MathsCBSE Class 12 · Probability

Conditional Probability and the Multiplication Theorem (Class 12)

Updated 15 October 2026 · Fact-checked

The conditional probability of E given F is P(E|F) = P(E ∩ F) / P(F), provided P(F) ≠ 0. It is the probability of E after you know F has happened, so F becomes the new sample space. Rearranged, it gives the multiplication theorem P(E ∩ F) = P(F) · P(E|F).

Understand Conditional Probability

Conditional probability answers: how likely is E, now that we know F has happened? Knowing F shrinks the sample space to F, so only the part of E inside F counts.

Definition. For events $E$ and $F$ of the same experiment with $P (F) \neq = 0$ ,
$P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )}$ .
With equally likely outcomes this is simply $P (E ∣ F) = \frac{n ( E \cap F )}{n ( F )}$ : count the favourable outcomes inside F and divide by the number of outcomes in F. This 'reduced sample space' view is usually the fastest way to solve dice, coin and card questions.

Properties (NCERT Section 13.2.1). For $P (F) \neq = 0$ :

$P (S ∣ F) = P (F ∣ F) = 1$ .
$P ((A \cup B) ∣ F) = P (A ∣ F) + P (B ∣ F) - P ((A \cap B) ∣ F)$ ; if $A$ and $B$ are disjoint the last term is 0.
$P (E^{'} ∣ F) = 1 - P (E ∣ F)$ .

Note that $P (E ∣ F)$ and $P (F ∣ E)$ are different in general: they have the same numerator but different denominators.

Multiplication theorem (NCERT Section 13.3). Rearranging the definition,
$P (E \cap F) = P (F) P (E ∣ F) = P (E) P (F ∣ E)$ .
For three events, $P (E \cap F \cap G) = P (E) P (F ∣ E) P (G ∣ E \cap F)$ . This is the rule behind draws without replacement: the probability of the first draw, times the probability of the second given the first, and so on, because each draw changes what is left.

Typical board questions.

Given $P (A)$ , $P (B)$ and one of $P (A \cap B)$ , $P (A \cup B)$ or a conditional probability: find the others.
Dice, coins or cards with a 'given that' condition: use the reduced sample space.
Draws without replacement: use the multiplication theorem.
Venn-diagram or table case studies: read $P (E ∣ F)$ as 'part of F that is also E'.

This is the base of the whole chapter: independence and Bayes' theorem are both built on $P (E ∣ F)$ .

Conditional Probability: formulas to remember

Definition: $P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )}, P (F) \neq = 0$; F is the condition (the new sample space).
Equally likely outcomes: $P (E ∣ F) = \frac{n ( E \cap F )}{n ( F )}$; Count inside F only.
Complement: $P (E^{'} ∣ F) = 1 - P (E ∣ F)$; The condition stays the same.
Union given F: $P ((A \cup B) ∣ F) = P (A ∣ F) + P (B ∣ F) - P ((A \cap B) ∣ F)$; Addition rule inside the reduced space.
Multiplication theorem: $P (E \cap F) = P (F) P (E ∣ F) = P (E) P (F ∣ E)$; Used for 'one after another' events.
Three events: $P (E \cap F \cap G) = P (E) P (F ∣ E) P (G ∣ E \cap F)$; Three draws without replacement.

How to solve Conditional Probability questions

Pick the route from what the question gives you.

1Name the events in words: E = the event asked about, F = the information given ('given that', 'if it is known that').
2If the outcomes are equally likely and easy to list, write F as a set and count the outcomes of E inside it: $P (E ∣ F) = \frac{n ( E \cap F )}{n ( F )}$ .
3If probabilities are given, find $P (E \cap F)$ first (often from $P (E \cup F) = P (E) + P (F) - P (E \cap F)$ ), then divide by $P (F)$ .
4For draws one after another without replacement, multiply: first draw, then second given the first, and so on.
5For questions about $E^{'}$ or unions given F, use the properties instead of starting again.
6Check that the answer lies between 0 and 1, and simplify the fraction.

Conditional Probability in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Updating a chance with information

Suppose 60% of a class passed the first unit test and 45% passed both unit tests.

A student is known to have passed the first test. What is the probability that the student passed the second?

Try it first, then show the solutionSolution

Let F = passed the first, E = passed the second. $P (F) = 0.6$ , $P (E \cap F) = 0.45$ .
$P (E ∣ F) = \frac{0.45}{0.6} = 0.75$ .

Answer: 0.75, that is 75%.

Memory hook: Given that = divide by the given.

Draws without replacement

Imagine a tray of 12 eggs in which 3 are cracked. You take out two eggs, one after the other, without putting any back.

Find the probability that both eggs are cracked.

Try it first, then show the solutionSolution

First egg cracked: $\frac{3}{12}$ .
Second cracked given the first was: $\frac{2}{11}$ (one cracked egg is gone).
Multiply: $\frac{3}{12} \times \frac{2}{11} = \frac{1}{22}$ .

Answer: $\frac{1}{22}$

Memory hook: Without replacement, the second fraction loses one from top and bottom.

Tricks to remember Conditional Probability

Short memory aids. Say each one aloud once, then use it on a question the same day.

Given goes down
In $P (E ∣ F)$ , the event after the bar (the given one) goes in the denominator.
Use it when: Never mixing up $P (E ∣ F)$ and $P (F ∣ E)$ .
Shrink the space
Cross out every outcome not in F; what is left is your new sample space.
Use it when: Dice, coins and card questions.
Same bar, same rules
With the same condition F, all the usual rules hold: complements add to 1, the addition rule works.
Use it when: $P (E^{'} ∣ F) = 1 - P (E ∣ F)$ and union questions.
Chain the draws
One after another without replacement: multiply fractions, reducing the totals each time.
Use it when: Multiplication theorem questions.

Conditional Probability: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: From a union to conditional probabilities · 2 marks · Very short answerBoard level

If $P (A) = 0.7$ , $P (B) = 0.5$ and $P (A \cup B) = 0.9$ , find $P (A ∣ B)$ and $P (B ∣ A)$ .

Try it first, then show the solutionSolution

$P (A \cap B) = 0.7 + 0.5 - 0.9 = 0.3$ . (1 mark)
$P (A ∣ B) = \frac{0.3}{0.5} = 0.6$ ; $P (B ∣ A) = \frac{0.3}{0.7} = \frac{3}{7}$ . (1 mark)

Answer: $P (A ∣ B) = 0.6$ , $P (B ∣ A) = \frac{3}{7}$

Example 2: Three coins with a condition · 2 marks · Very short answerBoard level

Three fair coins are tossed. Find the probability of getting three heads given that at least two heads appear, and the probability of at least two heads given that the first coin shows a head.

Try it first, then show the solutionSolution

At least two heads: ${H H T, H T H, T H H, H H H}$ , 4 outcomes; three heads is 1 of them: $\frac{1}{4}$ . (1 mark)
First coin head: ${H H H, H H T, H T H, H T T}$ ; at least two heads in 3 of them: $\frac{3}{4}$ . (1 mark)

Answer: $\frac{1}{4}$ and $\frac{3}{4}$

Example 3: Reduced sample space with two dice · 3 marks · Short answerHarder board level

Two dice are thrown. Find the probability that at least one die shows a 6, given that the sum of the numbers is at least 10.

Try it first, then show the solutionSolution

F (sum at least 10) $= {(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}$ , so $n (F) = 6$ . (1 mark)
E (at least one 6) inside F: $(4, 6), (6, 4), (5, 6), (6, 5), (6, 6)$ , so $n (E \cap F) = 5$ . (1 mark)
$P (E ∣ F) = \frac{5}{6}$ . (1 mark)

Answer: $\frac{5}{6}$

Example 4: Multiplication theorem for three draws · 3 marks · Short answerHarder board level

A bag has 4 red and 6 blue balls. Three balls are drawn one after another without replacement. Find the probability that they are red, blue, red in that order, and the probability that exactly two are red.

Try it first, then show the solutionSolution

$P (R_{1} \cap B_{2} \cap R_{3}) = \frac{4}{10} \times \frac{6}{9} \times \frac{3}{8} = \frac{72}{720} = \frac{1}{10}$ . (1½ marks)
Exactly two red can come in 3 orders (RRB, RBR, BRR), each with the same product $\frac{4 \cdot 3 \cdot 6}{10 \cdot 9 \cdot 8} = \frac{1}{10}$ . (1 mark)
So $P (exactly two red) = \frac{3}{10}$ . (½ mark)

Answer: $\frac{1}{10}$ ; $\frac{3}{10}$

Example 5: Using the complement inside a condition · 3 marks · Short answerHOTS / competency

For events $A$ and $B$ , $P (A) = 0.5$ , $P (B) = 0.4$ and $P (A^{'} ∣ B) = 0.3$ . Find $P (A \cap B)$ and $P (A ∣ B^{'})$ .

Try it first, then show the solutionSolution

$P (A ∣ B) = 1 - P (A^{'} ∣ B) = 0.7$ , so $P (A \cap B) = P (B) P (A ∣ B) = 0.4 \times 0.7 = 0.28$ . (1 mark)
$P (A \cap B^{'}) = P (A) - P (A \cap B) = 0.5 - 0.28 = 0.22$ , and $P (B^{'}) = 0.6$ . (1 mark)
$P (A ∣ B^{'}) = \frac{0.22}{0.6} = \frac{11}{30}$ . (1 mark)

Answer: $P (A \cap B) = 0.28$ , $P (A ∣ B^{'}) = \frac{11}{30}$

Example 6: Case study: games survey · 4 marks · Case-basedHOTS / competency

Suppose a survey of 200 students finds that 120 play cricket, 90 play football and 50 play both. A student is chosen at random.
(i) Find the probability that the student plays football, given that they play cricket. (1 mark)
(ii) Find the probability that the student plays cricket, given that they play football. (1 mark)
(iii) Find the probability that the student plays neither game. (1 mark)
(iv) Find the probability that the student plays football, given that they do not play cricket. (1 mark)

Try it first, then show the solutionSolution

(i) $P (F ∣ C) = \frac{50}{120} = \frac{5}{12}$ . (1 mark)
(ii) $P (C ∣ F) = \frac{50}{90} = \frac{5}{9}$ . (1 mark)
(iii) At least one game: $120 + 90 - 50 = 160$ ; neither: $40$ , so $\frac{40}{200} = \frac{1}{5}$ . (1 mark)
(iv) Not cricket: $80$ students, of whom football only: $90 - 50 = 40$ . $P (F ∣ C^{'}) = \frac{40}{80} = \frac{1}{2}$ . (1 mark)

Answer: (i) $\frac{5}{12}$ (ii) $\frac{5}{9}$ (iii) $\frac{1}{5}$ (iv) $\frac{1}{2}$

Where students lose marks in Conditional Probability

The slips examiners see most often, and what to write instead.

Mistake: Dividing by the wrong event
$P (E ∣ F)$ divides by $P (F)$ , the given event, not by $P (E)$ .
Wrong working
$P (A ∣ B) = \frac{0.3}{0.7}$ when $P (A) = 0.7$ , $P (B) = 0.5$ .
Right working
$P (A ∣ B) = \frac{0.3}{0.5} = 0.6$ .
Do this: Underline the event after the bar; it goes in the denominator.
Usually costs: The whole answer.
Mistake: Counting outside the reduced space
Once F is known, only outcomes in F are possible.
Wrong working
Two dice, sum at least 10: at least one 6 has 11 outcomes, so $\frac{11}{36}$ .
Right working
Only 6 outcomes have sum at least 10, and 5 of them contain a 6: $\frac{5}{6}$ .
Do this: List F first, then count inside it.
Usually costs: 1 to 2 marks.
Mistake: Not reducing totals without replacement
Each draw removes a ball, so both the favourable count and the total fall.
Wrong working
$\frac{4}{10} \times \frac{6}{10} \times \frac{4}{10}$ .
Right working
$\frac{4}{10} \times \frac{6}{9} \times \frac{3}{8} = \frac{1}{10}$ .
Do this: Write what is left in the bag before each fraction.
Usually costs: Most of the marks.
Mistake: Writing $P (E^{'} ∣ F) = 1 - P (E^{'} ∣ F^{'})$ or similar mixed rules
The complement rule only works with the same condition.
Wrong working
$P (A ∣ B^{'}) = 1 - P (A ∣ B)$ .
Right working
$P (A^{'} ∣ B) = 1 - P (A ∣ B)$ ; $P (A ∣ B^{'})$ must be found from $\frac{P ( A \cap B ^{'} )}{P ( B ^{'} )}$ .
Do this: Keep the bar's right side fixed when taking complements.
Usually costs: 1 mark.

Exam tips

A 1-mark conditional probability question appeared in every board paper from 2023 to 2026 (for example 2023 Q12, 2024 Q11, 2025 Q17, 2026 Q18); most need only $P (A \cap B)$ and one division.
Write the formula $P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )}$ before substituting; it earns the method mark.
In case studies with a table or Venn diagram, convert counts to probabilities only at the end; ratios of counts give the same answer.
For 'one after another' questions, say 'without replacement' and show the reduced totals in each factor.
Leave answers as fractions in lowest terms unless decimals are given in the question.

Practice questions on Conditional Probability

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $P (B) = 0.5$ and $P (A \cap B) = 0.2$ , then $P (A ∣ B)$ is:
(a) 0.1
(b) 0.4
(c) 0.7
(d) 2.5
Hint
Divide by the probability of the given event.
Answer
(b) $\frac{0.2}{0.5} = 0.4$ .
Q2 · 2 marks · Very short answerBoard level
A die is thrown. Given that the number shown is greater than 3, find the probability that it is (i) even (ii) prime.
Hint
The reduced sample space is ${4, 5, 6}$ .
Answer
(i) Even in ${4, 5, 6}$ : 4 and 6, so $\frac{2}{3}$ . (ii) Prime: only 5, so $\frac{1}{3}$ .
Q3 · 3 marks · Short answerHarder board level
For events $A$ and $B$ , $P (A) = 0.4$ , $P (B ∣ A) = 0.5$ and $P (B ∣ A^{'}) = 0.25$ . Find $P (B)$ and $P (A ∣ B)$ .
Hint
$P (B) = P (A \cap B) + P (A^{'} \cap B)$ .
Answer
$P (A \cap B) = 0.4 \times 0.5 = 0.2$ , $P (A^{'} \cap B) = 0.6 \times 0.25 = 0.15$ , so $P (B) = 0.35$ and $P (A ∣ B) = \frac{0.2}{0.35} = \frac{4}{7}$ .
Q4 · 2 marks · Very short answerHarder board level
Two cards are drawn one after the other without replacement from a well-shuffled pack of 52. Find the probability that the second card is an ace given that the first is an ace, and the probability that both are aces.
Hint
After one ace is drawn, 3 aces remain among 51 cards.
Answer
$P (2nd ace ∣ 1st ace) = \frac{3}{51} = \frac{1}{17}$ ; $P (both) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For any events E and F with $P (F) \neq = 0$ , $P (E ∣ F) + P (E^{'} ∣ F) = 1$ .
Reason (R): $P (E ∣ F) = \frac{P ( E )}{P ( F )}$ .
(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.
Hint
Check the numerator in R.
Answer
(c). A is a property of conditional probability. R is false: the numerator should be $P (E \cap F)$ , not $P (E)$ .
Q6 · 2 marks · Very short answerHarder board level
From a group of 5 boys and 4 girls, three children are chosen one after another for a relay team. Find the probability that the first two are boys and the third is a girl.
Hint
Multiplication theorem with shrinking totals.
Answer
$\frac{5}{9} \times \frac{4}{8} \times \frac{4}{7} = \frac{80}{504} = \frac{10}{63}$ .
Q7 · 3 marks · Short answerHOTS / competency
If $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{3}$ and $P (A ∣ B) = \frac{3}{4}$ , find $P (A \cup B)$ and $P (A^{'} ∣ B^{'})$ .
Hint
$P (A^{'} \cap B^{'}) = 1 - P (A \cup B)$ .
Answer
$P (A \cap B) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$ . $P (A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12}$ . $P (A^{'} ∣ B^{'}) = \frac{1 - \frac{7}{12}}{1 - \frac{1}{3}} = \frac{5/12}{2/3} = \frac{5}{8}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Formula for $P (E ∣ F)$ ?Show answer
$\frac{P ( E \cap F )}{P ( F )}$ , with $P (F) \neq = 0$ .
2.What is $P (F ∣ F)$ ?Show answer
1.
3. $P (E^{'} ∣ F)$ in terms of $P (E ∣ F)$ ?Show answer
$1 - P (E ∣ F)$ .
4.State the multiplication theorem for two events.Show answer
$P (E \cap F) = P (E) P (F ∣ E) = P (F) P (E ∣ F)$ .
5.A die shows an odd number. Probability it is 3?Show answer
$\frac{1}{3}$ (the reduced space is ${1, 3, 5}$ ).
6.If $P (A \cap B) = 0.2$ and $P (A) = 0.5$ , what is $P (B ∣ A)$ ?Show answer
0.4.

Conditional Probability: frequently asked questions

What is conditional probability in simple words?

It is the probability of an event after you are told that another event has happened. The known event becomes the new sample space, so you only count outcomes inside it.

Is P(A|B) the same as P(B|A)?

No. Both have P(A ∩ B) on top, but P(A|B) divides by P(B) and P(B|A) divides by P(A). They are equal only when P(A) = P(B) or P(A ∩ B) = 0.

When do I use the multiplication theorem?

Use it when events happen one after another and the earlier outcome affects the later one, such as drawing balls or cards without replacement. Multiply the first probability by the conditional probabilities that follow.

Which NCERT sections cover conditional probability?

Section 13.2 (Conditional Probability), including 13.2.1 on its properties, and Section 13.3 (Multiplication Theorem on Probability), in Chapter 13 of NCERT Mathematics Part II for Class 12.

Can a conditional probability be greater than 1?

No. Since E ∩ F is part of F, P(E ∩ F) is at most P(F), so P(E|F) always lies between 0 and 1. An answer above 1 means the wrong denominator was used.