Conditional Probability

Q: What is the difference between P(A and B) and P(A given B)?

P(A and B) is the chance that both events happen, measured against the whole sample space. P(A given B) measures A only inside B, so it divides by P(B): P(A given B) = P(A and B)/P(B). It is never smaller than P(A and B).

Q: How do you check whether two events are independent?

Compute P(A and B) directly from the sample space and compare it with P(A) times P(B). If they are equal, the events are independent; if not, they are dependent. Equivalently, check whether P(A given B) = P(A). Physically separate experiments, such as two different dice, are independent.

Q: What is the difference between pairwise and mutual independence?

Pairwise independence checks only pairs: P(A and B) = P(A)P(B) and so on. Mutual independence also needs every larger group to satisfy the product rule, such as P(A and B and C) = P(A)P(B)P(C). Two fair coins with exactly one head as the third event give the standard pairwise-only example.

MathsProbabilityFor JEE aspirants

Conditional probability $P (A ∣ B)$ is the probability that event $A$ occurs when event $B$ is known to have occurred. It is found from $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ , provided $P (B) > 0$ , because knowing $B$ happened shrinks the sample space to $B$ . This conditional probability page covers the multiplication theorem, independent events, pairwise versus mutual independence and why mutually exclusive events are dependent, with fully solved JEE Main and JEE Advanced problems.

On this page1Conditional probability2Multiplication theorem3Independent events4Three events5Strategy6Solved examples

Key Formulas - Quick Reference

★ Must learn $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ , $P (B) > 0$ ; for equally likely outcomes $P (A ∣ B) = \frac{n ( A \cap B )}{n ( B )}$
★ Must learnMultiplication theorem: $P (A \cap B) = P (A) P (B ∣ A) = P (B) P (A ∣ B)$
Three events: $P (A \cap B \cap C) = P (A) P (B ∣ A) P (C ∣ A \cap B)$
$P (S ∣ B) = P (B ∣ B) = 1$ and $P (A^{'} ∣ B) = 1 - P (A ∣ B)$
★ Must learnIndependent events: $P (A \cap B) = P (A) P (B)$ , equivalently $P (A ∣ B) = P (A)$
If $A$ , $B$ are independent, so are $A$ and $B^{'}$ , $A^{'}$ and $B$ , $A^{'}$ and $B^{'}$
★ Must learnIndependent $A_{1}, \dots, A_{n}$ : P(at least one) $= 1 - P (A_{1}^{'}) P (A_{2}^{'}) \dots P (A_{n}^{'})$
Independent, $P (A) = p_{1}$ , $P (B) = p_{2}$ : P(exactly one) $= p_{1} + p_{2} - 2 p_{1} p_{2}$
Mutually independent $A, B, C$ : all three pair products hold and $P (A \cap B \cap C) = P (A) P (B) P (C)$
★ Must learnReplay until decided: P(win eventually) $= \frac{w}{1 - t}$ , $w$ = win and $t$ = tie probability per round

1. Conditional Probability

The probability of event $A$ when it is known that event $B$ has occurred is called the conditional probability of $A$ given $B$ , written $P (A ∣ B)$ and read "probability of $A$ given $B$ ". Many books write it as $P (A / B)$ .

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (B) > 0

Why this formula? Once $B$ is known to have occurred, only the sample points of $B$ remain possible, so $B$ becomes the new sample space. Among these, $A$ happens only at points of $A \cap B$ . For equally likely outcomes:

P (A ∣ B) = \frac{n ( A \cap B )}{n ( B )} = \frac{n ( A \cap B ) / n ( S )}{n ( B ) / n ( S )} = \frac{P ( A \cap B )}{P ( B )}

Figure 1: Once

B

is known to have happened,

B

is the new sample space and only

A \cap B

still counts for

A

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

Quick illustration. A fair die is thrown. Let $B$ = "a 4 turns up" and $A$ = "a number greater than 3 turns up". Then $A = {4, 5, 6}$ and $B = {4}$ . Given $A$ , only 4, 5, 6 are possible, so $P (B ∣ A) = \frac{1}{3}$ . By formula: $P (A \cap B) = \frac{1}{6}$ , $P (A) = \frac{1}{2}$ , and $P (B ∣ A) = \frac{1/6}{1/2} = \frac{1}{3}$ .

1.1 Reading a two-way table

In a two-way table the condition picks a row or a column, and its total becomes the denominator (Figure 2).

Figure 2: The condition picks the row or column; its total is the denominator.

P (M ∣ E) = \frac{23}{30}

but

P (E ∣ M) = \frac{23}{25}

Exam Trick

With a two-way table, read the conditional probability straight off the table: numerator from the cell, denominator from the condition's total. No need to divide two probabilities.

1.2 Properties of conditional probability

$P (S ∣ B) = P (B ∣ B) = 1$ .
$P (A^{'} ∣ B) = 1 - P (A ∣ B)$ . The condition stays fixed; only the event is complemented.
$P (A \cup C ∣ B) = P (A ∣ B) + P (C ∣ B) - P (A \cap C ∣ B)$ .
If $A$ and $B$ are mutually exclusive, $P (A ∣ B) = 0$ .
$P (A ∣ B) \geq P (A \cap B)$ , since dividing by $P (B) \leq 1$ can only increase the value.

P (A ∣ B)

B

is known

Denominator $P (B)$ .

TV show: P(husband watches $∣$ wife watches) $= 0.7$ .

P (B ∣ A)

A

is known

Denominator $P (A)$ .

Same couple: P(wife watches $∣$ husband watches) $= \frac{0.35}{0.4} = \frac{7}{8}$ .

Key idea

Conditioning shrinks the sample space: the event after the bar is known, and its probability goes in the denominator.

Quick Recall: tap to check

P (A \cap B) = 0.2

and

P (B) = 0.5

, find

P (A ∣ B)

and

P (A^{'} ∣ B)

P (A ∣ B) = \frac{0.2}{0.5} = 0.4

;

P (A^{'} ∣ B) = 1 - 0.4 = 0.6

A die shows a number greater than 3. What is the chance it is even?

Given {4, 5, 6}, the even ones are {4, 6}:

\frac{2}{3}

A

and

B

are mutually exclusive with

P (B) > 0

, what is

P (A ∣ B)

0

: once

B

happens,

A

cannot.

2. Multiplication Theorem of Probability

Rearranging the definition gives the probability that two events occur together:

P (A \cap B) = P (A) P (B ∣ A), P (A) > 0

P (A \cap B) = P (B) P (A ∣ B), P (B) > 0

Whether the second event depends on the first decides which factor to use. Drawing with replacement keeps the second-draw probabilities fixed; drawing without replacement changes them (Figure 3).

Figure 3: Replacing the ball keeps the second-draw probabilities fixed (independent draws). Not replacing changes them to

\frac{5}{9}

\frac{6}{9}

, so the multiplication theorem needs

P (B ∣ A)

A probability tree is the multiplication theorem drawn out: multiply along a path, then add the paths that make up your event (Figure 4).

Figure 4: Multiply along each path. The second-draw probabilities change with the first result, yet

P (red on 2nd) = \frac{30 + 24}{90} = \frac{3}{5}

, the same as for the 1st draw.

2.1 Three or more events: the chain rule

For more events, keep conditioning on everything that has already happened:

P (A_{1} \cap A_{2} \cap \dots \cap A_{n}) = P (A_{1}) P (A_{2} ∣ A_{1}) P (A_{3} ∣ A_{1} \cap A_{2}) \dots P (A_{n} ∣ A_{1} \cap \dots \cap A_{n - 1})

Figure 5:

P (A_{1} \cap A_{2} \cap A_{3}) = P (A_{1}) P (A_{2} ∣ A_{1}) P (A_{3} ∣ A_{1} \cap A_{2}) = \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} = \frac{1}{5525}

Key idea

Multiply along the path, conditioning each step on what has already happened; add the paths that make up the event.

3. Independent Events

Two events $A$ and $B$ are independent if the occurrence or non-occurrence of one does not affect the probability of the other:

P (B ∣ A) = P (B), P (A) \neq = 0 and P (A ∣ B) = P (A), P (B) \neq = 0

Since $P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )}$ , this is equivalent to $P (A \cap B) = P (A) P (B)$ . Events that are not independent are called dependent.

Figure 6: Independence means the overlap has area exactly

P (A) P (B)

. Mutually exclusive events have no overlap at all, so they are dependent whenever

P (A), P (B) > 0

3.1 Remarks on independence

An impossible event is independent of every event. If $P (A) = 0$ , then $0 \leq P (A \cap B) \leq P (A) = 0$ , so $P (A \cap B) = 0 = P (A) P (B)$ .
Independent is not mutually exclusive. Independence is a property of probabilities; mutual exclusion is a set property. If $A$ , $B$ are mutually exclusive with $P (A) > 0$ , $P (B) > 0$ , then $P (A \cap B) = 0 \neq = P (A) P (B)$ : such events are strongly dependent.
Complements keep independence. $A$ , $B$ independent $⟺$ $A$ , $B^{'}$ independent $⟺$ $A^{'}$ , $B$ independent $⟺$ $A^{'}$ , $B^{'}$ independent.

Proof of the complement results. Given $P (A \cap B) = P (A) P (B)$ :

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

P (A^{'} \cap B^{'}) = 1 - P (A \cup B) = 1 - P (A) - P (B) + P (A) P (B)

= {1 - P (A)} {1 - P (B)} = P (A^{'}) P (B^{'})

3.2 At least one of several independent events

"At least one" is the complement of "none". For independent events the complements multiply, so $P (A_{1} \cup \dots \cup A_{n}) = 1 - P (A_{1}^{'}) P (A_{2}^{'}) \dots P (A_{n}^{'})$ . Switches in parallel are the classic picture (Figure 7).

Figure 7: Series needs "and" (multiply), parallel needs "or" (use the complement):

0.72

versus

0.98

Key idea

Independent is about probabilities, mutually exclusive is about sets: with positive probabilities, two events can never be both.

Quick Recall: tap to check

P (A) = 0.3

P (B) = 0.5

and

A

B

are independent. Find

P (A \cup B)

1 - 0.7 \times 0.5 = 0.65

(or

0.3 + 0.5 - 0.15

Can two events with

P (A) = 0.3

and

P (B) = 0.5

be both independent and mutually exclusive?

No: independence needs

P (A \cap B) = 0.15

, mutual exclusion needs

0

Three independent shooters hit with probabilities

\frac{1}{2}, \frac{1}{3}, \frac{1}{4}

. Find P(target is hit).

1 - \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{3}{4}

4. Pairwise and Mutual Independence

Three events $A$ , $B$ , $C$ are mutually independent if all four conditions hold:

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

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

They are pairwise independent if only the three pair conditions hold. Mutually independent events are pairwise independent, but the converse may fail (Figure 8).

Figure 8: Every pair shares one outcome (

\frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2}

), but the triple overlap is empty, so

P (A \cap B \cap C) = 0 \neq = \frac{1}{8}

Pairwise independent

Only the 3 pair products hold.

Two coins with $C$ = "exactly one head": every pair passes, the triple fails.

Mutually independent

All 4 conditions hold (for $n$ events, every sub-collection).

Only then may you multiply $P (A) P (B) P (C)$ for "all three".

Exam Trick

For $n$ events to be mutually independent, the product rule must hold for every sub-collection of 2 or more events: $^{n} C_{2} +^{n} C_{3} + \dots +^{n} C_{n} = 2^{n} - n - 1$ conditions. For 3 events that is 4 conditions; for 4 events it is 11.

JEE Advanced

Independence survives unions and intersections. If $A$ , $B$ , $C$ are mutually independent, then $A$ is independent of $B \cup C$ and of $B \cap C$ :

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

= P (A) {P (B) + P (C) - P (B) P (C)} = P (A) P (B \cup C)

and $P (A \cap B \cap C) = P (A) P (B \cap C)$ . With only pairwise independence this can fail: in Figure 8, $P (C ∣ A \cap B) = 0$ although $P (C) = \frac{1}{2}$ .

Quick Recall: tap to check

How many conditions must be checked for 4 events to be mutually independent?

2^{4} - 4 - 1 = 11

Do the three pair conditions alone prove that

A

B

C

are independent?

No. The triple condition

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

is also needed.

A coin is tossed thrice.

E

: first toss head,

F

: last toss tail. Are they independent?

Yes:

P (E \cap F) = \frac{2}{8} = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2}

5. Choosing the Right Formula

Name the events and write what is given: $P (A)$ , $P (B)$ , $P (A \cap B)$ or a conditional value.
If a conditional probability is given or natural (draws in sequence), use the multiplication theorem.
Use $P (A) P (B)$ only for events known to be independent: separate dice, coins, shooters, draws with replacement.
For "at least one" of independent events, subtract the product of the complements from 1.
For a game that restarts on a tie, sum the geometric series: $\frac{w}{1 - t}$ .

Figure 9: Choosing the right form of the multiplication theorem. Never use

P (A) P (B)

unless independence is given or proved.

Figure 10: Revision map of the page: definition, multiplication theorem, independence and the two classic traps (ME versus independent, pairwise versus mutual).

6. Solved Examples

6.1 Conditional probability

Solved Example 1

The probability that a married man watches a certain TV show is 0.4 and that a married woman watches it is 0.5. The probability that a man watches the show, given that his wife does, is 0.7. Find the probability that (a) the couple watch the show (b) a wife watches, given that her husband does (c) at least one of them watches.

Solution:

Let $H$ : husband watches, $W$ : wife watches. $P (H) = 0.4$ , $P (W) = 0.5$ , $P (H ∣ W) = 0.7$ .

(a) $P (H \cap W) = P (W) P (H ∣ W) = 0.5 \times 0.7 = 0.35$

(b) $P (W ∣ H) = \frac{P ( H \cap W )}{P ( H )} = \frac{0.35}{0.4} = \frac{7}{8}$

Solved Example 2

Adults in a small town who have completed a college degree are classified by gender and employment (Figure 2): Male 460 employed, 40 unemployed; Female 140 employed, 260 unemployed. An employed person is selected at random. Find the probability that the person is male.

Solution:

Let $M$ : a man is chosen, $E$ : the person is employed. Using the reduced sample space of 600 employed people:

$P (M ∣ E) = \frac{460}{600} = \frac{23}{30}$

By formula: $P (E) = \frac{600}{900} = \frac{2}{3}$ , $P (E \cap M) = \frac{460}{900} = \frac{23}{45}$ , so $P (M ∣ E) = \frac{23/45}{2/3} = \frac{23}{30}$ .

Solved Example 3

P (A ∣ B) = 0.2

P (B) = 0.5

and

P (A) = 0.2

, find

P (A \cap B^{'})

Solution:

$P (A \cap B) = P (B) P (A ∣ B) = 0.5 \times 0.2 = 0.1$ .

$P (A \cap B^{'}) = P (A) - P (A \cap B) = 0.2 - 0.1 = 0.1$ .

Solved Example 4

P (A) = 0.25

P (B) = 0.5

and

P (A \cap B) = 0.14

, find the probability that neither

A

nor

B

occurs. Also find

P (A \cap B^{'})

Solution:

$P (A \cup B) = 0.25 + 0.5 - 0.14 = 0.61$ .

By De Morgan's law, $P (A^{'} \cap B^{'}) = 1 - P (A \cup B) = 0.39$ .

$P (A \cap B^{'})$ is the part of $A$ outside $B$ : $P (A) - P (A \cap B) = 0.25 - 0.14 = 0.11$ .

6.2 Multiplication theorem

Solved Example 5

Two balls are drawn without replacement from a box of 6 red and 4 white balls. Is "red on the first draw" independent of (a) "red on the second draw" (b) "white on the second draw"?

Solution:

Let $E$ : red on first draw, $F$ : red on second draw, $G$ : white on second draw. From Figure 4, $P (E) = \frac{6}{10}$ , $P (F) = \frac{6}{10}$ , $P (G) = \frac{4}{10}$ .

(a) $P (E \cap F) = \frac{^{6} P _{2}}{^{10} P _{2}} = \frac{30}{90} = \frac{1}{3}$ , while $P (E) P (F) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} \neq = \frac{1}{3}$ . So $E$ and $F$ are not independent.

(b) $P (E \cap G) = \frac{^{6} P _{1} \times ^{4} P _{1}}{^{10} P _{2}} = \frac{24}{90} = \frac{4}{15}$ , while $P (E) P (G) = \frac{6}{25} \neq = \frac{4}{15}$ . So $E$ and $G$ are not independent either.

Solved Example 6

A bag contains 5 white and 8 red balls. Two draws of 3 balls each are made without replacement. Find the probability that the first draw gives 3 white balls and the second draw gives 3 red balls.

Solution:

Let $A$ : 3 white in the first draw, $B$ : 3 red in the second draw.

$P (A) = \frac{^{5} C _{3}}{^{13} C _{3}} = \frac{10}{286} = \frac{5}{143}$ .

After $A$ , 10 balls remain, all 8 red among them: $P (B ∣ A) = \frac{^{8} C _{3}}{^{10} C _{3}} = \frac{56}{120} = \frac{7}{15}$ .

Answer: $P (A \cap B) = P (A) P (B ∣ A) = \frac{5}{143} \times \frac{7}{15} = \frac{7}{429}$ .

Solved Example 7

A box contains 5 bulbs, 2 of them defective. Bulbs are tested one by one until both defective bulbs are found. Find the probability that the process stops after (i) the second test (ii) the third test.

Solution:

Tested bulbs are not put back. D = defective, N = not defective.

(i) Both of the first two must be D: $P = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$ .

(ii) The process stops after the third test in three mutually exclusive ways (Figure 11): DND, NDD and NNN, each with probability $\frac{1}{10}$ . In NNN all good bulbs are used up, so the last two are known to be defective without testing.

Answer: (i) $\frac{1}{10}$ (ii) $\frac{3}{10}$ .

Figure 11: The hidden case NNN also ends testing at step 3, because the last two bulbs are then known to be defective:

P = \frac{3}{10}

Solved Example 8

Cards are drawn one by one without replacement from a well-shuffled pack of 52 until an ace appears. Find the probability that the fourth card is the first ace.

Solution:

Method 1 (sequence). The draws must go non-ace, non-ace, non-ace, ace:

P = \frac{48}{52} \times \frac{47}{51} \times \frac{46}{50} \times \frac{4}{49} = \frac{17296}{270725}

Method 2 (selection then arrangement). Choose 3 non-aces and 1 ace; of the $4!$ orders, $3!$ put the ace last:

P = \frac{^{48} C _{3} \times ^{4} C _{1}}{^{52} C _{4}} \times \frac{3 !}{4 !} = \frac{17296}{270725} \approx 0.064

Solved Example 9

In a multiple-choice question with four options, one or more options may be correct, and marks are given only if all correct options are ticked. A candidate ticks at random and is allowed up to three chances. Find the probability that he gets the marks.

Solution:

Number of possible responses $=^{4} C_{1} +^{4} C_{2} +^{4} C_{3} +^{4} C_{4} = 15$ , of which exactly one is correct. Assume he never repeats a wrong response.

Correct at 1st chance: $\frac{1}{15}$ . At 2nd: $\frac{14}{15} \times \frac{1}{14} = \frac{1}{15}$ . At 3rd: $\frac{14}{15} \times \frac{13}{14} \times \frac{1}{13} = \frac{1}{15}$ .

Answer: $\frac{3}{15} = \frac{1}{5}$ .

Exam Trick

Guessing without repetition among $N$ equally likely options, the chance of success on any particular attempt is always the same, $\frac{1}{N}$ . So $k$ attempts give $\frac{k}{N}$ directly.

6.3 Independent events

Solved Example 10

Event

A_{1}

happens with probability

p_{1}

and an independent event

A_{2}

happens with probability

p_{2}

. Find the probability that (i) exactly one of them happens (ii) at least one of them happens.

Solution:

(i) $A_{1}$ happens and $A_{2}$ fails: $p_{1} (1 - p_{2})$ . $A_{1}$ fails and $A_{2}$ happens: $p_{2} (1 - p_{1})$ . These are mutually exclusive, so P(exactly one) $= p_{1} + p_{2} - 2 p_{1} p_{2}$ .

(ii) Both fail with probability $(1 - p_{1}) (1 - p_{2})$ , so P(at least one) $= 1 - (1 - p_{1}) (1 - p_{2}) = p_{1} + p_{2} - p_{1} p_{2}$ .

Solved Example 11

Two switches

S_{1}

and

S_{2}

work independently with probabilities 90% and 80%. Find the probability that the lamp glows when the switches are connected (i) in series (ii) in parallel.

Solution:

Let $A$ : $S_{1}$ works, $B$ : $S_{2}$ works. $P (A) = \frac{9}{10}$ , $P (B) = \frac{8}{10}$ .

(i) In series, current flows only if both switches work: $P (A \cap B) = \frac{9}{10} \times \frac{8}{10} = \frac{18}{25}$ .

(ii) In parallel, current flows if at least one works:

P (A \cup B) = 1 - P (A^{'}) P (B^{'}) = 1 - \frac{1}{10} \times \frac{2}{10} = \frac{49}{50}

Solved Example 12

A speaks the truth in 60% of cases and B in 90% of cases, independently. In what percentage of cases are they likely to contradict each other when stating the same fact?

Solution:

Let $E$ : A speaks the truth, $F$ : B speaks the truth. $P (E) = \frac{3}{5}$ , $P (F) = \frac{9}{10}$ . They contradict when exactly one tells the truth:

P = P (E) P (F^{'}) + P (E^{'}) P (F) = \frac{3}{5} \times \frac{1}{10} + \frac{2}{5} \times \frac{9}{10} = \frac{21}{50}

Answer: they contradict each other in 42% of cases.

Solved Example 13

E_{1}

and

E_{2}

are events with

P (E_{1}) = \frac{1}{4}

P (E_{2}) = \frac{1}{2}

and

P (E_{1} ∣ E_{2}) = \frac{1}{4}

. Which is correct: (i) independent (ii) exhaustive (iii) mutually exclusive (iv) dependent? Also find

P (E_{1}^{'} ∣ E_{2})

and

P (E_{2} ∣ E_{1}^{'})

Solution:

$P (E_{1} ∣ E_{2}) = \frac{1}{4} = P (E_{1})$ , so $E_{1}$ and $E_{2}$ are independent.

$P (E_{1} \cup E_{2}) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{5}{8} \neq = 1$ , so they are not exhaustive; independent events with non-zero probabilities cannot be mutually exclusive. Only (i) is correct.

Complements of independent events are independent: $P (E_{1}^{'} ∣ E_{2}) = P (E_{1}^{'}) = \frac{3}{4}$ and $P (E_{2} ∣ E_{1}^{'}) = P (E_{2}) = \frac{1}{2}$ .

Solved Example 14

Three students attempt a mathematics examination independently, with probabilities of success

\frac{1}{3}

\frac{1}{4}

and

\frac{1}{5}

. Find the probability that at least two succeed.

Solution:

At least two succeed in four mutually exclusive ways: $E_{1} E_{2} E_{3}^{'}$ , $E_{1}^{'} E_{2} E_{3}$ , $E_{1} E_{2}^{'} E_{3}$ , $E_{1} E_{2} E_{3}$ .

P = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{4}{5} + \frac{2}{3} \cdot \frac{1}{4} \cdot \frac{1}{5} + \frac{1}{3} \cdot \frac{3}{4} \cdot \frac{1}{5} + \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{5} = \frac{4 + 2 + 3 + 1}{60} = \frac{1}{6}

Solved Example 15

A tosses 2 fair coins and B tosses 3 fair coins. Whoever gets more heads wins; if there is a tie, the game is repeated under the same rules until someone wins. Find the probability that A wins.

Solution:

Let $A_{i}$ ( $B_{i}$ ) mean A (B) gets $i$ heads. $P (A_{0}) = \frac{1}{4}$ , $P (A_{1}) = \frac{2}{4}$ , $P (A_{2}) = \frac{1}{4}$ ; $P (B_{0}) = \frac{1}{8}$ , $P (B_{1}) = \frac{3}{8}$ , $P (B_{2}) = \frac{3}{8}$ , $P (B_{3}) = \frac{1}{8}$ .

A wins a round ( $A_{1} B_{0}$ , $A_{2} B_{0}$ , $A_{2} B_{1}$ ):

w = \frac{2}{4} \cdot \frac{1}{8} + \frac{1}{4} \cdot \frac{1}{8} + \frac{1}{4} \cdot \frac{3}{8} = \frac{3}{16}

Tie ( $A_{0} B_{0}$ , $A_{1} B_{1}$ , $A_{2} B_{2}$ ):

t = \frac{1}{4} \cdot \frac{1}{8} + \frac{2}{4} \cdot \frac{3}{8} + \frac{1}{4} \cdot \frac{3}{8} = \frac{5}{16}

A can win in round 1, or tie then win, or tie twice then win, and so on (Figure 12):

P = \frac{3}{16} + \frac{5}{16} \cdot \frac{3}{16} + (\frac{5}{16})^{2} \cdot \frac{3}{16} + \dots = \frac{3/16}{1 - 5/16} = \frac{3}{11}

Figure 12: A tie simply restarts the game, so

P (A wins) = \frac{w}{1 - t} = \frac{3/16}{11/16} = \frac{3}{11}

6.4 Pairwise and mutual independence

Solved Example 16

A lot has 50 defective and 50 non-defective bulbs. Two bulbs are drawn one at a time with replacement.

A

= {first bulb defective},

B

= {second bulb non-defective},

C

= {both defective or both non-defective}. Are

A

B

C

pairwise or mutually independent?

Solution:

Let $D_{i}$ ( $N_{i}$ ) denote a defective (non-defective) bulb on draw $i$ . $S = {D_{1} D_{2}, D_{1} N_{2}, N_{1} D_{2}, N_{1} N_{2}}$ , each with probability $\frac{1}{4}$ .

$A = {D_{1} D_{2}, D_{1} N_{2}}$ , $B = {D_{1} N_{2}, N_{1} N_{2}}$ , $C = {D_{1} D_{2}, N_{1} N_{2}}$ , so $P (A) = P (B) = P (C) = \frac{1}{2}$ .

$A \cap B = {D_{1} N_{2}}$ , $B \cap C = {N_{1} N_{2}}$ , $A \cap C = {D_{1} D_{2}}$ , each of probability $\frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2}$ : all pairs are independent.

But $A \cap B \cap C = ϕ$ , so $P (A \cap B \cap C) = 0 \neq = \frac{1}{8}$ . Pairwise independent, not mutually independent.

Solved Example 17

A pair of fair coins is tossed,

S = {H H, H T, T H, T T}

A

= {head on first coin} = {HH, HT},

B

= {head on second coin} = {HH, TH},

C

= {head on exactly one coin} = {HT, TH}. Are

A

B

C

independent?

Solution:

$P (A) = P (B) = P (C) = \frac{1}{2}$ .

$P (A \cap B) = P ({H H}) = \frac{1}{4} = P (A) P (B)$ ; $P (A \cap C) = P ({H T}) = \frac{1}{4} = P (A) P (C)$ ; $P (B \cap C) = P ({T H}) = \frac{1}{4} = P (B) P (C)$ .

But $P (A \cap B \cap C) = 0 \neq = P (A) P (B) P (C)$ . The events are pairwise independent but not mutually independent (Figure 8).

Solved Example 18

A family has three children, each equally likely to be a boy or a girl.

A

: at most one boy;

B

: at least one boy and one girl;

C

: at most one girl. Are

A

and

B

independent? Are

A

B

C

independent?

Solution:

P(3 boys) $= \frac{1}{8}$ ; P(2 boys, 1 girl) $= \frac{3}{8}$ (BBG, BGB, GBB); P(1 boy, 2 girls) $= \frac{3}{8}$ ; P(3 girls) $= \frac{1}{8}$ .

$A$ = {1 boy 2 girls, 3 girls}: $P (A) = \frac{1}{2}$ . $B$ = {2 boys 1 girl, 1 boy 2 girls}: $P (B) = \frac{3}{4}$ . $C$ = {3 boys, 2 boys 1 girl}: $P (C) = \frac{1}{2}$ .

$A \cap B$ = {1 boy 2 girls}: $P (A \cap B) = \frac{3}{8} = \frac{1}{2} \cdot \frac{3}{4} = P (A) P (B)$ , so $A$ and $B$ are independent.

$A \cap C = ϕ$ , so $P (A \cap C) = 0 \neq = P (A) P (C)$ . Hence $A$ , $B$ , $C$ are not independent.

6.5 JEE-style problems

Solved Example 19

A

and

B

are independent events with

P (A) = 0.4

and

P (B) = 0.3

. The probability that exactly one of them occurs is
(A)

0.46

(B)

0.58

(C)

0.12

(D)

0.70

Solution:

P(exactly one) $= P (A) P (B^{'}) + P (A^{'}) P (B) = 0.4 \times 0.7 + 0.6 \times 0.3 = 0.28 + 0.18 = 0.46$ . Check with the formula $p_{1} + p_{2} - 2 p_{1} p_{2} = 0.7 - 0.24 = 0.46$ .

Answer: (A) $0.46$ .

Solved Example 20

Two fair dice are thrown and the sum is 8. The probability that at least one die shows 3 is
(A)

1/5

(B)

2/5

(C)

1/3

(D)

5/36

Solution:

The condition "sum is 8" shrinks the sample space to {(2,6), (3,5), (4,4), (5,3), (6,2)}: 5 outcomes. Two of them contain a 3.

Answer: (B) $\frac{2}{5}$ . Option (D) is the unconditional $P (sum 8)$ , a common slip.

Solved Example 21

A family has two children. Find the probability that both are boys, given that (i) at least one is a boy (ii) the elder child is a boy.

Solution:

$S$ = {BB, BG, GB, GG} (elder first), equally likely.

(i) Given at least one boy: {BB, BG, GB}, so $P = \frac{1}{3}$ .

(ii) Given the elder is a boy: {BB, BG}, so $P = \frac{1}{2}$ .

Answer: $\frac{1}{3}$ and $\frac{1}{2}$ . A different condition gives a different reduced sample space, even though both sound like "we know about a boy".

Solved Example 22

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

P (B) = \frac{1}{3}

and

P (A \cap B) = \frac{1}{4}

, find

P (A^{'} ∣ B^{'})

Solution:

$P (A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12}$ .
$P (A^{'} \cap B^{'}) = 1 - \frac{7}{12} = \frac{5}{12}$ and $P (B^{'}) = \frac{2}{3}$ .
$P (A^{'} ∣ B^{'}) = \frac{5/12}{2/3} = \frac{5}{8}$ .

Answer: $\frac{5}{8}$ .

Solved Example 23

A and B throw a fair die alternately, A first. The first to throw a six wins. Find the probability that A wins.

Solution:

A wins on throw 1, 3, 5, ...: each earlier pair of throws must be non-sixes (probability $\frac{25}{36}$ per round).

P (A wins) = \frac{1}{6} + \frac{25}{36} \cdot \frac{1}{6} + (\frac{25}{36})^{2} \cdot \frac{1}{6} + \dots = \frac{1/6}{1 - 25/36} = \frac{6}{11}

Answer: $\frac{6}{11}$ , and B wins with $\frac{5}{11}$ . Moving first is an advantage.

Practice Questions

If $P (A^{'} ∣ B^{'}) = 0.2$ and $P (A \cup B) = 0.9$ , find $P (A \cap B^{'})$ .Answer: 0.4
An urn has 7 red and 4 blue balls. Two are drawn with replacement. Find P(2 red), P(2 blue), P(one of each).Answer: $\frac{49}{121}$ , $\frac{16}{121}$ , $\frac{56}{121}$
A and B solve a problem independently with probabilities $\frac{1}{2}$ and $\frac{1}{3}$ . Find P(problem solved) and P(exactly one solves it).Answer: $\frac{2}{3}$ , $\frac{1}{2}$
Two dice are thrown. Find the probability of an odd number on the first die and a total of 7.Answer: $\frac{1}{12}$
Two dice are thrown. Find the probability of a doublet or a total of 4.Answer: $\frac{2}{9}$
A bag has 3 blue and 5 red marbles. One is drawn, its colour noted, and it is replaced; a second is drawn. Find P(blue then red), P(blue and red in any order), P(same colour).Answer: $\frac{15}{64}$ , $\frac{15}{32}$ , $\frac{17}{32}$
A coin is tossed thrice. In which case are E and F independent? (i) E: first toss head, F: last toss tail (ii) E: exactly two heads, F: last toss head (iii) E: odd number of heads, F: odd number of tailsAnswer: (i) only

Common Mistakes to Avoid

Watch out

Swapping $P (A ∣ B)$ and $P (B ∣ A)$ . The event after the bar is the one that is known; it goes in the denominator.
Using $P (A \cap B) = P (A) P (B)$ without being told (or proving) independence. In general use $P (A) P (B ∣ A)$ .
Treating draws without replacement as independent. The second-draw probabilities depend on the first result.
Calling mutually exclusive events independent. If both have positive probability, they are dependent.
Stopping at the three pair checks. Mutual independence of three events also needs $P (A \cap B \cap C) = P (A) P (B) P (C)$ .
Checking only the triple product. $P (A \cap B \cap C) = P (A) P (B) P (C)$ alone does not give the three pair conditions.
Writing $P (A ∣ B^{'}) = 1 - P (A ∣ B)$ . Complementing the condition does not work; only $P (A^{'} ∣ B) = 1 - P (A ∣ B)$ is valid.
Missing hidden stopping cases, such as NNN in the bulb-testing problem, where the remaining items are forced.

Frequently Asked Questions

What is conditional probability in simple words?

Conditional probability is the chance of an event after you learn that another event has happened. The known event becomes the new sample space. For a die showing a number above 3, the chance that it is 4 becomes $\frac{1}{3}$ instead of $\frac{1}{6}$ .

What is the difference between P(A and B) and P(A given B)?

$P (A \cap B)$ is the chance that both events happen, measured against the whole sample space. $P (A ∣ B)$ measures $A$ only inside $B$ , so it divides by $P (B)$ : $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ . It is never smaller than $P (A \cap B)$ .

What is the multiplication theorem of probability?

The multiplication theorem gives the probability that two events both occur: $P (A \cap B) = P (A) P (B ∣ A)$ . For independent events it reduces to $P (A) P (B)$ . It extends to many events by conditioning each new event on all earlier ones, exactly like multiplying along a tree path.

How do you check whether two events are independent?

Compute $P (A \cap B)$ directly from the sample space and compare it with $P (A) \times P (B)$ . If they are equal, the events are independent; if not, they are dependent. Equivalently, check whether $P (A ∣ B) = P (A)$ . Physically separate experiments, such as two different dice, are independent.

Can two events be both independent and mutually exclusive?

Only if at least one of them has probability zero. Mutually exclusive means $P (A \cap B) = 0$ , while independence needs $P (A \cap B) = P (A) P (B)$ . Both hold together only when $P (A) = 0$ or $P (B) = 0$ . For events that can actually happen, the two ideas are incompatible.

What is the difference between pairwise and mutual independence?

Pairwise independence checks only pairs: $P (A \cap B) = P (A) P (B)$ and so on. Mutual independence also needs every larger group to satisfy the product rule, such as $P (A \cap B \cap C) = P (A) P (B) P (C)$ . Two fair coins with "exactly one head" as the third event give the standard pairwise-only example.

How is conditional probability asked in JEE Main?

JEE Main lists the addition and multiplication theorems of probability, so questions typically give some of $P (A)$ , $P (B)$ , $P (A \cap B)$ or a conditional value and ask for another. Draws without replacement, independent trials such as switches or shooters, and two-way tables are the most common settings.

Does the JEE Advanced syllabus include independence of events?

Yes. The JEE Advanced 2026 syllabus lists conditional probability, the multiplication rule and independence of events. Expect multi-step problems: proving or disproving independence of three events, repeated games that need a geometric series, and conditions hidden inside counting arguments.

Previous year questions on Conditional Probability

7 questions from past papers, each with a step-by-step solution.

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