Hard

24 min read

Conditional Probability & Bayes' Theorem

Master Conditional Probability and Bayes Theorem for JEE Main and Board Exams. Covers conditional probability definition P(B|A)=P(A∩B)/P(A), properties of conditional probability, law of total probability for partition of sample space, Bayes theorem formula prior likelihood posterior probabilities, step-by-step method for bags and balls problems, tabular method for exam efficiency, manufacturing defect problems, medical testing examples with fully verified JEE Board-level practice questions.

New

Many real-world probability questions involve updated information: "Given that I already know this happened, what is the probability that something else also happened (or will happen)?" This is the domain of Conditional Probability — one of the most profound and counterintuitive ideas in mathematics. Bayes' Theorem builds on this: it provides a systematic way to reverse the direction of conditioning, allowing us to go from P(evidence | hypothesis) to P(hypothesis | evidence). From medical testing to spam filters to legal reasoning, Bayes' Theorem is among the most powerful and widely applied results in all of probability. For JEE and CBSE, Bayes' Theorem problems (especially involving bags and balls, defective items, and disease testing) appear almost every year.

1. Conditional Probability

The conditional probability of event B occurring given that event A has already occurred is:

$P (B | A) = \frac{P (A \cap B)}{P (A)}, P (A) \neq 0$

Interpretation: Restricts the sample space from S to A, then measures B within that restricted space.

Properties of Conditional Probability

0 ≤ P(B|A) ≤ 1
P(B'|A) = 1 − P(B|A)
P(B₁ ∪ B₂ | A) = P(B₁|A) + P(B₂|A) − P(B₁ ∩ B₂|A)
If A and B are independent: P(B|A) = P(B) and P(A|B) = P(A)

Worked Example

P(A) = 1/2, P(A ∩ B) = 1/6. Find P(B|A).

P(B|A) = P(A∩B)/P(A) = (1/6)/(1/2) = 1/3

2. Law of Total Probability

If B₁, B₂, …, Bₙ are mutually exclusive and exhaustive events (they partition the sample space S), then for any event A:

$P (A) = \sum_{i = 1}^{n} P (B_{i}) \cdot P (A | B_{i})$

This decomposes a complex probability into weighted conditional probabilities over all possible "routes" to A.

Worked Example

Bag 1 has 3 red, 2 blue. Bag 2 has 2 red, 3 blue. A bag is chosen randomly and a ball drawn. Find P(red).

P(red) = P(Bag1)·P(red|Bag1) + P(Bag2)·P(red|Bag2) = (1/2)(3/5) + (1/2)(2/5) = 3/10 + 2/10 = 1/2

3. Bayes' Theorem

If B₁, B₂, …, Bₙ partition S, and A is any event with P(A) > 0, then:

$P (B_{k} | A) = \frac{P (B_{k}) \cdot P (A | B_{k})}{\sum_{i = 1}^{n} P (B_{i}) \cdot P (A | B_{i})}$

Terminology

Term	Meaning
Prior probability	P(Bₖ) — probability of hypothesis before observing evidence
Likelihood	P(A\|Bₖ) — probability of evidence given hypothesis
Posterior probability	P(Bₖ\|A) — updated probability of hypothesis after observing evidence

Step-by-Step Method for Bayes' Theorem Problems

Identify the partition: B₁, B₂, …, Bₙ (the "routes" or "causes")
Write down the prior probabilities P(Bᵢ)
Write down the likelihoods P(A|Bᵢ) for each route
Compute P(A) using the Law of Total Probability
Apply Bayes' formula for the required posterior P(Bₖ|A)

Worked Example — Bags and Balls

Continuing from above: A bag is chosen randomly (Bag 1: 3R 2B; Bag 2: 2R 3B) and a red ball is drawn. Find P(it came from Bag 1).

P(Bag1) = P(Bag2) = 1/2; P(red|Bag1) = 3/5; P(red|Bag2) = 2/5

P(red) = (1/2)(3/5) + (1/2)(2/5) = 1/2

P(Bag1|red) = [(1/2)(3/5)] / (1/2) = (3/10)/(1/2) = 3/5

4. Bayes' Theorem — 3-Bag Problem

Bag 1: 2W 3B (5 balls); Bag 2: 4W 2B (6 balls); Bag 3: 3W 4B (7 balls). A bag is selected at random and a white ball is drawn. Find P(selected from Bag 2).

P(each bag) = 1/3; P(W|B1) = 2/5; P(W|B2) = 4/6 = 2/3; P(W|B3) = 3/7

P(W) = (1/3)(2/5) + (1/3)(2/3) + (1/3)(3/7) = 2/15 + 2/9 + 1/7

LCM(15,9,7) = 315: = 42/315 + 70/315 + 45/315 = 157/315

P(B2|W) = [(1/3)(2/3)] / (157/315) = (2/9) / (157/315) = (2/9)(315/157) = 70/157 ≈ 0.446

Bayes' Theorem — Tabular Method (Faster for Exams)

For the 3-bag problem above, organise as a table to avoid arithmetic errors:

Hypothesis Bᵢ	Prior P(Bᵢ)	Likelihood P(A\|Bᵢ)	P(Bᵢ)·P(A\|Bᵢ)
Bag 1	1/3	2/5	2/15 = 42/315
Bag 2	1/3	2/3	2/9 = 70/315
Bag 3	1/3	3/7	1/7 = 45/315
Total	1	—	157/315 = P(A)

P(Bag2|W) = 70/315 ÷ 157/315 = 70/157. The table makes the calculation mechanical — no formula needed beyond dividing the row's joint probability by the total.

JEE/Board Power Tips

P(B|A) ≠ P(A|B) in general. This is the most fundamental fact about conditional probability. P(being wet | it rained) is high; P(it rained | being wet) is much lower — you might have showered.
In Bayes' problems, always identify what the "observed evidence" A is and what the "possible causes" Bᵢ are. The evidence is what you've seen; the causes are what you want to figure out.
Use the tabular method in exams — list each Bᵢ, its P(Bᵢ), its P(A|Bᵢ), and their product. Sum the products to get P(A). Divide for each posterior. This is faster and more reliable than memorising the formula.
Law of Total Probability is always implicit in Bayes' Theorem. The denominator of Bayes' formula IS the Law of Total Probability. Always compute it first.

Practice Questions

Q1 (Board): A box contains 4 red and 6 white balls. Two balls are drawn in succession without replacement. Given that the first ball was red, find the probability that the second is also red.

Explanation:

Initially, the box contains $4 red + 6 white = 10 balls$ .

After drawing 1 red ball on the first draw, the box now contains $3 red$ and $6 white$ balls, making a total of $9 balls$ .

Therefore, the conditional probability of drawing a second red ball is:
$P (2nd red ∣ 1st red) = \frac{3}{9} = \frac{1}{3}$

Note: The total probability of drawing both red balls would be $P (both red) = (\frac{4}{10}) (\frac{3}{9}) = \frac{12}{90} = \frac{2}{15}$ .

Q2 (JEE Main / Board): There are two bags. Bag A contains 3 white and 2 red balls; Bag B contains 2 white and 4 red balls. A bag is selected randomly and a ball drawn at random. It turns out to be white. Find the probability that Bag A was selected.

Explanation:

Let $A$ and $B$ be the events of selecting Bag A and Bag B.
Let $W$ be the event of drawing a white ball.
$P (A) = P (B) = \frac{1}{2}$

The conditional probabilities of drawing a white ball from each bag are:
$P (W | A) = \frac{3}{5}$
$P (W | B) = \frac{2}{6} = \frac{1}{3}$

Using the Law of Total Probability to find $P (W)$ :
$P (W) = P (A) P (W | A) + P (B) P (W | B)$
$P (W) = (\frac{1}{2}) (\frac{3}{5}) + (\frac{1}{2}) (\frac{1}{3}) = \frac{3}{10} + \frac{1}{6} = \frac{9}{30} + \frac{5}{30} = \frac{14}{30} = \frac{7}{15}$

Using Bayes' Theorem to find $P (A | W)$ :
$P (A | W) = \frac{P (A) P (W | A)}{P (W)}$
$P (A | W) = \frac{3 / 10}{7 / 15} = (\frac{3}{10}) (\frac{15}{7}) = \frac{45}{70} = \frac{9}{14}$

Q3 (JEE / Board): In a factory, machine A produces 60% of items and machine B produces 40%. Machine A produces 5% defective items and machine B produces 3% defective. An item is chosen at random and found defective. What is the probability it was produced by machine A?

Explanation:

Let $A$ and $B$ be the events that the item is produced by Machine A and Machine B respectively.
Let $D$ be the event that the item is defective.
$P (A) = 0.6$ , $P (B) = 0.4$
$P (D | A) = 0.05$ , $P (D | B) = 0.03$

Total probability of a defective item $P (D)$ :
$P (D) = P (A) P (D | A) + P (B) P (D | B)$
$P (D) = (0.6) (0.05) + (0.4) (0.03) = 0.030 + 0.012 = 0.042$

Using Bayes' Theorem to find $P (A | D)$ :
$P (A | D) = \frac{P (A) P (D | A)}{P (D)} = \frac{0.030}{0.042} = \frac{30}{42} = \frac{5}{7} \approx 0.714$

Q4 (MCQ): Given $P (A) = \frac{1}{4}$ , $P (B | A) = \frac{1}{2}$ , and $P (A | B) = \frac{1}{4}$ . Find $P (B)$ .

Explanation:

Using the conditional probability formula, we can find the intersection $P (A \cap B)$ :
$P (A \cap B) = P (A) \cdot P (B | A) = (\frac{1}{4}) (\frac{1}{2}) = \frac{1}{8}$

We also know that the intersection can be expressed using $P (A | B)$ :
$P (A \cap B) = P (B) \cdot P (A | B)$
$\frac{1}{8} = P (B) \cdot (\frac{1}{4})$
$P (B) = \frac{1 / 8}{1 / 4} = \frac{4}{8} = \frac{1}{2}$

Q5 (JEE Advanced): Let $E^{c}$ denote the complement of an event $E$ . Let $E_{1}, E_{2}$ , and $E_{3}$ be any pairwise independent events such that $P (E_{1}) > 0$ and $P (E_{1} \cap E_{2} \cap E_{3}) = 0$ . Then $P (E_{2}^{c} \cap E_{3}^{c} ∣ E_{1})$ is equal to:

A) $P (E_{2}^{c}) - P (E_{3})$
B) $P (E_{2}^{c}) + P (E_{3})$
C) $P (E_{2}^{c}) - P (E_{3}^{c})$
D) $P (E_{3}) - P (E_{2}^{c})$

Answer: A) $P (E_{2}^{c}) - P (E_{3})$

Explanation:

By the definition of conditional probability:
$P (E_{2}^{c} \cap E_{3}^{c} ∣ E_{1}) = \frac{P (E_{1} \cap E_{2}^{c} \cap E_{3}^{c})}{P (E_{1})}$

Using set theory and De Morgan's Laws, the numerator can be expanded as the probability of $E_{1}$ occurring MINUS the probability of $E_{1}$ intersecting with either $E_{2}$ or $E_{3}$ :
$P (E_{1} \cap E_{2}^{c} \cap E_{3}^{c}) = P (E_{1}) - P (E_{1} \cap E_{2}) - P (E_{1} \cap E_{3}) + P (E_{1} \cap E_{2} \cap E_{3})$

We are given that the events are pairwise independent, meaning $P (E_{1} \cap E_{2}) = P (E_{1}) P (E_{2})$ and $P (E_{1} \cap E_{3}) = P (E_{1}) P (E_{3})$ . We are also given $P (E_{1} \cap E_{2} \cap E_{3}) = 0$ . Substituting these in:
$P (E_{1} \cap E_{2}^{c} \cap E_{3}^{c}) = P (E_{1}) - P (E_{1}) P (E_{2}) - P (E_{1}) P (E_{3}) + 0$
$= P (E_{1}) [1 - P (E_{2}) - P (E_{3})]$

Substitute this back into the conditional probability fraction:
$P (E_{2}^{c} \cap E_{3}^{c} ∣ E_{1}) = \frac{P (E_{1}) [1 - P (E_{2}) - P (E_{3})]}{P (E_{1})}$
$= 1 - P (E_{2}) - P (E_{3})$

Since $1 - P (E_{2}) = P (E_{2}^{c})$ , this simplifies to:
$= P (E_{2}^{c}) - P (E_{3})$

Q6 (JEE Advanced): A laboratory blood test is 99% effective in detecting a certain disease when it is, in fact, present. However, the test also yields a false positive result for 0.5% of the healthy persons tested. If 0.1% of the population actually has the disease, what is the probability that a person actually has the disease given that their test result is positive?

Explanation:

Let $D$ be the event that a person has the disease, and $H$ be the event that they are healthy (no disease).
$P (D) = 0.001$
$P (H) = 1 - 0.001 = 0.999$

Let $T$ be the event that the test result is positive.
True Positive Rate: $P (T | D) = 0.99$
False Positive Rate: $P (T | H) = 0.005$

We need to find the posterior probability $P (D | T)$ . Using Bayes' Theorem:
$P (D | T) = \frac{P (D) P (T | D)}{P (D) P (T | D) + P (H) P (T | H)}$

Substitute the given values:
$P (D | T) = \frac{(0.001) (0.99)}{(0.001) (0.99) + (0.999) (0.005)}$
$P (D | T) = \frac{0.00099}{0.00099 + 0.004995}$
$P (D | T) = \frac{0.00099}{0.005985}$

To simplify, multiply the numerator and denominator by 100,000:
$P (D | T) = \frac{99}{598.5} = \frac{990}{5985}$

Dividing both by 5:
$P (D | T) = \frac{198}{1197}$
Dividing both by 9:
$P (D | T) = \frac{22}{133} \approx 16.5 %$

(Note: This is a classic demonstration of the Base Rate Fallacy. Despite the test being 99% accurate, because the disease is so rare (0.1%), a person with a positive result still only has a ~16.5% chance of actually having the disease!)

Q7 (JEE Advanced): Three randomly chosen non-negative integers $x, y,$ and $z$ satisfy the condition $x + y + z = 10$ . Find the probability that $z$ is strictly greater than $x + y$ .

Explanation:

Step 1: Find total outcomes
The total number of non-negative integer solutions to $x + y + z = 10$ is given by the stars and bars formula $^{n + r - 1} C_{r - 1}$ where $n = 10$ and $r = 3$ :
Total cases = $^{10 + 3 - 1} C_{3 - 1} =^{12} C_{2} = \frac{12 \times 11}{2} = 66$ .

Step 2: Find favorable outcomes
We require $z > x + y$ .
We know $x + y + z = 10 ⟹ x + y = 10 - z$ .
Substitute this into the inequality:
$z > 10 - z ⟹ 2 z > 10 ⟹ z > 5$ .

Since $x, y, z$ are non-negative integers summing to 10, the possible values for $z$ that satisfy $z > 5$ are $6, 7, 8, 9, 10$ . We must count the number of non-negative integer solutions for $x + y = 10 - z$ for each valid $z$ :

If $z = 6$ , then $x + y = 4$ → $5$ solutions (from $^{5} C_{1}$ )
If $z = 7$ , then $x + y = 3$ → $4$ solutions
If $z = 8$ , then $x + y = 2$ → $3$ solutions
If $z = 9$ , then $x + y = 1$ → $2$ solutions
If $z = 10$ , then $x + y = 0$ → $1$ solution

Total favorable cases = $5 + 4 + 3 + 2 + 1 = 15$ .

Step 3: Calculate final probability
$P (z > x + y) = \frac{Favorable cases}{Total cases} = \frac{15}{66} = \frac{5}{22}$

Q8 (JEE Advanced): A player plays a game against a machine. At each turn, the player has a probability of $p = \frac{1}{3}$ of winning and $q = \frac{2}{3}$ of losing. The game ends either when the player wins two consecutive turns or loses two consecutive turns. Find the probability that the player wins the game (i.e., the game ends with two consecutive wins).

Explanation:

Let $W$ be the event of winning a turn and $L$ be losing. The game can end in a win via paths starting with a Win ( $W W, W L W W, W L W L W W \dots$ ) or starting with a Loss ( $L W W, L W L W W, L W L W L W W \dots$ ).

Case 1: Game starts with a Win
The sequence must alternate $W L W L \dots$ and suddenly end in $W W$ .
Probability sum $P_{1} = P (W W) + P (W L W W) + P (W L W L W W) + \dots$
$P_{1} = p^{2} + p q (p^{2}) + (p q)^{2} (p^{2}) + \dots$
This is an infinite geometric progression with first term $a = p^{2}$ and common ratio $r = p q$ .
$P_{1} = \frac{p^{2}}{1 - p q}$

Case 2: Game starts with a Loss
The sequence must be exactly the same alternating pattern, just prefaced by one loss: $L W W, L W L W W \dots$
Probability sum $P_{2} = P (L W W) + P (L W L W W) + P (L W L W L W W) + \dots$
$P_{2} = q p^{2} + q (p q) p^{2} + q (p q)^{2} (p^{2}) + \dots$
This is exactly $q$ times the series $P_{1}$ .
$P_{2} = q \times (\frac{p^{2}}{1 - p q})$

Total Probability of Winning the Game:
$P (Win Game) = P_{1} + P_{2} = \frac{p^{2}}{1 - p q} + \frac{q p^{2}}{1 - p q} = \frac{p^{2} (1 + q)}{1 - p q}$

Substitute $p = \frac{1}{3}$ and $q = \frac{2}{3}$ :
$P (Win Game) = \frac{(1 / 9) (1 + 2 / 3)}{1 - (1 / 3) (2 / 3)} = \frac{(1 / 9) (5 / 3)}{1 - 2 / 9} = \frac{5 / 27}{7 / 9} = (\frac{5}{27}) \times (\frac{9}{7}) = \frac{5}{21}$

Related Study Material

More Mathematics Notes More Statistics And Probability Notes All Probability Topics

Practice Questions

Mathematics Practice Probability Questions

Topic Information

DifficultyHard

Est. Read Time24 minutes

CategoryProbability

Quick Actions

Track Your Learning

Study Tips

Take notes of key points while reading

Practice related questions after studying

Review topics marked as "Learning" regularly

Test your knowledge with practice questions

Edvaya Target

Detailed Notes & Solved Examples

Master complex concepts with our rich, topic-wise study notes. Access hundreds of step-by-step solved examples and interactive flashcards right from your phone.

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Conditional Probability & Bayes' Theorem

1. Conditional Probability

Properties of Conditional Probability

Worked Example

2. Law of Total Probability

Worked Example

3. Bayes' Theorem

Terminology

Step-by-Step Method for Bayes' Theorem Problems

Worked Example — Bags and Balls

4. Bayes' Theorem — 3-Bag Problem

Bayes' Theorem — Tabular Method (Faster for Exams)

Practice Questions

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

Edvaya Target

We Value Your Privacy