Chapter 13 · Probability

13.2 The Multiplication Theorem and Independent Events

Turn the definition of conditional probability round and you can find the chance that several things all happen, one stage at a time: $P(E\cap F)=P(E)\,P(F\mid E)$. When the information about one event does not change the chance of the other, the events are independent and the rule becomes $P(E\cap F)=P(E)\,P(F)$.

Two cards are dealt from a well-shuffled pack. What is the chance that both are aces? You could count pairs: $^{4}C_2=6$ pairs of aces out of $^{52}C_2=1326$ pairs, so $\tfrac{6}{1326}=\tfrac1{221}$. But there is a more natural way, the way a card player thinks: the first card is an ace with probability $\tfrac4{52}$; given that it was an ace, $51$ cards remain and $3$ of them are aces, so the second is an ace with probability $\tfrac3{51}$. Multiplying, $\tfrac4{52}\cdot\tfrac3{51}=\tfrac{12}{2652}=\tfrac1{221}$ — the same answer, found one stage at a time.

That stage-by-stage product is the multiplication theorem, and it is how almost every multi-step probability is computed: a batch of components passing several inspections, a patient going through two tests, a team winning three matches in a row. Sometimes the second stage does not care what happened in the first — a coin has no memory, and a card that is put back leaves the pack as it was. Then the conditional probability equals the ordinary one and the events are called independent. This section makes both ideas precise and shows you how to tell independence apart from a very different idea that students often confuse with it: events being mutually exclusive.

The Multiplication Theorem

Definition 13.1 says $P(F\mid E)=\dfrac{P(E\cap F)}{P(E)}$ when $P(E)\ne0$. Multiply both sides by $P(E)$ and you have a formula for the probability that both events happen.

📐 Theorem 13.2 — Multiplication theorem on probability

Let $E$, $F$, $G$ be events of a sample space $S$.

(a) If $P(E)\ne0$, then $P(E\cap F)=P(E)\,P(F\mid E)$. If $P(F)\ne0$, then also $P(E\cap F)=P(F)\,P(E\mid F)$.

(b) If $P(E\cap F)\ne0$, then $$P(E\cap F\cap G)=P(E)\,P(F\mid E)\,P(G\mid E\cap F).$$

Proof. (a) is Definition 13.1 multiplied out, once with $E$ and once with $F$ as the condition. For (b), note that $E\cap F\subset E$, so $P(E)\ge P(E\cap F)\gt0$ and all the conditional probabilities are defined. Apply (a) to the two events $E\cap F$ and $G$: $P\big((E\cap F)\cap G\big)=P(E\cap F)\,P(G\mid E\cap F)$, and then apply (a) again to $P(E\cap F)=P(E)\,P(F\mid E)$. $\square$

The same pattern continues for four or more events: multiply the probability of the first stage by the probability of each later stage given everything that happened before it.

🧭 Problem-solving strategy — tree diagrams for multi-stage experiments
  1. Draw one level of branches per stage. From each point, draw a branch for every possible result of the next stage.
  2. Label each branch with a conditional probability — the chance of that result given the path that leads to it. The branches leaving any point add up to $1$.
  3. Multiply along a path to get the probability of that whole sequence of results (Theorem 13.2).
  4. Add across paths when the event you want can happen along several paths: different paths are mutually exclusive.
  5. Check that the probabilities of all the complete paths add up to $1$.
✏️ Example 1 — Two aces, two ways
Two cards are drawn one after the other, without replacement, from a well-shuffled pack of $52$. Find the probability that (a) both are aces; (b) the first is a king and the second is a queen.
(a) Let $E$: "first card an ace", $F$: "second card an ace". $P(E)=\tfrac4{52}$ and $P(F\mid E)=\tfrac3{51}$. By Theorem 13.2, $P(E\cap F)=\tfrac4{52}\cdot\tfrac3{51}=\tfrac1{221}$.
Check By counting, as in the introduction: $\tfrac{^4C_2}{^{52}C_2}=\tfrac6{1326}=\tfrac1{221}$ ✓.
(b) $P(\text{king first})=\tfrac4{52}$; given that, all $4$ queens are among the $51$ cards left: $P(\text{queen second}\mid\text{king first})=\tfrac4{51}$. Product: $\tfrac{16}{2652}=\tfrac4{663}$.
✏️ Example 2 — Three draws from a bag
A bag contains $4$ red and $6$ black balls. Three balls are drawn one after another without replacement. Find the probability that (a) the first two are red and the third is black; (b) all three have the same colour.
(a) Theorem 13.2(b): $P(R_1)\,P(R_2\mid R_1)\,P(B_3\mid R_1\cap R_2)$ $=\tfrac4{10}\cdot\tfrac39\cdot\tfrac68$ $=\tfrac{72}{720}=\tfrac1{10}$.
(b) All red $\tfrac4{10}\cdot\tfrac39\cdot\tfrac28=\tfrac{24}{720}$.
(b) All black $\tfrac6{10}\cdot\tfrac59\cdot\tfrac48=\tfrac{120}{720}$.
(b) Add The two paths are mutually exclusive: $\tfrac{24+120}{720}=\tfrac{144}{720}=\tfrac15$.
Check By counting: $\tfrac{^4C_3+{}^6C_3}{^{10}C_3}=\tfrac{4+20}{120}=\tfrac15$ ✓.
✏️ Example 3 — A tree diagram
A box holds $3$ red and $2$ blue counters. Two counters are drawn one after the other without replacement. Draw the tree diagram and find the probability that (a) the second counter is red; (b) both counters have the same colour.
Tree First stage: R with $\tfrac35$, B with $\tfrac25$. After R: R with $\tfrac24$, B with $\tfrac24$. After B: R with $\tfrac34$, B with $\tfrac14$. Paths: $RR=\tfrac6{20}$, $RB=\tfrac6{20}$, $BR=\tfrac6{20}$, $BB=\tfrac2{20}$ (total $1$ ✓). Interactive 1 below draws this tree.
(a) Second red along $RR$ or $BR$: $\tfrac6{20}+\tfrac6{20}=\tfrac35$.
(b) $RR$ or $BB$: $\tfrac6{20}+\tfrac2{20}=\tfrac25$.
Notice $P(\text{second red})=\tfrac35$ is the same as $P(\text{first red})$. Before you look at the first counter, the second is just as likely to be red as the first — each position is equally likely to hold any counter. But given that the first was red, the chance falls to $\tfrac24$: the two draws are not independent.
Interactive 1 — Probability Tree Builder

A box holds red and blue counters; two are drawn in turn. Set the numbers of counters and choose whether the first counter is put back. Each branch carries a conditional probability; the lab multiplies along each path and adds across the paths that make up the chosen event (shown thick). Watch the second-stage branches: without replacement they depend on the first draw, with replacement they do not.

$P(R_1)$—
$P(R_2\mid R_1)$—
$P(R_2)$—
P(event)—

Independent Events

Throw a die and let $E$: "an even number" and $F$: "a multiple of $3$". Then $P(F)=\tfrac26=\tfrac13$. If you are told that $E$ has happened, the reduced sample space is $\{2,4,6\}$, and $F$ takes up one of its three outcomes: $P(F\mid E)=\tfrac13$ again. The information "even" has not changed the chance of "multiple of $3$" at all. In that case the multiplication theorem reads $P(E\cap F)=P(E)\,P(F)$, and this product rule is used as the definition, because it is symmetric in $E$ and $F$ and makes sense even when a probability is $0$.

📘 Definition 13.2 — Independent events
Two events $E$ and $F$ of the same sample space are independent if $$P(E\cap F)=P(E)\,P(F).$$ Otherwise they are dependent. When $P(E)\ne0$, independence is the same as $P(F\mid E)=P(F)$; when $P(F)\ne0$, it is the same as $P(E\mid F)=P(E)$.

The equivalence follows from Theorem 13.2: if $P(E)\ne0$, then $P(E\cap F)=P(E)P(F)$ exactly when $\dfrac{P(E\cap F)}{P(E)}=P(F)$, i.e. $P(F\mid E)=P(F)$. Independence is a property of the probabilities, not of whether the events "look connected": in the die example, "even" and "multiple of $3$" share the outcome $6$, yet they are independent.

✏️ Example 4 — Testing for independence
A fair die is thrown. Let $E=\{2,4,6\}$ (even), $F=\{3,6\}$ (multiple of $3$), $G=\{1,2,3,4\}$ (at most $4$) and $H=\{2,3,5\}$ (prime). Which of the pairs $E,F$; $E,G$; $E,H$ are independent?
$E,F$ $E\cap F=\{6\}$: $P(E\cap F)=\tfrac16$ and $P(E)P(F)=\tfrac12\cdot\tfrac13=\tfrac16$. Independent.
$E,G$ $E\cap G=\{2,4\}$: $P(E\cap G)=\tfrac13$ and $P(E)P(G)=\tfrac12\cdot\tfrac23=\tfrac13$. Independent.
$E,H$ $E\cap H=\{2\}$: $P(E\cap H)=\tfrac16$ but $P(E)P(H)=\tfrac12\cdot\tfrac12=\tfrac14$. Dependent: knowing the number is even lowers the chance of a prime from $\tfrac12$ to $\tfrac13$.
⚠️ Independent is not the same as mutually exclusive

Mutually exclusive ($E\cap F=\phi$) is a statement about the sets: the events cannot happen together. Independent ($P(E\cap F)=P(E)P(F)$) is a statement about the probabilities: one event carries no information about the other.

If $P(E)\gt0$ and $P(F)\gt0$, the two ideas can never hold together. Mutually exclusive events give $P(E\cap F)=0$, while $P(E)P(F)\gt0$. Indeed mutually exclusive events are strongly dependent: if you know that $E$ happened, you know for certain that $F$ did not.

Also, "independent trials" (a coin tossed again, a card put back and reshuffled) is a statement about how the experiment is run; it is what makes events about different trials independent.

For three events we need more than one product rule, because three events can be independent in pairs and still be linked as a trio (Example 5).

📘 Definition 13.3 — Mutually independent events
Three events $A$, $B$, $C$ are mutually independent if all four of the following hold: $$P(A\cap B)=P(A)P(B),\quad P(A\cap C)=P(A)P(C),$$ $$P(B\cap C)=P(B)P(C),\quad P(A\cap B\cap C)=P(A)P(B)P(C).$$ If only the first three hold, the events are pairwise independent.
✏️ Example 5 — Independent in pairs, but not all together
Two fair coins are tossed. Let $A$: "the first coin shows a head", $B$: "the second coin shows a head", $C$: "the two coins show different faces". Show that $A$, $B$, $C$ are pairwise independent but not mutually independent.
Events $S=\{HH,HT,TH,TT\}$; $A=\{HH,HT\}$, $B=\{HH,TH\}$, $C=\{HT,TH\}$, each with probability $\tfrac12$.
Pairs $A\cap B=\{HH\}$, $A\cap C=\{HT\}$, $B\cap C=\{TH\}$: each has probability $\tfrac14=\tfrac12\cdot\tfrac12$. Pairwise independent ✓.
Triple $A\cap B\cap C=\phi$, so $P(A\cap B\cap C)=0\ne\tfrac18$. Not mutually independent.
Meaning Any one event tells you nothing about any other one, but any two of them decide the third: if both coins show heads, the faces are certainly not different.

If knowing $E$ tells you nothing about $F$, it should tell you nothing about "not $F$" either. The next theorem confirms this, and gives the most useful formula in independence problems.

📐 Theorem 13.3 — Complements of independent events

If $E$ and $F$ are independent events, then so are (i) $E$ and $F'$; (ii) $E'$ and $F$; (iii) $E'$ and $F'$. Consequently $$P(\text{at least one of }E,F)=1-P(E')\,P(F').$$ The same holds for three mutually independent events: $P(A\cup B\cup C)=1-P(A')P(B')P(C')$.

Proof. (i) $E$ is the union of the mutually exclusive events $E\cap F$ and $E\cap F'$, so

$$P(E\cap F')=P(E)-P(E\cap F)=P(E)-P(E)P(F)=P(E)\,P(F').$$

(ii) is (i) with the roles of $E$ and $F$ exchanged. (iii) Apply (i) to the independent pair $E'$, $F$ from (ii). Finally, "at least one of $E$, $F$" is the complement of "neither", and $P(E'\cap F')=P(E')P(F')$ by (iii). (The three-event version is proved the same way; we use it without proof.) $\square$

✏️ Example 6 — Will the problem be solved?
Two students try a problem independently. The probability that Asha solves it is $\tfrac12$ and that Bilal solves it is $\tfrac13$. Find the probability that (a) the problem is solved; (b) exactly one of them solves it; (c) Bilal solves it, given that it is solved.
(a) "Solved" means at least one solves it. By Theorem 13.3: $1-P(A')P(B')=1-\tfrac12\cdot\tfrac23=1-\tfrac13=\tfrac23$.
(b) "Asha but not Bilal" or "Bilal but not Asha", which are mutually exclusive: $\tfrac12\cdot\tfrac23+\tfrac12\cdot\tfrac13=\tfrac13+\tfrac16=\tfrac12$.
(c) Bilal solving it is part of "solved": $P(B\mid\text{solved})=\dfrac{P(B)}{P(\text{solved})}=\dfrac{1/3}{2/3}=\dfrac12$.
Check Both solve it: $\tfrac16$. Exactly one plus both: $\tfrac12+\tfrac16=\tfrac23$, the answer to (a) ✓. A common error is $\tfrac12+\tfrac13=\tfrac56$, which counts "both" twice.
Interactive 2 — Independence Tester

Choose two events for a throw of two dice. On the left, the outcomes in $A$, in $B$ and in both are coloured. On the right, a square of area $1$ is cut into a strip of width $P(A)$ and a strip of width $P(A')$; the blue part of each strip has height $P(B\mid A)$ and $P(B\mid A')$. The events are independent exactly when the two heights match — the blue region is then a single rectangle of height $P(B)$, and its part inside $A$ has area $P(A)P(B)$. Press Play to tour some pairs.

$P(A)$—
$P(B)$—
$P(A\cap B)$—
$P(A)\,P(B)$—
⚠️ Common mistakes when studying alone

1. Multiplying without checking independence. $P(E\cap F)=P(E)P(F)$ only for independent events. For draws without replacement use $P(E)\,P(F\mid E)$.

2. Adding for "at least one". $P(E)+P(F)$ counts "both" twice. Use $1-P(E')P(F')$ for independent events, or the addition rule.

3. Calling mutually exclusive events independent. With positive probabilities they are never independent.

4. Checking only pairs. Three events need all four product rules of Definition 13.3.

5. Forgetting to update the counts. Without replacement, both the number of favourable objects and the total go down after each draw.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026 as an application of independent events and combinations (neither syllabus names the binomial distribution); not in the CBSE Class 12 syllabus. Suppose an experiment with two results, "success" (probability $p$) and "failure" (probability $q=1-p$), is repeated $n$ times, the repetitions being independent and $p$ the same each time. Such repetitions are often called Bernoulli trials. A particular sequence with $k$ successes and $n-k$ failures, such as $SSF\dots F$, has probability $p^kq^{n-k}$ by the product rule for independent events, and there are $^nC_k$ ways to choose the positions of the successes. These sequences are mutually exclusive, so

$$P(\text{exactly }k\text{ successes})={}^nC_k\,p^k\,q^{\,n-k}.$$

Worked example. A fair die is thrown $5$ times. Find the probability of exactly two sixes. Solution. $p=\tfrac16$, $q=\tfrac56$: ${}^5C_2\left(\tfrac16\right)^2\left(\tfrac56\right)^3=10\cdot\tfrac{125}{7776}=\tfrac{625}{3888}\approx0.161$. Check by listing: of the $6^5=7776$ equally likely sequences, the number with exactly two sixes is ${}^5C_2\cdot5^3=1250$, and $\tfrac{1250}{7776}=\tfrac{625}{3888}$ ✓.

Exercises

I
The multiplication theorem. Packs have $52$ cards and are well shuffled.
  • 1 $P(A)=0.4$ and $P(B\mid A)=0.3$. Find $P(A\cap B)$.
  • 2 Two cards are drawn without replacement. Find the probability that (a) both are kings; (b) both are hearts.
  • 3 A bag has $5$ white and $3$ black balls. Two are drawn one after the other without replacement. Find the probability that (a) both are black; (b) one is white and one is black.
  • 4 Repeat part 3(a) when the first ball is put back before the second draw.
  • 5 Three cards are drawn one after another without replacement. Find the probability that all three are spades.
  • 6 A box of $10$ bulbs contains $3$ defective ones. Two bulbs are taken out without replacement. Find the probability that neither is defective.
  • 7 $P(E)=0.5$, $P(F\mid E)=0.4$ and $P(G\mid E\cap F)=0.25$. Find $P(E\cap F\cap G)$.
  • 8 An urn has $3$ red and $2$ blue balls. A ball is drawn, its colour noted, and it is returned together with $2$ more balls of the same colour. A second ball is then drawn. Find $P(\text{both red})$ and $P(\text{second red})$.
1 $0.4\times0.3=0.12$.
2 (a) $\tfrac4{52}\cdot\tfrac3{51}=\tfrac1{221}$; (b) $\tfrac{13}{52}\cdot\tfrac{12}{51}=\tfrac1{17}$.
3 (a) $\tfrac38\cdot\tfrac27=\tfrac3{28}$; (b) $WB$ or $BW$: $\tfrac58\cdot\tfrac37+\tfrac38\cdot\tfrac57=\tfrac{30}{56}=\tfrac{15}{28}$.
4 With replacement the second draw is from the full bag again: $\tfrac38\cdot\tfrac38=\tfrac9{64}$.
5 $\tfrac{13}{52}\cdot\tfrac{12}{51}\cdot\tfrac{11}{50}=\tfrac{1716}{132600}=\tfrac{11}{850}$.
6 $\tfrac7{10}\cdot\tfrac69=\tfrac{42}{90}=\tfrac7{15}$.
7 Theorem 13.2(b): $0.5\times0.4\times0.25=0.05$.
8 After a red draw the urn has $5$ red, $2$ blue; after a blue draw, $3$ red, $4$ blue. $P(RR)=\tfrac35\cdot\tfrac57=\tfrac37$. $P(R_2)=\tfrac35\cdot\tfrac57+\tfrac25\cdot\tfrac37=\tfrac{15+6}{35}=\tfrac35$.
II
Independent or not?
  • 1 $P(A)=0.3$, $P(B)=0.6$, $P(A\cap B)=0.18$. Are $A$ and $B$ independent?
  • 2 $P(A)=0.5$, $P(B)=0.4$, $P(A\cup B)=0.7$. Are $A$ and $B$ independent?
  • 3 $P(A)=\tfrac14$, $P(B)=\tfrac13$, $P(A\cup B)=\tfrac5{12}$. Are $A$ and $B$ independent?
  • 4 A die is thrown; $E=\{1,3,5\}$, $F=\{1,2,3\}$. Are $E$ and $F$ independent?
  • 5 Two dice are thrown; $A$: "first die shows $4$", $B$: "sum is $7$", $C$: "sum is $6$". Is $A$ independent of $B$? Of $C$?
  • 6 One card is drawn. Are "the card is a spade" and "the card is an ace" independent?
  • 7 A coin is tossed three times. Are "the first toss is a head" and "exactly two heads" independent?
  • 8 $A$ and $B$ are independent with $P(A)=0.3$, $P(B)=0.4$. Find $P(A\cap B)$, $P(A\cup B)$, $P(A\mid B)$ and $P(A'\cap B')$.
  • 9 $A$ and $B$ are independent, $P(A)=0.2$ and $P(A\cup B)=0.6$. Find $P(B)$.
Check: are $A$ and $B$ independent?
—
1 $0.3\times0.6=0.18=P(A\cap B)$: independent.
2 $P(A\cap B)=0.5+0.4-0.7=0.2=0.5\times0.4$: independent.
3 $P(A\cap B)=\tfrac14+\tfrac13-\tfrac5{12}=\tfrac16$, but $P(A)P(B)=\tfrac1{12}$: not independent.
4 $E\cap F=\{1,3\}$: $\tfrac13\ne\tfrac12\cdot\tfrac12$: not independent.
5 $A\cap B=\{(4,3)\}$: $\tfrac1{36}=\tfrac16\cdot\tfrac16$, independent. $A\cap C=\{(4,2)\}$: $\tfrac1{36}$, but $P(A)P(C)=\tfrac16\cdot\tfrac5{36}=\tfrac5{216}$: dependent.
6 $P(\text{ace of spades})=\tfrac1{52}=\tfrac14\cdot\tfrac1{13}$: independent.
7 $P(A)=\tfrac12$, $P(B)=\tfrac38$, $A\cap B=\{HHT,HTH\}$ with probability $\tfrac14$, while $P(A)P(B)=\tfrac3{16}$: dependent.
8 $P(A\cap B)=0.12$; $P(A\cup B)=0.3+0.4-0.12=0.58$; $P(A\mid B)=P(A)=0.3$; $P(A'\cap B')=0.7\times0.6=0.42$ $(=1-0.58$ ✓$)$.
9 Let $P(B)=p$: $0.6=0.2+p-0.2p$, so $0.8p=0.4$ and $P(B)=0.5$.
III
Using independence. Where a probability is given for a person or device, assume independence as stated; the data are illustrative.
  • 1 Two students solve a problem independently with probabilities $0.6$ and $0.5$. Find the probability that the problem is solved, and that exactly one of them solves it.
  • 2 Three archers aim at a target independently; their probabilities of hitting it are $\tfrac12$, $\tfrac13$ and $\tfrac14$. Find the probability that (a) the target is hit; (b) exactly one arrow hits it.
  • 3 A water system has two pumps, each working on a given day with probability $0.9$, independently. Find the probability that the system works if (a) both pumps are needed (pumps in series); (b) one pump is enough (pumps in parallel).
  • 4 A die is thrown repeatedly. Find the probability that the first six appears on the third throw.
  • 5 Asha and Bilal toss a coin in turn, Asha first; whoever first gets a head wins. Find the probability that Asha wins.
  • 6 Prove part (iii) of Theorem 13.3 directly: if $E$, $F$ are independent, then $P(E'\cap F')=P(E')P(F')$. (Use $E'\cap F'=(E\cup F)'$.)
  • 7 Show that an event $E$ with $P(E)=0$ or $P(E)=1$ is independent of every event $F$.
1 Solved: $1-0.4\times0.5=0.8$. Exactly one: $0.6\times0.5+0.4\times0.5=0.5$.
2 (a) $1-\tfrac12\cdot\tfrac23\cdot\tfrac34=1-\tfrac14=\tfrac34$. (b) $\tfrac12\cdot\tfrac23\cdot\tfrac34+\tfrac12\cdot\tfrac13\cdot\tfrac34$ $+\tfrac12\cdot\tfrac23\cdot\tfrac14$ $=\tfrac{6+3+2}{24}=\tfrac{11}{24}$.
3 (a) $0.9\times0.9=0.81$; (b) $1-0.1\times0.1=0.99$. A spare pump in parallel cuts the failure probability from $0.19$ to $0.01$.
4 Not six, not six, six: $\tfrac56\cdot\tfrac56\cdot\tfrac16=\tfrac{25}{216}$.
5 Asha wins on toss $1$, $3$, $5$, …: $\tfrac12+\left(\tfrac12\right)^3+\left(\tfrac12\right)^5+\cdots$, a geometric series with first term $\tfrac12$ and ratio $\tfrac14$: $\dfrac{1/2}{1-1/4}=\dfrac23$.
6 $P(E'\cap F')=1-P(E\cup F)$ $=1-P(E)-P(F)+P(E)P(F)$ $=(1-P(E))(1-P(F))$ $=P(E')P(F')$.
7 If $P(E)=0$: $E\cap F\subset E$, so $P(E\cap F)=0=P(E)P(F)$. If $P(E)=1$: $P(E')=0$, so $E'$ is independent of $F'$ by the first case, hence $E$ is independent of $F$ by Theorem 13.3. (Directly: $P(E\cap F)=P(F)-P(E'\cap F)$ $=P(F)=P(E)P(F)$.)
IV
Mixed problems.
  • 1 A phone call drops with probability $0.05$, independently of other calls (illustrative). Find the probability that none of $3$ calls drops, and that at least one drops.
  • 2 A bag has $4$ red and $6$ black balls; two are drawn. Find $P(\text{first red and second black})$ (a) with replacement; (b) without replacement. In each case decide whether "first red" and "second black" are independent.
  • 3 A coin shows a head with probability $p$, where $0\lt p\lt1$, and is tossed twice. Show that "the first toss is a head" and "both tosses show the same face" are independent if and only if $p=\tfrac12$.
  • 4 A bowler takes a wicket in an over with probability $0.2$, independently of other overs (illustrative). Find the probability that the bowler takes at least one wicket in a spell of $4$ overs.
1 None: $0.95^3=0.857375$; at least one: $1-0.857375=0.142625\approx0.143$.
2 (a) $\tfrac4{10}\cdot\tfrac6{10}=0.24=P(R_1)P(B_2)$: independent. (b) $\tfrac4{10}\cdot\tfrac69=\tfrac4{15}$. Here $P(B_2)=\tfrac4{10}\cdot\tfrac69+\tfrac6{10}\cdot\tfrac59=\tfrac35$, and $P(R_1)P(B_2)=\tfrac25\cdot\tfrac35=\tfrac6{25}\ne\tfrac4{15}$: dependent.
3 $P(A)=p$, $P(B)=p^2+q^2$ with $q=1-p$, and $A\cap B=\{HH\}$ with probability $p^2$. Independence: $p^2=p(p^2+q^2)$, i.e. $p=p^2+(1-p)^2$ (as $p\ne0$), i.e. $2p^2-3p+1=0$, $(2p-1)(p-1)=0$. Since $p\ne1$, $p=\tfrac12$.
4 $1-0.8^4=1-0.4096=0.5904$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksJEE MainJEE Advanced

How this is examined: Board papers test the product rule for independent events and its complement form in 1-mark MCQs and Assertion–Reason items, ask for draws without replacement for 2 marks, and set "problem solved by at least one / exactly one" questions for 3 marks. JEE adds equations for unknown probabilities and games where players take turns.

E1. Board · 1 mark $A$ and $B$ are independent events with $P(A)=0.3$ and $P(B)=0.4$. Then $P(A'\cap B')$ is  (A) $0.42$  (B) $0.12$  (C) $0.58$  (D) $0.88$

Idea By Theorem 13.3, $A'$ and $B'$ are independent: $P(A'\cap B')=0.7\times0.6=0.42$.
Answer (A). Trap (B) is $P(A\cap B)$; (C) is $P(A\cup B)$; (D) is $1-P(A\cap B)$, the probability of "not both", which is not "neither".

E2. Board · 1 mark Assertion–Reason. Assertion (A): two mutually exclusive events $A$, $B$ with $P(A)\gt0$ and $P(B)\gt0$ cannot be independent. Reason (R): for mutually exclusive events, $P(A\cap B)=0$.  (A) Both A and R are true and R is the correct explanation of A  (B) Both A and R are true but R is not the correct explanation of A  (C) A is true but R is false  (D) A is false but R is true

Idea R is true since $A\cap B=\phi$. Then $P(A\cap B)=0\ne P(A)P(B)$, because the product of two positive numbers is positive — so R is exactly why A holds.
Answer (A).

E3. Board · 2 marks Three cards are drawn one after another, without replacement, from a well-shuffled pack of $52$. Find the probability that the first two are kings and the third is an ace.

Method Theorem 13.2(b): $P(K_1)=\tfrac4{52}$, $P(K_2\mid K_1)=\tfrac3{51}$, and after two kings all $4$ aces are among the $50$ cards left: $P(A_3\mid K_1\cap K_2)=\tfrac4{50}$. (1 mark)
Answer $\tfrac4{52}\cdot\tfrac3{51}\cdot\tfrac4{50}=\tfrac{48}{132600}=\tfrac2{5525}$. (1 mark)

E4. Board · 3 marks The probabilities that three students $P$, $Q$, $R$ can solve a problem are $\tfrac13$, $\tfrac14$ and $\tfrac15$, and they work independently. Find the probability that (a) the problem is solved; (b) exactly one of them solves it.

(a) None solves it: $\tfrac23\cdot\tfrac34\cdot\tfrac45=\tfrac{24}{60}=\tfrac25$, by independence of the complements. So $P(\text{solved})=1-\tfrac25=\tfrac35$. (1 mark)
(b) Three mutually exclusive cases: only $P$: $\tfrac13\cdot\tfrac34\cdot\tfrac45=\tfrac{12}{60}$; only $Q$: $\tfrac23\cdot\tfrac14\cdot\tfrac45=\tfrac8{60}$; only $R$: $\tfrac23\cdot\tfrac34\cdot\tfrac15=\tfrac6{60}$. (1½ marks)
Answer (a) $\tfrac35$; (b) $\tfrac{26}{60}=\tfrac{13}{30}$. (½ mark)

E5. JEE Main $A$ and $B$ are independent events with $P(A\cap B)=\tfrac3{20}$ and $P(A'\cap B')=\tfrac3{10}$. If $P(A)\gt P(B)$, then $P(A)$ is  (A) $\tfrac35$  (B) $\tfrac14$  (C) $\tfrac34$  (D) $\tfrac25$

Method Let $a=P(A)$, $b=P(B)$. Then $ab=\tfrac3{20}$ and $(1-a)(1-b)=\tfrac3{10}$, i.e. $1-(a+b)+ab=\tfrac3{10}$, so $a+b=1+\tfrac3{20}-\tfrac3{10}=\tfrac{17}{20}$.
Solve $a,b$ are the roots of $t^2-\tfrac{17}{20}t+\tfrac3{20}=0$, i.e. $20t^2-17t+3=0$, $(5t-3)(4t-1)=0$: $t=\tfrac35$ or $\tfrac14$. With $a\gt b$, $a=\tfrac35$.
Answer (A). Trap (B) is $P(B)$; (D) $=P(A')$ and (C) $=P(B')$, which come from solving for the complements by mistake.

E6. JEE Advanced (Numerical answer.) Players $A$ and $B$ throw a pair of fair dice in turn, $A$ first. $A$ wins if he throws a sum of $9$ before $B$ throws a sum of $10$ or more; $B$ wins if he does so first. If the probability that $A$ wins is $p$, find $7p$.

Single throws Sum $9$: $(3,6),(4,5),(5,4),(6,3)$, so $a=\tfrac4{36}=\tfrac19$. Sum $\ge10$: $6$ outcomes, so $b=\tfrac16$.
Rounds A round in which neither succeeds has probability $(1-a)(1-b)=\tfrac89\cdot\tfrac56=\tfrac{20}{27}$, by independence of the throws. $A$ wins on his first throw, or after one empty round, or after two, …: $p=a\left(1+\tfrac{20}{27}+\left(\tfrac{20}{27}\right)^2+\cdots\right)$ $=\dfrac{1/9}{1-20/27}=\dfrac{1/9}{7/27}=\dfrac37$.
Answer $7p=\mathbf 3$.
⏱ Fast route In each round, $A$ wins with weight $a$ and $B$ with weight $(1-a)b$; the rounds just repeat, so $p=\dfrac{a}{a+(1-a)b}$ $=\dfrac{1/9}{1/9+\tfrac89\cdot\tfrac16}=\dfrac37$.
🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. Can people fake randomness? If you write down a "random" list of $100$ heads and tails from your head, can a classmate tell it apart from $100$ real coin tosses — using only the idea of independence?

Why it matters. Real coin tosses are independent: the coin has no memory, so after five heads in a row the next toss is still a head with probability $\tfrac12$. People who invent sequences usually avoid long runs because they "feel" non-random. The same intuition misleads gamblers (the "gambler's fallacy"), sports fans who believe in streaks, and even auditors — while statisticians use exactly this weakness to detect invented data in surveys and accounts.

What to do.

  1. Each student secretly makes two lists of $100$ H/T: one invented, one from real tosses of a coin (or a phone's random number generator). Label them X and Y in random order.
  2. For each list record (a) the longest run of identical results; (b) the number of times the result changes from one toss to the next.
  3. Using independence, find the probability that a particular block of $6$ consecutive real tosses is all the same ($2\cdot\left(\tfrac12\right)^6=\tfrac1{32}$), and the expected number of changes in $100$ real tosses ($99\times\tfrac12\approx50$).
  4. Swap lists with a partner and try to identify the real one. Collect the class's success rate.
  5. Pool all the real lists: what fraction have a run of $6$ or more? Compare with the invented lists.

Tools: A coin or a random-number app; a spreadsheet to count runs and changes.

Hand in: Your two lists with the counts, the class results in a table, and a paragraph explaining what independence predicts and how the invented lists differed.

Going further. Look up the "hot hand" debate in basketball. Why is it hard to decide from data whether successive shots are independent?

Instructor note: in $100$ fair independent tosses the longest run is usually $5$ to $8$; a run of length at least $6$ occurs with probability about $0.8$ (a quick simulation confirms this), while invented lists rarely contain one, and they usually show far more than $50$ changes. The detection rate in class is typically well above $\tfrac12$.

🧪 Self-check quiz

0/5 correct

Q1. Two cards are drawn without replacement from a pack of $52$. The probability that both are red is

Q2. $E$ and $F$ are independent with $P(E)=0.4$ and $P(F)=0.5$. Then $P(E\cup F)$ is

Q3. Two events with positive probabilities are mutually exclusive. Then they are

Q4. A die is thrown; $E=\{2,4,6\}$ and $F=\{1,2,3,4\}$. The events $E$ and $F$ are

Q5. Independent events have probabilities $0.2$ and $0.3$. The probability that at least one of them happens is

← 13.1 Conditional Probability13.3 The Theorem of Total Probability and Bayes' Theorem →