,6\}$ — $(1,4)$ is genuinely different from $(4,1)$. This gives $6 \times 6 = 36$ equally likely outcomes, laid out below with the five outcomes favourable to "the sum is $8$" highlighted.

Beyond this chapter's syllabus, the same favourable-to-total idea extends from counting discrete outcomes to measuring continuous ones. If a person is told to stop music at a random instant within a $2$-minute interval, the probability of stopping within the first half-minute is the ratio of favourable length to total length, $\tfrac{0.5}{2} = \tfrac14$; similarly, a dart landing inside a rectangular board gives probabilities as ratios of areas. NCERT marks these geometric-probability examples as not required for examination — included here only to show how far $\dfrac{\text{favourable}}{\text{total}}$ ultimately reaches.

Board level - probability from a deck of cards (NCERT Example 4, extended)

One card is drawn from a well-shuffled deck of $52$ cards. Calculate the probability that the card will be (i) an ace, (ii) not an ace, (iii) a face card.

Step 1. Well-shuffling makes every card equally likely, so the total number of outcomes is $52$.

Step 2. There are $4$ aces in a deck (one per suit). Let $E$ be "the card is an ace"; the favourable count is $4$, so $P(E) = \dfrac{4}{52} = \dfrac{1}{13}$.

Step 3. Let $F$ be "the card is not an ace," the complement of $E$. Use the complement rule instead of counting all $48$ non-aces directly: $P(F) = 1 - P(E) = 1 - \dfrac{1}{13} = \dfrac{12}{13}$.

Step 4. As an extension, find $P(\text{face card})$. Kings, queens, and jacks are face cards, giving $3$ per suit and $12$ in the deck: $P(\text{face card}) = \dfrac{12}{52} = \dfrac{3}{13}$. Answer: $P(\text{ace}) = \dfrac{1}{13}$, $P(\text{not ace}) = \dfrac{12}{13}$, $P(\text{face card}) = \dfrac{3}{13}$.

Board level - probability from two dice thrown together (NCERT Example 13, extended)

Two dice, one blue and one grey, are thrown together. Find the probability that the sum of the numbers on top is (i) $8$, (ii) $13$, (iii) less than or equal to $12$, (iv) $7$.

Step 1. Build the sample space using the fundamental counting principle: each die has $6$ equally likely faces, so the pair has $6 \times 6 = 36$ equally likely ordered outcomes.

Step 2. For sum $= 8$, the favourable outcomes are $(2,6), (3,5), (4,4), (5,3), (6,2)$ — $5$ outcomes, so $P(\text{sum}=8) = \dfrac{5}{36}$.

Step 3. For sum $= 13$, no pair from $1$ to $6$ can add to more than $12$, so $P(\text{sum}=13) = \dfrac{0}{36} = 0$, an impossible event. For sum $\leq 12$, every one of the $36$ outcomes qualifies: $P(\text{sum} \leq 12) = \dfrac{36}{36} = 1$, a sure event.

Step 4. As an extension, sum $=7$ has the favourable outcomes $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — $6$ outcomes, the largest for any single sum, so $P(\text{sum}=7) = \dfrac{6}{36} = \dfrac{1}{6}$, making $7$ the single most likely sum. Answer: $P(\text{sum}=8) = \dfrac{5}{36}$, $P(\text{sum}=13) = 0$, $P(\text{sum} \leq 12) = 1$, $P(\text{sum}=7) = \dfrac{1}{6}$.

Board HOTS / foundation JEE-NEET - probability on a shrinking sample space (extends NCERT Exercise 14.1, Q15)

Five cards — the ten, jack, queen, king, and ace of diamonds — are well-shuffled with their faces downward. One card is picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and kept aside (not replaced), what is the probability that the next card picked from the remaining four is (a) an ace, (b) the queen?

Step 1. For the first pick, all five cards are equally likely, so the total is $5$. Exactly one, the queen, favours "card is the queen," so $P(\text{queen, first pick}) = \dfrac{1}{5}$.

Step 2. Once the queen is drawn and set aside, the remaining cards are the ten, jack, king, and ace — only $4$ cards, all still equally likely to be picked next. The total for the second pick is therefore $4$, not $5$; recomputing the sample space after an item is removed is the step students most often skip.

Step 3. Among these $4$ cards, exactly one is the ace, so $P(\text{ace, second pick}) = \dfrac{1}{4}$.

Step 4. The queen is no longer among the $4$ remaining cards, so no outcome favours "second card is the queen": $P(\text{queen, second pick}) = \dfrac{0}{4} = 0$ — an impossible event on this reduced sample space, even though the queen was perfectly possible on the first pick. Answer: $P(\text{queen first}) = \dfrac{1}{5}$, $P(\text{ace second}) = \dfrac{1}{4}$, $P(\text{queen second}) = 0$. Timing note: aim for under three minutes. The whole problem hinges on noticing the sample space shrinks from $5$ to $4$ once a card is removed without replacement — forgetting this and writing $P(\text{ace second}) = \dfrac{1}{5}$ is the single most common error. Recomputing probability on a reduced, still-equally-likely sample space is exactly the seed of conditional probability you will meet formally in later study; Class 10 only asks you to redo the count carefully.

• Assuming outcomes are equally likely without checking. "A car starts or does not start" is not automatically $50$-$50$ just because there are two named outcomes — equally likely-ness must be justified by the physical setup, never assumed from the wording alone.
• Confusing empirical and theoretical probability. Theoretical probability is computed in advance from counting; empirical probability is measured after actually running trials, and small numbers of trials can give quite different values.
• Miscounting ordered outcomes in compound experiments. For two coins or two dice, $(H,T)$ and $(T,H)$, or $(1,4)$ and $(4,1)$, are different outcomes — treating them as the same undercounts the sample space.
• Dropping the boundary in "less than or equal to" or "greater than." "Less than or equal to $4$" on a die includes $4$ itself; "greater than $4$" excludes it — read inequality wording literally.
• Card-deck slips. Forgetting there are $4$ suits of $13$ cards each; confusing "face card" ($12$: kings, queens, jacks) with "red face card" (only $6$); or miscounting a specific card as $1$ out of $13$ instead of $1$ out of $52$.
• Forgetting to shrink the sample space after removal without replacement. Once an item is drawn and not put back, later probabilities must be computed over the smaller remaining set, not the original total.

• State the classical definition of probability and correctly identify the favourable and total outcomes in a given experiment.
• Judge, from the description of an experiment, whether its outcomes are genuinely equally likely, and explain why when they are not.
• List the elementary events of a simple experiment and verify their probabilities sum to exactly $1$.
• Compute probabilities confidently for coins, dice (single and two together), a $52$-card deck, and bags or boxes of coloured objects or numbered tickets.
• Build the $36$-outcome sample space for two dice and count favourable outcomes for a given sum accurately.
• Apply the complement rule $P(\overline{E}) = 1 - P(E)$ to shortcut "not" and "at least one" problems.
• Correctly identify and justify impossible events ($P=0$) and sure events ($P=1$).
• Recompute probability correctly on a reduced sample space after an item has been removed without replacement.

Ask the tutor: if you are unsure whether an experiment's outcomes are truly equally likely, or want to work through a trickier deck-of-cards or two-dice counting problem step by step, ask the tutor to walk through it with you.