(a) The probability distribution of a random variable $X$ is given below: $X$: 1, 2, 3 with $P(X)$: $\frac{k}{2}$, $\frac{k}{3}$, $\frac{k}{6}$ respectively. (i) Find the value of $k$. (ii) Find $P(1 \leq X < 3)$. (iii) Find $E(X)$, the mean of $X$. OR (b) $A$ and $B$ are independent events such that $P(A \cap \bar{B}) = \frac{1}{4}$ and $P(\bar{A} \cap B) = \frac{1}{6}$. Find $P(A)$ and $P(B)$.

(a) The probability distribution of a random variable X is given below: X: 1, 2, 3 with P(X): (k)/(2), (k)/(3)

Question

X

(i) Find the value of

k

(ii) Find

P(1 \leq X < 3)

(iii) Find

E(X)

, the mean of

X

(OR)

(b)

A

and

B

are independent events such that

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

and

P(\bar{A} \cap B) = \frac{1}{6}

. Find

P(A)

and

P(B)

CBSECBSE Class XII Board 2023Subjective· 3mImportance★★★★★

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Part (a)Concept understanding — Probability Complement Rule

The Probability Complement Rule

Every event $A$ splits the sample space in two: outcomes where $A$ happens, and outcomes where it does not. The second group is the complement of $A$ , written $A'$ (also $A^c$ or $\bar{A}$ ). Because exactly one of the two must occur, their probabilities together fill the whole space:

P(A) + P(A') = 1 \quad\Longrightarrow\quad P(A') = 1 - P(A).

That is the complement rule: the probability that $A$ does not happen is $1$ minus the probability that it does.

Why It Holds

$A$ and $A'$ are mutually exclusive (no outcome lies in both) and exhaustive (together they are the entire sample space $S$ , with $P(S) = 1$ ). So $P(A) + P(A') = P(S) = 1$ , and rearranging gives the rule.

A Simple Example

For a fair die, $P(\text{six}) = \tfrac{1}{6}$ , so $P(\text{not six}) = 1 - \tfrac{1}{6} = \tfrac{5}{6}$ .

Why It Is So Useful: the "At Least One" Trick

Counting "at least one" directly often means adding many separate cases, while its complement, "none," is a single easy case. For instance, the probability of at least one head in three tosses of a fair coin:

P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \left(\tfrac{1}{2}\right)^3 = 1 - \tfrac{1}{8} = \tfrac{7}{8}.

Computing "no heads" once is far quicker than summing the one-head, two-head and three-head cases separately. …

Part (b)Concept understanding — Event Independence

Event Independence

Two events are independent when the occurrence of one does not change the probability of the other. Toss a coin and roll a die: the coin landing heads tells you nothing about whether the die shows a six. Contrast this with drawing cards without replacement, where the first draw does change the odds for the second — those events are dependent.

From Conditional Probability to a Clean Test

"Knowing $B$ doesn't change $A$ " means $P(A \mid B) = P(A)$ . Substituting the definition $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ and clearing the fraction gives the symmetric form used in practice:

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

Events $A$ and $B$ are independent exactly when the probability of both occurring equals the product of their individual probabilities. This version is preferred because it needs no non-zero condition and treats $A$ and $B$ alike.

A Quick Check

Roll a fair die. Let $A = \{2,4,6\}$ (even) and $B = \{4,5,6\}$ (greater than 3). Then $P(A) = P(B) = \tfrac{1}{2}$ , and $A \cap B = \{4,6\}$ so $P(A \cap B) = \tfrac{1}{3}$ . Since $\tfrac{1}{3} \neq \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4}$ , these events are not independent.

Three or More Events

Events $A, B, C$ are mutually independent only if all four conditions hold: the three pairwise products and

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

Pairwise independence alone is not enough to guarantee mutual independence. …

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$k=1$ , $P(1\le X<3)=\tfrac56$ , $E(X)=\tfrac53$ .
$P(A)=\tfrac34,P(B)=\tfrac23$ or $P(A)=\tfrac13,P(B)=\tfrac14$ .

Part (a)

A probability distribution obeys $\sum P=1$ , and $E(X)=\sum x\,P(x)$ .

Find $k$ : $\dfrac{k}{2}+\dfrac{k}{3}+\dfrac{k}{6}=\dfrac{3k+2k+k}{6}=k=1\Rightarrow k=1.$ So $P(1)=\tfrac12,P(2)=\tfrac13,P(3)=\tfrac16$ .

$P(1\le X<3)$ means $X=1$ or $2$ : $\tfrac12+\tfrac13=\tfrac56.$ …

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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 67 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$A$
›Reveal solutionSolution
We need to find $P(A'|B')$ . By applying the definition of conditional probability, De Morgan's Law, and the complement rule, we can express this as $\frac{1 - P(A \cup B)}{P(B')}$ , which corresponds to option (D).

Let's break down this problem by first understanding the core concepts involved: conditional probability and the complement rule.

Conditional probability, $P(X|Y)$ , represents the probability of event $X$ occurring given that event $Y$ has already occurred. Its definition is fundamental:
$P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$
The complement rule states that the probability of an event not happening is $1$ minus the probability of it happening. If $X'$ denotes the complement of event $X$ (i.e., $X$ does not occur), then:
$P(X') = 1 - P(X)$
We are asked to find $P(A'|B')$ , which means "the probability that event $A$ does not occur, given that event $B$ does not occur."

Now, let's work through the problem step-by-step.

Apply the definition of conditional probability. Using the formula $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$ , we replace $X$ with $A'$ and $Y$ with $B'$ .

$P(A'|B') = \frac{P(A' \cap B')}{P(B')}$
The problem states $P(B) \ne 1$. This is important because it implies $P(B') = 1 - P(B) \ne 0$, ensuring that the denominator is not zero and the conditional probability is well-defined.

2. Simplify the numerator using De Morgan's Law.

The term $A' \cap B'$ represents the event where neither $A$ nor $B$ occurs. This is equivalent to the event that $A \cup B$ (either $A$ or $B$ or both occur) does not occur. This is a direct application of De Morgan's Law for sets:
$(A \cup B)' = A' \cap B'$
Therefore, we can rewrite the numerator:
$P(A' \cap B') = P((A \cup B)')$

Apply the complement rule to the numerator. Now we have $P((A \cup B)')$ . Using the complement rule $P(X') = 1 - P(X)$ , where $X$ is the event $(A \cup B)$ :

$P((A \cup B)') = 1 - P(A \cup B)$

Substitute back into the conditional probability formula. Substitute the simplified numerator back into the expression from Step 1: …
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CBSE 2026Set 65/3/11 markMCQ
$E$
›Reveal solutionSolution
We use the property of independent events, $P(E \cap F) = P(E)P(F)$ , along with the union formula to first find $P(F)$ . Then, we use the fact that for independent events, $P(E|F) = P(E)$ and $P(F|E) = P(F)$ , to calculate the required difference. The final result is $\boxed{\frac{1}{70}}$ .

The core of this problem lies in understanding how the concept of "independent events" simplifies probability calculations, especially when dealing with unions and conditional probabilities.

When two events, $E$ and $F$ , are independent, it means that the occurrence of one event does not affect the probability of the other event occurring. This has two crucial implications:

Intersection Probability: The probability of both $E$ and $F$ happening, $P(E \cap F)$ , is simply the product of their individual probabilities: $P(E \cap F) = P(E)P(F)$ .

Conditional Probability: The probability of $E$ happening given that $F$ has already happened, $P(E|F)$ , is just the probability of $E$ , because $F$ 's occurrence doesn't change $E$ 's likelihood. So, $P(E|F) = P(E)$ . Similarly, $P(F|E) = P(F)$ .

We are given $P(E)$ , $P(E \cup F)$ , and that $E$ and $F$ are independent. Our strategy will be to first use the formula for the union of events, combined with the independence property, to find $P(F)$ . Once we have $P(F)$ , we can directly use the independence property to find $P(E|F)$ and $P(F|E)$ , and then calculate their difference.

Find $P(F)$ using the union formula and independence.

The general formula for the probability of the union of two events is:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
Since $E$ and $F$ are independent, we can substitute $P(E \cap F)$ with $P(E)P(F)$ :

$P(E \cup F) = P(E) + P(F) - P(E)P(F)$

Now, substitute the given values: $P(E) = \frac{3}{10}$ and $P(E \cup F) = \frac{1}{2}$ .

$\frac{1}{2} = \frac{3}{10} + P(F) - \frac{3}{10} P(F)$

To solve for $P(F)$ , group the terms involving $P(F)$ :

$\frac{1}{2} = \frac{3}{10} + P(F) \left(1 - \frac{3}{10}\right)$

$\frac{1}{2} = \frac{3}{10} + P(F) \left(\frac{7}{10}\right)$

Subtract $\frac{3}{10}$ from both sides:

$\frac{1}{2} - \frac{3}{10} = P(F) \left(\frac{7}{10}\right)$

To subtract the fractions on the left, find a common denominator, which is $10$ :

$\frac{5}{10} - \frac{3}{10} = P(F) \left(\frac{7}{10}\right)$

$\frac{2}{10} = P(F) \left(\frac{7}{10}\right)$

Now, isolate $P(F)$ by multiplying both sides by $\frac{10}{7}$ : …
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CBSE 2026Set V11 markMCQ
$Q. The probability of obtaining an even prime number on each die when a pair of dice is rolled$
(a) $\frac{1}{36}$
(b) $\frac{1}{6}$
(c) $\frac{1}{18}$
(d) $\frac{1}{4}$
›Reveal solutionSolution

The even prime is $2$ ; $P(2\text{ on each of two dice})=\tfrac{1}{6}\cdot\tfrac{1}{6}=\tfrac{1}{36}$ ; answer (a).

The only even prime number is $2$ . For one die, $P(\text{show }2)=\dfrac{1}{6}$ . The two dice are independent, so …
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CBSE 2026Set V11 markMCQ
$A$
(a) $1.2$
(b) $0.12$
(c) $0.7$
(d) $\frac{3}{4}$
›Reveal solutionSolution

Independence gives $P(A\cap B)=P(A)P(B)=0.12$ ; answer (b).

For independent events $A$ and $B$ ,

$P(A\cap B)=P(A)\cdot P(B)=0.3\times0.4=0.12.$ …
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CBSE 2026Set A1 markMCQ
$1 - P(A' \cap B') =$
(a) $P(A \cap B)$
(b) $P(A \cup B)$
(c) $P(A)$
(d) $P(B)$
›Reveal solutionSolution

$1 - P(A'\cap B') = P(A\cup B)$ .

By De Morgan's law,

$A' \cap B' = (A\cup B)'.$

So …

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CBSE 2026Set A1 markMCQ
$A$
(a) $P(A) + P(B) + P(C)$
(b) $P(A) - P(B) - P(C)$
(c) $P(A) \cdot P(B) \cdot P(C)$
(d) None of these
›Reveal solutionSolution

For independent events, $P(A\cap B\cap C) = P(A)P(B)P(C)$ .

By definition, events $A$ , $B$ , $C$ are (mutually) independent when the probability of their joint occurrence equals the product of their individual probabilities:

$P(ABC) = P(A)\cdot P(B)\cdot P(C).$

…

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CBSE 2026Set ANNUAL1 markMCQ
$P(A) = 0.3$
(a) $0.58$
(b) $0.70$
(c) $0.12$
(d) $0.10$
›Reveal solutionSolution

For independent events, $P(A\cap B)=P(A)P(B)$ , and the addition rule gives $P(A\cup B)$ .

$P(A\cap B) = 0.3\times0.4=0.12$ (independence).

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The probability of obtaining an even prime number on each dice, when a pair of dice is rolled, is:$
(a) 0
(b) $\dfrac{1}{3}$
(c) $\dfrac{1}{12}$
(d) $\dfrac{1}{36}$
›Reveal solutionSolution

The only even prime number is 2, so we need a 2 on each die.

Among $\{1,2,3,4,5,6\}$ , the only even prime is $2$ .

$P(\text{2 on one die})=\dfrac{1}{6}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Ajay and Meera are contesting for two vacancies in a company. Probability of selection of Ajay is 7/9 and that of Meera is 4/7. What is the probability that both will be rejected?$
(a) 61/63
(b) 6/63
(c) 41/63
(d) 28/63
›Reveal solutionSolution

The rejection probabilities of each candidate are complements of their selection probabilities; since the two events are independent, multiply them.

$P(\text{Ajay selected}) = \dfrac79 \implies P(\text{Ajay rejected}) = 1-\dfrac79 = \dfrac29$

$P(\text{Meera selected}) = \dfrac47 \implies P(\text{Meera rejected}) = 1-\dfrac47 = \dfrac37$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): Two events A and B are such that P(A) = 1/4, P(B) = 1/2 and P(A\capB) = 1/8 then two events A and B are independent. Reason (R): Two events are independent if the probability of occurrence of one does not affect the probability of occurrence of other and P(A\capB) = P(A) + P(B) - P(A\cupB)$
(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

The independence check in A is numerically correct, but R states the wrong criterion — it gives the addition-rule identity (true for ANY two events), not the actual independence condition $P(A\cap B)=P(A)\cdot P(B)$ .

Checking Assertion (A): Independence requires $P(A\cap B) = P(A)\cdot P(B)$ .

$P(A)\cdot P(B) = \frac14 \times \frac12 = \frac18,$

which equals the given $P(A\cap B)=1/8$ . So A and B are independent — A is true.

Checking Reason (R): R correctly describes independence in words ("occurrence of one does not affect the other"), but then states the test as

$P(A\cap B) = P(A)+P(B)-P(A\cup B).$ …

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CBSE 2026Set ANNUAL1 markMCQ
$E$
(a) $E$ and $F$ are independent events
(b) $E$ and $F$ are mutually exclusive events
(c) $E$ and $F$ are disjoint events
(d) None of the above
›Reveal solutionSolution

Two events are independent exactly when $P(E\cap F)=P(E)\cdot P(F)$ ; check whether the given numbers satisfy this.

Given $P(E)=\dfrac13$ , $P(F)=\dfrac12$ , $P(E\cap F)=\dfrac16$ .

Test for independence:

$P(E)\cdot P(F) = \frac13\times\frac12 = \frac16$

This equals the given $P(E\cap F)=\dfrac16$ . Since $P(E\cap F)=P(E)P(F)$ , $E$ and $F$ are independent.

…

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CBSE 2026Set ANNUAL1 mark
$Q. If A and B are two independent events with P(A) = 1/2 and P(B) = 1/3, then find P(A \cup B).$
›Reveal solutionSolution

For independent events, $P(A\cap B)=P(A)P(B)$ ; then apply the addition rule.

…

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Q.(a) The probability distribution of a random variable XXX is given below: XXX: 1, 2, 3 with P(X)P(X)P(X): k2\frac{k}{2}2k​, k3\frac{k}{3}3k​, k6\frac{k}{6}6k​ respectively.

Part (a)

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Q.(a) The probability distribution of a random variable $X$ is given below: $X$ : 1, 2, 3 with $P(X)$ : $\frac{k}{2}$ , $\frac{k}{3}$ , $\frac{k}{6}$ respectively.

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