Probability

This companion follows the chapter's own build-up, section by section. It opens with conditional probability, defining $P(E|F)$ from first principles by restricting the sample space, then records the three properties that follow immediately from that definition and are used constantly afterward. It then proves the multiplication theorem of probability, the algebraic rearrangement of the conditional-probability formula that lets you compute the probability of several events happening together, and extends it from two events to three (and beyond). Next comes independence: the precise definition $P(E \cap F) = P(E)P(F)$, the crucial distinction from mutually exclusive events, and the extension to mutual independence of three or more events. The chapter then builds the machinery needed for its capstone result: a partition of the sample space, the theorem of total probability (computing $P(A)$ by summing over every possible "cause" in a partition), and finally Bayes' theorem itself, which reverses conditioning to answer "given that $A$ happened, which cause $E_i$ was responsible." Work through the sections in this order — total probability is a special case hiding inside every Bayes' theorem calculation, and independence is what lets many probability trees (coin-then-die, machine-then-defect) be built at all. Every worked example below reuses NCERT's own example numbers, extended with a Board level and Competitive level version so you can see the same ideas at CBSE and at JEE/NEET depth.

Conditional probability: restricting the sample space

Suppose two events $E$ and $F$ are associated with the same random experiment. If you are told that $F$ has occurred, every outcome outside $F$ is now impossible, and the sample space effectively shrinks from $S$ down to $F$. The conditional probability of $E$ given that $F$ has occurred, written $P(E|F)$, measures how likely $E$ is inside this smaller universe:

where: $E \cap F$ is the event that both $E$ and $F$ occur, and the condition $P(F) \neq 0$ (equivalently $F \neq \phi$) is essential — conditioning on an impossible event is simply not defined. When outcomes are equally likely, dividing both numerator and denominator of $n(E \cap F)/n(F)$ by $n(S)$ recovers exactly this formula, which is how NCERT first motivates it with the three-coins example (tossing three fair coins, $E$: at least two heads, $F$: first coin shows tail, giving $P(E|F) = \tfrac14$). Three properties follow directly from the definition and are used everywhere downstream: $P(S|F) = P(F|F) = 1$ (the restricted universe itself, and $F$ inside it, both have full probability $1$); for any events $A, B$ and any $F$ with $P(F) \neq 0$,

and the complement rule $P(E'|F) = 1 - P(E|F)$, the direct analogue of the ordinary complement rule but now inside the restricted sample space $F$. In particular, if $A$ and $B$ are mutually exclusive, $P((A \cup B)|F) = P(A|F) + P(B|F)$, exactly like ordinary probability of a union of disjoint events.

The multiplication theorem of probability

Rearranging the conditional-probability formula gives a way to compute the probability of two events occurring together, $P(E \cap F)$, when the conditional probabilities are what you can actually read off a problem. Since $P(E|F) = P(E \cap F)/P(F)$ and, symmetrically, $P(F|E) = P(E \cap F)/P(E)$, cross-multiplying both gives the multiplication theorem of probability:

where: $P(F|E)$ is read as "the probability that $F$ occurs, given that $E$ has already occurred" — the natural quantity when events unfold in a sequence, such as drawing balls from an urn one after another without replacement. The theorem extends to three events $E, F, G$: $P(E \cap F \cap G) = P(E)\,P(F|E)\,P(G|E \cap F)$, chaining the conditioning one step at a time, and it generalises the same way to four or more events. This is the exact tool NCERT uses for its urn and card-drawing examples (worked out below), and it is the single most common calculation engine in "without replacement" probability problems on both boards and competitive exams.

Independent events

Sometimes learning that $F$ has occurred changes nothing about how likely $E$ is — the classic example is drawing the ace of spades from a shuffled deck, where knowing the card is a spade does not change the probability it is an ace, and vice versa. Two events $E$ and $F$ are called independent if

where: this single condition is equivalent, whenever $P(E) \neq 0$ and $P(F) \neq 0$, to $P(F|E) = P(F)$ and $P(E|F) = P(E)$ — substituting either into the multiplication theorem (3) collapses it to equation (4), which is how NCERT first presents independence before showing the two definitions coincide. Events that are not independent are called dependent events. It is worth stating plainly what independence is not: two mutually exclusive events with nonzero probability can never be independent (if $E \cap F = \phi$ then $P(E \cap F) = 0$, but $P(E)P(F) > 0$), and two independent events with nonzero probability are never mutually exclusive (they must share at least the possibility of joint occurrence). "Independent" is a statement about probabilities; "mutually exclusive" is a statement about the sets themselves — confusing the two is the single most common conceptual slip in this chapter, addressed again in the warning below. The idea extends to mutually independent events: three events $A, B, C$ are mutually independent only if all four conditions hold simultaneously,

where: the three pairwise equalities alone are not enough — a triple can satisfy all three pairwise conditions ("pairwise independent") yet still fail $P(A \cap B \cap C) = P(A)P(B)P(C)$, so all four equalities must be checked before declaring three events mutually independent; if even one fails, the events are not (mutually) independent.

Partition of a sample space, and the theorem of total probability

Bayes' theorem needs one more piece of vocabulary. A collection of events $E_1, E_2, \ldots, E_n$ is called a partition of the sample space $S$ if it satisfies three conditions together: $E_i \cap E_j = \phi$ for every $i \neq j$ (pairwise disjoint), $E_1 \cup E_2 \cup \cdots \cup E_n = S$ (exhaustive), and $P(E_i) > 0$ for every $i$. The simplest partition is any nonempty event $E$ together with its complement $E'$, since $E \cap E' = \phi$ and $E \cup E' = S$. Given such a partition, and any other event $A$ associated with $S$, the theorem of total probability states

where: the idea is to break event $A$ into the $n$ mutually exclusive pieces $A \cap E_1, A \cap E_2, \ldots, A \cap E_n$ (the part of $A$ occurring alongside each cause in turn), add their probabilities, then rewrite each $P(A \cap E_j)$ using the multiplication theorem as $P(E_j)P(A|E_j)$ — this is proved formally in the first derivation below. Physically, the theorem says: to find the total chance of $A$, weight the chance of $A$ under each possible "cause" $E_j$ by how likely that cause is, and add. This is exactly how NCERT computes, for instance, the overall chance a construction job finishes on time given the chance of a strike and the conditional completion rates in each case.

Bayes' theorem: reversing conditional probability

The theorem of total probability answers "given the cause, what is the chance of the effect?" Bayes' theorem answers the reverse, much harder question: "given that the effect has been observed, what is the chance it came from a particular cause?" If $\{E_1, E_2, \ldots, E_n\}$ is a partition of $S$ with every $P(E_i) > 0$, and $A$ is any event with $P(A) \neq 0$, then for any $i$,

where: the events $E_1, \ldots, E_n$ are called the hypotheses; $P(E_i)$, the probability of a hypothesis before any evidence is considered, is called its priori probability; and $P(E_i|A)$, the updated probability once the evidence $A$ is known, is called the posteriori probability. The denominator is nothing but $P(A)$ from the theorem of total probability, so Bayes' theorem is really the multiplication-theorem numerator $P(E_i \cap A) = P(E_i)P(A|E_i)$ divided by the total-probability denominator — a proof carried out in full in the second derivation below. Bayes' theorem is sometimes called the formula for the "probability of causes," because since the $E_i$ form a partition, exactly one of them must actually have occurred, and the formula weighs how much each one is implicated by the observed evidence $A$.

Board level - A family's two children: conditional probability from a small sample space

A family has two children. What is the probability that both children are boys, given that at least one of them is a boy?

Step 1. Let $b$ stand for boy and $g$ for girl; with birth order recorded, the sample space is $S = \{(b,b), (g,b), (b,g), (g,g)\}$, four equally likely outcomes.

Step 2. Let $E$: "both children are boys" and $F$: "at least one child is a boy." Then $E = \{(b,b)\}$, so $P(E) = \tfrac14$, and $F = \{(b,b), (g,b), (b,g)\}$, so $P(F) = \tfrac34$.

Step 3. Find $E \cap F$: the only outcome in both sets is $(b,b)$, so $E \cap F = \{(b,b)\}$ and $P(E \cap F) = \tfrac14$.

Step 4. Apply the definition of conditional probability: $P(E|F) = \dfrac{P(E \cap F)}{P(F)} = \dfrac{1/4}{3/4} = \dfrac{1}{3}$. Answer: $P(E|F) = \dfrac{1}{3}$. Notice this is smaller than the naive guess of $\tfrac12$ — knowing "at least one boy" rules out $(g,g)$ but leaves two of the three remaining outcomes with a girl in them, so the extra information is less favourable to "both boys" than intuition first suggests.

Board level - Drawing two black balls without replacement, using the multiplication theorem

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other, without replacement. What is the probability that both drawn balls are black?

Step 1. Let $E$: the first ball drawn is black, and $F$: the second ball drawn is black. We need $P(E \cap F)$, the probability both draws are black, and the multiplication theorem gives $P(E \cap F) = P(E)\,P(F|E)$.

Step 2. Compute $P(E)$: there are $10$ black balls out of $15$ total, so $P(E) = \dfrac{10}{15} = \dfrac{2}{3}$.

Step 3. Compute $P(F|E)$: given that the first ball drawn was black, the urn now holds $9$ black and $5$ white balls, $14$ total, so $P(F|E) = \dfrac{9}{14}$.

Step 4. Multiply: $P(E \cap F) = \dfrac{10}{15} \times \dfrac{9}{14} = \dfrac{90}{210} = \dfrac{3}{7}$. Answer: the probability both balls drawn are black is $\dfrac{3}{7}$.

Competitive level (JEE / NEET) - Which machine made the defective bolt: Bayes' theorem across three causes

In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5%, 4% and 2% are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured by machine B?

Step 1. Define the partition: $B_1$: the bolt is from machine A, $B_2$: from machine B, $B_3$: from machine C. Since every bolt comes from exactly one machine, $\{B_1, B_2, B_3\}$ is a partition of the sample space, with priori probabilities $P(B_1) = 0.25$, $P(B_2) = 0.35$, $P(B_3) = 0.40$.

Step 2. Let $E$ be the event "the bolt is defective." The given percentages are conditional probabilities: $P(E|B_1) = 0.05$, $P(E|B_2) = 0.04$, $P(E|B_3) = 0.02$.

Step 3. Compute the denominator using the theorem of total probability: $P(E) = P(B_1)P(E|B_1) + P(B_2)P(E|B_2) + P(B_3)P(E|B_3) = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02) = 0.0125 + 0.0140 + 0.0080 = 0.0345$.

Step 4. Apply Bayes' theorem for hypothesis $B_2$: $P(B_2|E) = \dfrac{P(B_2)P(E|B_2)}{P(E)} = \dfrac{0.0140}{0.0345} = \dfrac{28}{69} \approx 0.406$. Answer: the probability the defective bolt came from machine B is $\dfrac{28}{69} \approx 0.406$. Timing note: about ninety seconds once the setup is recognised. The entire calculation is two clean stages — first total probability for the denominator (one multiply-and-add per machine), then a single Bayes' theorem fraction reusing the numerator you already computed for machine B. The time sink is not the arithmetic but correctly identifying the partition and matching each percentage to the right conditional probability; write $P(B_i)$ and $P(E|B_i)$ as a two-row table for all three machines before touching Bayes' theorem, so nothing gets mismatched under time pressure.

Competitive level (JEE / NEET) - Alternating dice, an infinite game, and a geometric series

A and B throw a die alternately till one of them gets a six and wins the game. Find their respective probabilities of winning, if A starts first.

Step 1. Let $S$ denote success (getting a six) on a single throw and $F$ denote failure. Since the die is fair, $P(S) = \dfrac{1}{6}$ and $P(F) = \dfrac{5}{6}$.

Step 2. List the ways A can win. A wins on the first throw with probability $P(S) = \dfrac{1}{6}$. A wins on the third throw only if the sequence is failure (A), failure (B), success (A) — using the multiplication theorem for independent trials, $P(FFS) = P(F)P(F)P(S) = \left(\dfrac{5}{6}\right)^2 \dfrac{1}{6}$. In general A wins on the $(2k+1)$th throw with probability $\left(\dfrac{5}{6}\right)^{2k}\dfrac{1}{6}$, for $k = 0, 1, 2, \ldots$

Step 3. Sum the infinite geometric series for $P(A \text{ wins})$, whose terms have first term $a = \dfrac{1}{6}$ and common ratio $r = \left(\dfrac{5}{6}\right)^2 = \dfrac{25}{36}$: $P(A \text{ wins}) = \dfrac{1}{6} + \dfrac{1}{6}\left(\dfrac{25}{36}\right) + \dfrac{1}{6}\left(\dfrac{25}{36}\right)^2 + \cdots = \dfrac{a}{1-r} = \dfrac{1/6}{1 - 25/36} = \dfrac{1/6}{11/36} = \dfrac{36}{6 \times 11} = \dfrac{6}{11}$.

Step 4. B's winning probability is the complement, since exactly one of A or B eventually wins (the game ends with probability $1$): $P(B \text{ wins}) = 1 - P(A \text{ wins}) = 1 - \dfrac{6}{11} = \dfrac{5}{11}$. Answer: $P(A \text{ wins}) = \dfrac{6}{11}$ and $P(B \text{ wins}) = \dfrac{5}{11}$. Timing note: about three minutes. The two traps are algebraic, not conceptual: writing the ratio of the geometric series as $\left(\dfrac{5}{6}\right)^2$, not $\dfrac{5}{6}$ (each "round" that A fails to win consumes two throws, one by each player), and remembering $\left|r\right| < 1$ so the standard sum formula $\dfrac{a}{1-r}$ applies. Once $P(A \text{ wins})$ is found, get $P(B \text{ wins})$ by subtracting from $1$ rather than summing a second series — it is faster and self-checking, since the two probabilities must add to exactly $1$.

• Writing $P(E|F)$ when $P(F|E)$ was intended, or vice versa. These are generally different numbers; always identify which event is the "given" (denominator) event before writing the formula, and match it to what the question actually asks.
• Applying the conditional probability formula when $P(F) = 0$. The formula $P(E|F) = P(E \cap F)/P(F)$ is simply undefined if $F$ is impossible; check $P(F) \neq 0$ before dividing.
• Confusing independent events with mutually exclusive events. These are different, near-opposite ideas — see the warning below; this is the single most tested conceptual trap in the chapter.
• Assuming pairwise independence proves mutual independence for three events. All three pairwise conditions plus $P(A \cap B \cap C) = P(A)P(B)P(C)$ must hold together; a counterexample can satisfy the three pairwise equations yet fail the fourth.
• Forgetting to check that a proposed collection of events is actually a partition before applying total probability or Bayes' theorem. The events must be pairwise disjoint, their union must be the whole sample space, and each must have nonzero probability — skipping this check risks summing over overlapping or incomplete "causes."
• Using $P(E_i)$ (the priori probability) where $P(E_i|A)$ (the posteriori probability) was wanted, or vice versa. The priori probability is what you know before the evidence; the posteriori is the whole point of running Bayes' theorem, and the two are usually numerically quite different, as the HIV-test and bolt-factory style examples show vividly.
• Forgetting the denominator in Bayes' theorem is the full theorem-of-total-probability sum, not just the term for the hypothesis in question. Every path to $A$ must be added, not just the path through the hypothesis you are solving for.
• Treating "with replacement" and "without replacement" draws identically. Without replacement, conditional probabilities on later draws change (fewer balls or cards remain, and the composition shifts); the multiplication theorem is precisely the tool built for this, and misapplying independence (multiplying unconditional probabilities) when draws are without replacement gives a wrong answer.
• Rounding intermediate decimals too early in a Bayes' theorem calculation. These problems often involve small numbers, like $0.001$ or $0.005$, multiplied together; rounding an intermediate product to two decimal places can wipe out the entire numerator and silently produce a wildly wrong final answer, so carry full precision, or exact fractions, until the very last step.
• Writing $P(E'|F) = 1 - P(E)$ instead of $1 - P(E|F)$. The complement rule for conditional probability complements only within the restricted sample space $F$; do not complement against the unconditioned probability $P(E)$.

• State the definition of conditional probability $P(E|F) = P(E \cap F)/P(F)$, explain why $P(F) \neq 0$ is required, and use the three basic properties confidently.
• Derive and apply the multiplication theorem of probability for two and for three or more events, especially in "without replacement" drawing problems.
• State the definition of independent events, and explain clearly why independence and mutually exclusive events are different (and, for nonzero probabilities, incompatible) properties.
• Check mutual independence of three events by verifying all four required conditions, not just the three pairwise ones.
• Identify whether a given collection of events forms a valid partition of the sample space, checking pairwise disjointness, exhaustiveness and nonzero probability.
• State and apply the theorem of total probability to compute an overall probability from probabilities conditioned on each cause in a partition.
• State and apply Bayes' theorem to compute a posteriori probability from priori probabilities and conditional probabilities, correctly assembling the total-probability denominator.
• Recognise, from the wording of a problem, whether it is asking for a forward (total probability) or reverse (Bayes' theorem) calculation.
• Prove short results about conditional probability and independence (such as independence of complements) using only the definitions and the multiplication theorem.
• Combine conditional probability with an infinite geometric series to solve "alternating trials until someone succeeds" problems.

Ask the tutor: Want to work through a few more Bayes' theorem or independence problems together, building up to the JEE-style multi-stage tree diagrams before moving on?