NDA Maths · Probability

Independent Events & the Multiplication Rule

When one event tells you nothing about another, probabilities multiply — the engine behind 'all of', 'at least one', and problem-solving questions.

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Why this matters

Independence is the multiplication counterpart of the addition rule: when events do not influence each other, the probability that all of them happen is the product of the individual probabilities. This 16-question subtopic is dominated by two archetypes — 'a problem is given to several students' and 'several independent trials' — both solved with the multiplication rule and the 'at least one = 1 - none' complement. The recurring exam trap is confusing independent with mutually exclusive, so that distinction is drilled here.

Concept 1 of 4

Independence and the multiplication rule

Intuition

Two events are independent when knowing one happened does not change the probability of the other — separate coins, separate machines, separate people. For independent events, the probability that BOTH happen is the product of their probabilities.

Definition

$A$ and $B$ are independent if $P(A \cap B) = P(A)\,P(B)$ . This extends to any number: $P(A_1 \cap \dots \cap A_n) = P(A_1)\cdots P(A_n)$ . If $A, B$ are independent then so are $A, B'$ (and $A', B'$ ) — independence carries over to complements.

Multiplication rule (independent events)

P(A \cap B) = P(A)\,P(B)

$P(A \cap B)$ probability both occur — a product, only when independent

Diagram · mutually exclusive ≠ independent

Mutually exclusive events can't both happen, so they don't overlap and P(A∩B) = 0. Independent events do overlap — one happening doesn't change the other, so P(A∩B) = P(A)·P(B). Disjoint events with non-zero probability are therefore never independent.

Worked example

A switch is closed with probability

0.8

. Two such switches operate independently. What is the probability that both are closed?

Independent, so multiply: $P(\text{both closed}) = 0.8 \times 0.8$ .
$= 0.64$ .

Answer:

0.64

Practice this conceptself-check · 4 quick reps

Try it yourself

The probability that a seed germinates is

0.7

. Two seeds are sown independently. What is the probability that both germinate?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Independent $P(A)=0.5, P(B)=0.6$ . $P(A\cap B)$ ?
2.
Fair coin twice: $P(\text{two heads})$ ?
3.
Independent $P(A)=\tfrac{1}{3},P(B)=\tfrac{1}{4}$ . $P(A\cap B)$ ?
4.
If $A,B$ independent and $P(A)=0.4$ , is $A$ independent of $B'$ ?

From the bank · past-year question

Example 1ProbabilityEASY

In a sample survey of a village, the probability that a farmer is in debt is 0.60. What is the probability that three randomly selected farmers are all in debt (assume independence of events)?

[Q102 · Sep · 2025]

Independent $\ne$ mutually exclusive

Independent means

P(A \cap B) = P(A)P(B)

(both can happen, just unrelated); mutually exclusive means

P(A \cap B) = 0

(both cannot happen). They are opposite conditions for positive-probability events.

Only multiply when independence is given or physically clear

Drawing without replacement, or events from the same experiment, are usually NOT independent — use conditional probability there. Multiply only when the trials genuinely don't influence each other.

Drill 6 more on independence and the multiplication rule

Concept 2 of 4

"At least one" via the complement

Intuition

For independent trials, "at least one success" is messy to count directly but easy through the complement: the only way to get no successes is for every trial to fail.

Definition

For independent events $A_1, \dots, A_n$ , $P(\text{at least one occurs}) = 1 - P(\text{none occurs}) = 1 - \prod_{i}\big(1 - P(A_i)\big)$ . When all $n$ probabilities equal $p$ , this is $1 - (1-p)^n$ .

At least one (independent trials)

P(\text{at least one}) = 1 - \prod_{i}\big(1 - P(A_i)\big)

$1 - P(A_i)$ probability trial $i$ fails
$\prod$ product over all trials — probability all fail

Worked example

Two independent alarms fire (when needed) with probabilities

0.9

and

0.8

. What is the probability that at least one fires?

Complement: both fail with probability $(1 - 0.9)(1 - 0.8) = 0.1 \times 0.2 = 0.02$ .
At least one fires: $1 - 0.02 = 0.98$ .

Answer:

0.98

Practice this conceptself-check · 4 quick reps

Try it yourself

A marksman hits a target with probability

\dfrac{1}{3}

on each shot. He fires two independent shots. What is the probability he hits at least once?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Independent $P(A)=0.4,P(B)=0.5$ . $P(\text{at least one})$ ?
2.
Fair coin 3 times: $P(\text{at least one head})$ ?
3.
$p=\tfrac{1}{2}$ each, 2 trials: $P(\text{at least one})$ ?
4.
Two trials fail with probs $0.2, 0.5$ . $P(\text{at least one succeeds})$ ?

From the bank · past-year question

Example 2ProbabilityEASY

P(solves A)=0.4, P(solves B)=0.5. P(solves at least one)?

[Q104 · Apr · 2018]

"At least one" = 1 - (all fail), not the sum of individual probabilities

Adding

P(A) + P(B)

double-counts the both-succeed case and can exceed 1. Always go through the complement of "none".

Multiply the FAILURE probabilities, not the success ones, for "none"

"None occurs" means every trial fails, so multiply

(1-p_i)

. A common slip is multiplying the

p_i

(that is "all succeed").

Drill 1 more on "at least one" via the complement

Concept 3 of 4

The "problem solved by students" archetype

Intuition

A whole family of PYQs gives several people who each solve a problem (or hit a target) independently with their own probability, and asks for the chance the problem is solved at all. "Solved by at least one" is again

1 - P(\text{nobody solves it})

Definition

If solvers have independent success probabilities $p_1, \dots, p_n$ , then $P(\text{problem solved}) = 1 - (1-p_1)(1-p_2)\cdots(1-p_n)$ . Variants ask for "solved by exactly one" (sum of one-succeeds-rest-fail products) or "by both/all" (product of the $p_i$ ).

Problem solved by at least one solver

P(\text{solved}) = 1 - \prod_{i}(1 - p_i)

$p_i$ probability solver $i$ solves it, independently

Worked example

Two students

A

and

B

can solve a problem independently with probabilities

\dfrac{1}{2}

and

\dfrac{1}{3}

. What is the probability the problem is solved?

Neither solves it: $\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{3}\right) = \dfrac{1}{2} \cdot \dfrac{2}{3} = \dfrac{1}{3}$ .
Solved by at least one: $1 - \dfrac{1}{3} = \dfrac{2}{3}$ .

Answer:

\dfrac{2}{3}

Practice this conceptself-check · 4 quick reps

Try it yourself

Three students solve a problem independently with probabilities

\dfrac{2}{3}, \dfrac{3}{4}, \dfrac{4}{5}

. What is the probability that at least one solves it?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Solvers $\tfrac{1}{2}, \tfrac{1}{2}$ . $P(\text{solved})$ ?
2.
Solvers $\tfrac{1}{2}, \tfrac{1}{3}$ . $P(\text{both solve})$ ?
3.
Solvers $\tfrac{1}{3}, \tfrac{1}{4}$ . $P(\text{none solves})$ ?
4.
Solvers $\tfrac{2}{3}, \tfrac{3}{4}$ . $P(\text{solved})$ ?

From the bank · past-year question

Example 3ProbabilityEASY

A question is given to three students

A

B

and

C

whose chances of solving it are

\dfrac{1}{2}

\dfrac{1}{3}

and

\dfrac{1}{4}

respectively. What is the probability that the question will be solved?

[Q104 · Apr · 2017]

"Problem solved" means at least one solver — use the complement

Do not add the solvers' probabilities (that overcounts and can exceed 1). Compute

1 - \prod(1-p_i)

"Exactly one solves" is a different computation

For exactly one, sum the cases where one succeeds and the rest fail:

p_1(1-p_2) + (1-p_1)p_2

for two solvers. Don't confuse it with "at least one".

Drill 2 more on the "problem solved by students" archetype

Concept 4 of 4

Finding an unknown probability using independence

Intuition

Some questions give the union (or a complement) plus independence and ask you to back out a missing probability. Combine the addition rule with

P(A \cap B) = P(A)P(B)

and solve the resulting equation.

Definition

For independent $A, B$ : $P(A \cup B) = P(A) + P(B) - P(A)P(B)$ . Given any three of $\{P(A), P(B), P(A\cup B)\}$ you can solve for the fourth. Equivalently, since complements of independent events are independent, $P(A \cup B) = 1 - P(A')P(B')$ .

Union of independent events

P(A \cup B) = P(A) + P(B) - P(A)P(B) = 1 - P(A')\,P(B')

$P(A')P(B')$ probability neither occurs (independent complements)

Worked example

A

and

B

are independent with

P(A) = 0.5

and

P(A \cup B) = 0.7

. Find

P(B)

Independent union: $P(A \cup B) = P(A) + P(B) - P(A)P(B)$ .
Substitute: $0.7 = 0.5 + P(B) - 0.5\,P(B) = 0.5 + 0.5\,P(B)$ .
Solve: $0.5\,P(B) = 0.2 \Rightarrow P(B) = 0.4$ .

Answer:

0.4

Practice this conceptself-check · 4 quick reps

Try it yourself

A

and

B

are independent with

P(A) = \dfrac{1}{3}

and

P(B) = \dfrac{1}{4}

. Find

P(A \cup B)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Independent $P(A)=0.6,P(B)=0.5$ . $P(A\cup B)$ ?
2.
Independent, $P(A')=0.7,P(B')=0.4$ . $P(A\cup B)$ ?
3.
Independent $P(A)=0.5,P(A\cup B)=0.75$ . $P(B)$ ?
4.
Independent $P(A)=\tfrac{1}{2},P(B)=\tfrac{1}{2}$ . $P(A\cup B)$ ?

From the bank · past-year question

Example 4ProbabilityMODERATE

Let

A

and

B

be two independent events such that

P(\bar{A})=0.7

P(\bar{B})=k

P(A\cup B)=0.8

. What is the value of

k

[Q102 · Apr · 2023]

For independent events the union is NOT $P(A) + P(B)$

You must subtract

P(A)P(B)

for the overlap.

P(A) + P(B)

(no subtraction) is the mutually-exclusive formula — the wrong model for independent events.

Use $1 - P(A')P(B')$ as a fast check

Computing the union as

1

minus "neither" often avoids a sign slip; both routes must agree.

Drill 3 more on finding an unknown probability using independence

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Independence and the multiplication rule
Multiplication rule (independent events)
$P(A \cap B) = P(A)\,P(B)$
"At least one" via the complement
At least one (independent trials)
$P(\text{at least one}) = 1 - \prod_{i}\big(1 - P(A_i)\big)$
The "problem solved by students" archetype
Problem solved by at least one solver
$P(\text{solved}) = 1 - \prod_{i}(1 - p_i)$
Finding an unknown probability using independence
Union of independent events
$P(A \cup B) = P(A) + P(B) - P(A)P(B) = 1 - P(A')\,P(B')$

Watch out for (8)

Independent $\ne$ mutually exclusive
→ Independence and the multiplication rule
Only multiply when independence is given or physically clear
→ Independence and the multiplication rule
"At least one" = 1 - (all fail), not the sum of individual probabilities
→ "At least one" via the complement
Multiply the FAILURE probabilities, not the success ones, for "none"
→ "At least one" via the complement
"Problem solved" means at least one solver — use the complement
→ The "problem solved by students" archetype
"Exactly one solves" is a different computation
→ The "problem solved by students" archetype
For independent events the union is NOT $P(A) + P(B)$
→ Finding an unknown probability using independence
Use $1 - P(A')P(B')$ as a fast check
→ Finding an unknown probability using independence

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1ProbabilityMODERATE

A husband and wife appear in an interview for two vacancies for the same post. The probability of the husband's selection is

\frac{1}{7}

and that of the wife's selection is

\frac{1}{5}

. If the events are independent, then the probability of which one of the following is

\frac{11}{35}

[Q113 · Apr · 2020]

Example 2ProbabilityEASY

A coin is tossed twice. If

E

and

F

denote occurrence of head on first toss and second toss respectively, then what is

P(E\cup F)

equal to?

[Q117 · Apr · 2021]

Example 3ProbabilityMODERATE

A problem is given to three students A, B and C whose probabilities of solving the problem are

\dfrac{1}{2}

\dfrac{3}{4}

and

\dfrac{1}{4}

respectively. What is the probability that the problem will be solved if they all solve the problem independently?

[Q103 · Apr · 2019]

Example 4ProbabilityHARD

Three events A, B and C are such that A and B are disjoint, A and C are independent, B and C are independent. If

4P(A)=2P(B)=P(C)

and

P(A\cup B\cup C)=5P(A)

, then what is the value of P(C)?

[Q118 · Apr · 2026]

Example 5ProbabilityHARD

If A, B, C are three events, then what is the probability that at least two of these events occur together?

[Q106 · Apr · 2019]

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