An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the

Exercise 13.3 · Q1

Q. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Punjab PsebTextbookSubjective· 3mImportance★★★★★

Appeared in past exams:COMEDK 2021· Set 2021-B· 1mreworded

27% · 45/165 Questions

✓ Free question

Concept understanding — Conditional Probability

Conditional Probability

Roll a die and ask "what is the chance of an even number?" — that is $3/6$ . But suppose someone tells you the result is greater than 3. Now you are no longer looking at all six faces, only at $\{4,5,6\}$ , and two of those (4 and 6) are even, so the probability becomes $2/3$ . That change — from the probability of $A$ to the probability of $A$ given that $B$ has already occurred — is conditional probability.

The Idea: Shrink the Sample Space

Conditioning on $B$ throws away every outcome where $B$ is false and treats $B$ as the new "whole world." You measure $A$ only against what is still possible.

Note

Think of filtering a table of data: unconditional probability uses every row; conditional probability keeps only the rows where the condition is true.

The Definition

For events $A$ and $B$ with $P(B) > 0$ ,

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

We divide by $P(B)$ to rescale so that $B$ itself has probability $1$ ; the surviving part of $A$ is the overlap $A \cap B$ . Checking the die: $P(A \cap B) = P(\{4,6\}) = \tfrac{2}{6}$ and $P(B) = \tfrac{3}{6}$ , so $P(A \mid B) = \dfrac{2/6}{3/6} = \dfrac{2}{3}$ , matching the intuition.

Tip

Rearranging gives the multiplication rule $P(A \cap B) = P(A \mid B)\,P(B)$ , which is usually the easier way to compute a joint probability when a problem says "given that."

Two Cautions

$P(A \mid B)$ and $P(B \mid A)$ are generally not equal; swapping them is the classic mistake. They are linked by Bayes' theorem, $P(A \mid B) = \dfrac{P(B \mid A)\,P(A)}{P(B)}$ .
If $P(A \mid B) = P(A)$ , then knowing $B$ tells you nothing about $A$ — the events are independent. That is a special case, not the general rule.

Conditional probability is the foundation of the multiplication theorem, independence, and Bayes' theorem — every "given that" question in this chapter rests on it.

Conditional Probability opens the CBSE Class 12 Probability chapter and is foundational for everything that follows in that unit, including Bayes' theorem and the multiplication rule — making "conditional probability formula class 12 with examples" one of the most searched topics in Class 12 Mathematics. It is equally important for JEE Main and CUET, where conditional probability questions are set almost every year.

Condition on the first draw's colour: the returned ball plus $2$ same-colour balls make $12$ in the urn, giving $P(R_2\mid R_1)=\tfrac{7}{12}$ and $P(R_2\mid B_1)=\tfrac{5}{12}$ ; the total probability is $\tfrac12$ .

Why we condition

The urn's make-up before the second draw depends on the colour of the first draw, so we handle the two cases separately and combine them with the law of total probability.

First draw

The urn starts with $5$ red and $5$ black ( $10$ balls), so

P(R_1)=\frac{5}{10}=\frac12,\qquad P(B_1)=\frac12.

Rebuild the urn (the drawn ball is returned)

The drawn ball is put back, and then $2$ extra balls of the same colour are added. Either way the urn now holds $10+2=12$ balls.

First red: back to $5$ red and $5$ black, then $+2$ red $\Rightarrow 7$ red, $5$ black.

P(R_2\mid R_1)=\frac{7}{12}.

First black: back to $5$ red and $5$ black, then $+2$ black $\Rightarrow 5$ red, $7$ black.

P(R_2\mid B_1)=\frac{5}{12}.

Law of total probability

P(R_2)=P(R_1)P(R_2\mid R_1)+P(B_1)P(R_2\mid B_1)=\frac12\cdot\frac{7}{12}+\frac12\cdot\frac{5}{12}=\frac{7+5}{24}=\frac{12}{24}=\frac12.

Tip

Because the urn starts with equal colours, the two conditional probabilities $\tfrac{7}{12}$ and $\tfrac{5}{12}$ are symmetric about $\tfrac12$ and average back to $\tfrac12$ .

✓Final answer

P(\text{second ball is red})=\dfrac12

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CA Foundation 2026Set jan-20261 markMCQ
$\frac{3}{7}$
›Reveal solutionSolution

Conditional probability $P(S\mid M) = \dfrac{P(M \cap S)}{P(M)}$ .

Step 1 — identify the probabilities

50% study both maths and science, so $P(M \cap S) = 0.5$ ; 70% study maths, so $P(M) = 0.7$ .

Step 2 — apply the conditional-probability formula
$P(S \mid M) = \frac{P(M \cap S)}{P(M)} = \frac{0.5}{0.7} = \frac{5}{7}.$
Watch out

Divide by the given event's probability: since maths is given, the denominator is $P(M)=0.7$ , not $P(S)$ or the total. Dividing the other way ( $0.7/0.5$ ) gives a value above 1, which is impossible for a probability.

Tip

"Given that" tells you the denominator. Here it is "given studying mathematics," so put $P(M)$ on the bottom: $0.5/0.7 = 5/7$ .

✓Final answer
$(D) 5/7$
CA Foundation 2026Set jan-20261 markMCQ
$\frac{1}{2}$
›Reveal solutionSolution

Conditional probability on a reduced sample space: $P(A\mid B) = \dfrac{n(A \cap B)}{n(B)}$ .

Step 1 — list the outcomes with sum 7.
$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \Rightarrow n(B) = 6.$
Step 2 — count first die greater than second, among those.

$(4,3),(5,2),(6,1)$ → 3 outcomes.

Step 3 — conditional probability.
$P = \frac{3}{6} = \frac{1}{2}.$
Watch out

The condition "sum = 7" shrinks the sample space to those 6 outcomes — divide by 6, not by the full 36. Using $3/36$ gives $1/12$ , which isn't even an option.

Tip

None of the sum-7 pairs are ties, so by symmetry "first > second" and "first < second" split the 6 outcomes evenly — the answer is simply half.

✓Final answer
$(A) 1/2$
CA Foundation 2023Set jun-20231 markMCQ
$P(A)=\dfrac{1}{3}, P(B)=\dfrac{1}{4}, P(A/B)=\dfrac{1}{6}$
›Reveal solutionSolution

P(B/A) = P(A∩B)/P(A) = (1/24)/(1/3) = 1/8.

Step 1 — Find the joint probability
$P(A\cap B) = P(A/B)\,P(B) = \frac{1}{6}\times\frac{1}{4} = \frac{1}{24}$
Step 2 — Apply the definition of conditional probability
$P(B/A) = \frac{P(A\cap B)}{P(A)} = \frac{1/24}{1/3} = \frac{3}{24} = \frac{1}{8}$
Watch out

P(A/B) and P(B/A) are not equal — you must recompute the joint probability first, then divide by P(A), not P(B).

Tip

Anchor everything on P(A∩B): both conditionals flow from it via division by the conditioning event's probability.

✓Final answer
$\dfrac{1}{8}$

Note

This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
CA Foundation 2022Set dec-20221 markMCQ
$P(A)=\dfrac{1}{3}$
›Reveal solutionSolution

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

Step 1 — Intersection via the addition rule
$P(A\cap B)=P(A)+P(B)-P(A\cup B)=\frac13+\frac34-\frac{11}{12}=\frac{4+9-11}{12}=\frac{2}{12}=\frac16$
Step 2 — Apply the conditional-probability formula
$P\!\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}=\frac{1/6}{1/3}=\frac12$
Watch out

Option (A) 1/6 is just P(A∩B) — you must still divide by P(A) to get the conditional probability.

Tip

Conditional probability always divides the joint probability by the probability of the given (conditioning) event.

✓Final answer
$\dfrac{1}{2}$

Note

This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
CA Foundation 2021Set dec-20211 markMCQ
$A$
›Reveal solutionSolution

Divide the joint probability by the conditioning event's probability: $P(A\mid B)=5/9$ , $P(B\mid A)=6/11$ .

Step 1 — Apply the conditional probability formula for $P(A\mid B)$
$P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{10/33}{6/11} = \frac{10}{33}\times\frac{11}{6} = \frac{110}{198} = \frac{5}{9}$
Step 2 — Apply it for $P(B\mid A)$
$P(B\mid A) = \frac{P(A\cap B)}{P(A)} = \frac{10/33}{5/9} = \frac{10}{33}\times\frac{9}{5} = \frac{90}{165} = \frac{6}{11}$
Step 3 — Sanity check

Since $P(A)P(B) = (5/9)(6/11) = 10/33 = P(A\cap B)$ , the conditionals collapse to the marginals — consistent with the computed values.

Watch out

The trap is to multiply $P(A\cap B)$ by $P(B)$ instead of dividing (giving small fractions like 1/9, 2/9 in options C/D). Always divide by the given/conditioning event.

Tip

$P(A\mid B)$ = joint over the second letter's probability; $P(B\mid A)$ = joint over the first letter's probability.

✓Final answer
$(A) 5/9, 6/11$

Note

This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.
CA Foundation 2021Set dec-20211 markMCQ
$Q. In a group of 20 males and 15 females, 12 males and 8 females are service holders. What is the probability that a person selected at random from the group is a service holder given that the selected person is a male? (A) 0.40 (B) 0.60 (C) 0.45 (D) 0.55$
›Reveal solutionSolution

Condition on males only: 12 service holders out of 20 males = 0.60.

Step 1 — Identify the reduced sample space

Given the person is male, only the 20 males matter.

Step 2 — Apply the conditional formula
$P(\text{service}\mid \text{male}) = \frac{\text{male service holders}}{\text{total males}} = \frac{12}{20} = 0.60$
Watch out

Do not divide by the full group of 35 — the condition 'given male' shrinks the denominator to 20.

Tip

'Given that ...' problems: throw away everyone outside the given category, then take the simple fraction.

✓Final answer
$(B) 0.60$

Note

This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.

←Previous question · 13.5.2Example 21 — A man is known to speak truth 3 out of 4 times. He throws a die and re…→Next question · 13.5.2Q2 — A bag contains 4 red and 4 black balls, another bag contains 2 red and…

Q.An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

Why we condition

First draw

Rebuild the urn (the drawn ball is returned)

Law of total probability

Method: Law of Total Probability (conditioning on the first stage)

Steps

Common Mistakes

Step 1 — identify the probabilities

Step 2 — apply the conditional-probability formula

Step 1 — list the outcomes with sum 7.

Step 2 — count first die greater than second, among those.

Step 3 — conditional probability.

Step 1 — Find the joint probability

Step 2 — Apply the definition of conditional probability

Step 1 — Intersection via the addition rule

Step 2 — Apply the conditional-probability formula

Step 1 — Apply the conditional probability formula for $P(A\mid B)$

Step 2 — Apply it for $P(B\mid A)$

Step 3 — Sanity check

Step 1 — Identify the reduced sample space

Step 2 — Apply the conditional formula

Q.An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Why we condition

First draw

Rebuild the urn (the drawn ball is returned)

Law of total probability

🎓Unlock everything free for 14 days

Unlock everything free for 14 days