JEE Mains Maths · Probability

Addition, Conditional Probability and Independence

The rules that combine probabilities: the addition rule for unions, the conditional-probability formula, independence as a product, and repeated independent trials until a success.

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Why this matters

Twenty-five PYQs, twenty of them multiple choice. Ten work with the rules themselves — unions, complements, conditional probabilities — eight with independent events, and seven with trials repeated until a success, which sum a geometric series. Three ideas cover the page.

Concept 1 of 3: Addition rule and conditional probability

Most of these questions give two or three of

P(A)

P(B)

P(A\cap B)

P(A\mid B)

and ask for another. Write everything in terms of

P(A)

P(B)

and

P(A\cap B)

P(A\cup B)=P(A)+P(B)-P(A\cap B)

and

P(A\mid B)=\frac{P(A\cap B)}{P(B)}

. Complements follow:

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

P(A'\cap B')=1-P(A\cup B)

Definition

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$ .
$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ .
$P(A\cap B')=P(A)-P(A\cap B)$ ; $P(A'\cap B')=1-P(A\cup B)$ .
Probabilities lie in $[0,1]$ , and mutually exclusive events' probabilities add to at most 1.
Geometric probability: a ratio of lengths or areas.

Conditional probability

P(A\mid B)=\frac{P(A\cap B)}{P(B)}

Worked example

P(A)=0.5

P(B)=0.4

P(A\cup B)=0.7

. Find

P(A\mid B)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 25 July 2022 · Q158Moderate

Example 1 · Probability · Addition, Conditional Probability and Independence

A

and

B

are two events such that

P(A) =\frac{1}{3},P(B) =\frac{1}{5}

and

\ P(A \cup B) =\frac{1}{2}

, then

P\left( A \mid B^{'} \right)+ P\left( B \mid A^{'} \right)

is equal to

Condition on the complement

P(A\mid B')=\frac{P(A)-P(A\cap B)}{1-P(B)}

. Dividing by

P(B)

out of habit, or forgetting to remove

A\cap B

, are the two usual slips.

Drill 9 more on addition rule and conditional probability

Concept 2 of 3: Independent events

Independent means

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

, and then every combination of the events and their complements also multiplies. 'Exactly one of

A,B

' is

P(A)P(B')+P(A')P(B)

. With three independent events, 'only

E_1

' is

P(E_1)P(E_2')P(E_3')

, and dividing by 'none' leaves

\frac{P(E_1)}{1-P(E_1)}

Definition

Independent: $P(A\cap B)=P(A)P(B)$ ; so are $A,B'$ , $A',B$ , $A',B'$ .
Exactly one: $P(A)(1-P(B))+(1-P(A))P(B)$ .
At least one of $n$ independent events: $1-\prod(1-P(E_i))$ .
Independent is not the same as mutually exclusive.

Independence

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

Worked example

Independent events have

P(A)=0.3

P(B)=0.4

. Find

P(

exactly one

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 18 · Q65Moderate

Example 2 · Probability · Addition, Conditional Probability and Independence

Let

A

and

B

be independent events such that

P(A) = p,P(B) = 2p

. The largest value of

p

, for which

P

(exactly one of

A,B

occurs)

=\frac{5}{9}

, is:

Take the root that is a probability

Equations from 'exactly one' are quadratic in

p

. Both roots may lie in

(0,1)

, but each event's probability must too: with

P(B)=2p

, need

p\le\frac12

Drill 7 more on independent events

Concept 3 of 3: Trials repeated until a success

If each trial succeeds with probability

p

, the first success is on trial

k

with probability

q^{k-1}p

. Questions about an even-numbered trial, or about two players throwing in turn, add a geometric series: player A (first) wins with

\frac{p_A}{1-q_Aq_B}

. 'At least

k

trials' is

q^{k-1}

, which makes conditional questions short.

Definition

$P(X=k)=q^{k-1}p$ ; $P(X\ge k)=q^{k-1}$ .
$P(X>m+n\mid X>n)=P(X>m)=q^{m}$ (memoryless).
First success on an even trial: $\frac{qp}{1-q^2}=\frac{q}{1+q}$ .
A and B alternate, A first: $P(A\text{ wins})=\frac{p_A}{1-q_Aq_B}$ .

Alternate turns

P(A\text{ wins})=p_A+q_Aq_Bp_A+(q_Aq_B)^2p_A+\cdots=\frac{p_A}{1-q_Aq_B}

Worked example

A coin is tossed until a head appears. Find the probability that it takes an odd number of tosses.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q64Moderate

Example 3 · Probability · Addition, Conditional Probability and Independence

A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is

Who throws first

The player who throws first has the extra head start: their series starts at

p_A

, the other's at

q_Ap_B

. Swapping them gives the other player's chance.

Drill 6 more on trials repeated until a success

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 (3)

Addition rule and conditional probability
Conditional probability
$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$
Independent events
Independence
$P(A\cap B)=P(A)\,P(B)$
Trials repeated until a success
Alternate turns
$P(A\text{ wins})=p_A+q_Aq_Bp_A+(q_Aq_B)^2p_A+\cdots=\frac{p_A}{1-q_Aq_B}$

Watch out for (3)

Condition on the complement
→ Addition rule and conditional probability
Take the root that is a probability
→ Independent events
Who throws first
→ Trials repeated until a success

Test yourself on Probability

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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