(a) Out of two bags, Bag I contains 3 red and 4 white balls and Bag II contains 8 red and 6 white balls. A die

Question

Q. (a) Out of two bags, Bag I contains 3 red and 4 white balls and Bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3, then a ball is drawn at random from Bag I, otherwise a ball is drawn at random from Bag II. Find the probability that the ball drawn is a red ball.

(OR)

(b) The probability of simultaneous occurrence of at least one of two events

X

and

Y

a

. If the probability that exactly one of

X

Y

occurs is

b

, then prove that

P(X') + P(Y') = 2 - 2a + b

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

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Part (a)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, $P(A \mid B) = \dfrac{P(B \mid A)\,P(A)}{P(B)}$ . …

Part (b)Concept understanding — Inclusion Exclusion Principle

The Inclusion–Exclusion Principle

Thirty students play cricket and twenty-five play football — so does "cricket or football" total $55$ ? Not if some play both. If ten play both, they have been counted twice, so the real total is $30 + 25 - 10 = 45$ . That single correction is the whole principle: add everything, then subtract what you double-counted.

Why "Inclusion" and "Exclusion"

Include every set by adding its size.
Exclude the overlaps counted more than once by subtracting the intersections.

For three or more sets the signs keep alternating, because each correction slightly over-corrects and must itself be fixed.

The General Statement

For finite sets $A_1, A_2, \dots, A_n$ ,

\left| \bigcup_{i=1}^n A_i \right| = \sum_{i} |A_i| - \sum_{i<j} |A_i \cap A_j| + \sum_{i<j<k} |A_i \cap A_j \cap A_k| - \cdots + (-1)^{n+1}\,|A_1 \cap \cdots \cap A_n|.

Add all singles, subtract all pairwise intersections, add all triples, and so on — the sign alternates, starting with $+$ .

Why It Works

An element lying in exactly $m$ of the sets is counted $\binom{m}{1}$ times among the singles, subtracted $\binom{m}{2}$ times among the pairs, added $\binom{m}{3}$ times among the triples, and so on. Its net count is

\binom{m}{1} - \binom{m}{2} + \binom{m}{3} - \cdots + (-1)^{m+1}\binom{m}{m} = 1,

using $(1-1)^m = 0$ from the binomial theorem. So every element is counted exactly once.

A Three-Set Example

Of $100$ students: $40$ take Physics, $30$ Chemistry, $25$ Biology; $15$ take P and C, $10$ P and B, $8$ C and B; $5$ take all three.

|P \cup C \cup B| = (40+30+25) - (15+10+8) + 5 = 95 - 33 + 5 = 67.

So $67$ take at least one subject and $33$ take none. …

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$P(\text{red})=\dfrac{11}{21}$ .
Proved $P(X')+P(Y')=2-2a+b$ .

Part (a)

The die decides the bag. "Number less than $3$ " means $\{1,2\}$ , so $P(\text{Bag I})=\dfrac{2}{6}=\dfrac13$ and $P(\text{Bag II})=\dfrac46=\dfrac23$ .

Bag I has $3$ red of $7$ : $P(\text{red}\mid I)=\dfrac37$ .
Bag II has $8$ red of $14$ : $P(\text{red}\mid II)=\dfrac{8}{14}=\dfrac47$ .

By the law of total probability: …

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

Showing the 12 most recent of 105 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\dfrac{2}{3}$
›Reveal solutionSolution
The assertion is true: given the outcome is odd, the probability it is a prime is $\frac{2}{3}$ . The reason states the correct conditional probability formula. Since the reason directly justifies the calculation in the assertion, both are true and the reason is the correct explanation.

Concept first — Conditional probability asks: If we already know that event $B$ has occurred, what is the probability that event $A$ also occurs? The sample space shrinks from all possible outcomes to just those in $B$ . The formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ is the precise way to compute this reduced probability.

Here, the die is unbiased, so each face $\{1,2,3,4,5,6\}$ has probability $\frac{1}{6}$ . The assertion involves two events:

$A$ : the number is prime. On a die, the primes are $2,3,5$ .

$B$ : the number is odd. The odd numbers are $1,3,5$ .

The condition "given that the number is odd" means we restrict attention to $B = \{1,3,5\}$ . Among these three equally likely outcomes, the primes are $3$ and $5$ — that's two out of three. So the conditional probability is $\frac{2}{3}$ .

Now let's verify step by step using the formula in Reason (R).

Define the events precisely.

$A = \{2,3,5\}$ , $B = \{1,3,5\}$ .

The sample space $S = \{1,2,3,4,5,6\}$ .

Compute $P(B)$ .

$B$ has 3 outcomes, each with probability $\frac{1}{6}$ , so $P(B) = \frac{3}{6} = \frac{1}{2}$ .

Compute $P(A \cap B)$ .

$A \cap B$ = numbers that are both prime and odd = $\{3,5\}$ . That's 2 outcomes, so $P(A \cap B) = \frac{2}{6} = \frac{1}{3}$ .

Apply the formula from Reason (R).

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/3}{1/2} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3}.$
This matches the assertion exactly. …
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CBSE 2026Set V11 mark
$[0, 3, -1, 2, -2, 1]$
›Reveal solutionSolution

Since $S\cap F=F$ , the conditional probability $P(S\mid F)=1$ .

By the definition of conditional probability (with $P(F)\neq0$ ),

$P(S\mid F)=\frac{P(S\cap F)}{P(F)}.$ …
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CBSE 2026Set CX1 markMCQ
$3P(A)=P(B)=\dfrac{5}{13}$
(a) $\dfrac{20}{39}$
(b) $\dfrac{16}{39}$
(c) $\dfrac{11}{39}$
(d) $\dfrac{14}{39}$
›Reveal solutionSolution

Using $P(A\cap B)=P(A/B)P(B)$ and the addition rule gives $P(A\cup B)=\dfrac{14}{39}$ — option (d).

Given: $3P(A)=P(B)=\dfrac{5}{13}$ and $P(A/B)=\dfrac{2}{5}$ .

So $P(B)=\dfrac{5}{13}$ and $P(A)=\dfrac{1}{3}\cdot\dfrac{5}{13}=\dfrac{5}{39}$ .

Intersection (multiplication rule):

$P(A\cap B)=P(A/B)\,P(B)=\dfrac{2}{5}\cdot\dfrac{5}{13}=\dfrac{2}{13}.$

…

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CBSE 2026Set A1 markMCQ
$P(A) = \frac{7}{13},\ P(B) = \frac{9}{13},\ P(A \cap B) = \frac{4}{13} \Rightarrow P(A/B) =$
(a) $\frac{4}{9}$
(b) $\frac{4}{7}$
(c) $\frac{12}{13}$
(d) $\frac{1}{6}$
›Reveal solutionSolution

$P(A|B) = \frac{4}{9}$ .

Use the conditional-probability definition:

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

Substitute the given values: …

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CBSE 2026Set ANNUAL1 markMCQ
$P(B)=0.5$
(a) $\frac{15}{23}$
(b) $\frac{16}{25}$
(c) $\frac{16}{27}$
(d) $\frac{16}{23}$
›Reveal solutionSolution

By the definition of conditional probability, $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{16}{25}$ .

The conditional probability of $A$ given $B$ is defined as

$P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B) eq0$

…

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CBSE 2026Set ANNUAL1 mark
$Q. A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?$
›Reveal solutionSolution

List the equally likely outcomes for two children, restrict to those with at least one boy, then find the fraction that are both boys.

Sample space $=\{BB, BG, GB, GG\}$ , each equally likely.

Given at least one boy: reduced sample space $=\{BB,BG,GB\}$ (3 outcomes).

…

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CBSE 2026Set ANNUAL1 markMCQ
$P(A) = 0.8$
(a) 0.8
(b) 0.5
(c) 0.32
(d) 0.4
›Reveal solutionSolution

Use the multiplication rule of conditional probability: $P(A\cap B) = P(B\mid A)\cdot P(A)$ .

Given $P(A)=0.8$ , $P(B\mid A)=0.4$ .

$P(A\cap B) = P(B\mid A)\cdot P(A) = 0.4 \times 0.8 = 0.32$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$P(A) = \dfrac{3}{10}$
(a) $\dfrac{1}{4}$
(b) $\dfrac{1}{3}$
(c) $\dfrac{5}{12}$
(d) $\dfrac{7}{12}$
›Reveal solutionSolution

Find $P(A\cap B)$ from the addition rule, then use the conditional probability formula.

Given $P(A)=\dfrac{3}{10}$ , $P(B)=\dfrac{2}{5}$ , $P(A\cup B)=\dfrac{3}{5}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$A$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution

The value $P=0.2$ is correctly computed, but the stated definition of "mutually exclusive" is wrong — that condition actually defines independence.

Assertion (A): For mutually exclusive events, $P(A\cup B)=P(A)+P(B)$ .

$0.6=0.4+P \Rightarrow P=0.2$ . So A is true.

…

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CBSE 2026Set ANNUAL1 mark
$\dfrac{3}{10}, \dfrac{1}{5}, \dfrac{1}{10}$
›Reveal solutionSolution

The problem statement itself states that if the doctor comes by other means, he will never be late.

Let $E_1,E_2,E_3,E_4$ denote the events that the doctor comes by train, bus, scooter, or other means, with $P(E_1)=\dfrac{3}{10}$ , $P(E_2)=\dfrac{1}{5}$ , $P(E_3)=\dfrac{1}{10}$ , $P(E_4)=\dfrac{2}{5}$ (these sum to 1, confirming they form a complete set of cases).

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If P(A) = 1/2, P(B) = 3/8 and P(A\cupB) = 27/40 then P(A/B) is equal to:$
(a) 2/5
(b) 8/15
(c) 2/3
(d) 5/8
›Reveal solutionSolution

First find $P(A\cap B)$ using the addition rule, then apply the conditional probability formula $P(A/B)=\frac{P(A\cap B)}{P(B)}$ .

Given $P(A) = \frac12$ , $P(B)=\frac38$ , $P(A\cup B) = \frac{27}{40}$ .

Using the addition rule $P(A\cup B) = P(A)+P(B)-P(A\cap B)$ :

$P(A\cap B) = P(A)+P(B)-P(A\cup B) = \frac{20}{40}+\frac{15}{40}-\frac{27}{40} = \frac{8}{40} = \frac15$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4 then P(A\capB) = 0.32 Reason (R): Conditional Probability of 'B' when event A has occurred is given by P(B/A) = P(A\capB) / P(A)$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Both Assertion (A) and Reason (R) are false.
›Reveal solutionSolution

Applying the conditional-probability formula in R directly to the given numbers reproduces the value claimed in A.

Reason (R): $P(B/A) = \dfrac{P(A\cap B)}{P(A)}$ is the standard definition of conditional probability — true.

Checking Assertion (A): Given $P(A)=0.8,\ P(B/A)=0.4$ . Rearranging R's formula:

$P(A\cap B) = P(B/A)\times P(A) = 0.4 \times 0.8 = 0.32.$ …

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Part (a)

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