If 2/11 is the probability of an event then the probability of the event "not A" is: (a) 0 (b) 2/11 (c) 9/11 (

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Q. If 2/11 is the probability of an event then the probability of the event "not A" is:

(a) 0

(b) 2/11

(d) -2/11

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2021MCQ· 1mImportance★★★★★

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Concept understanding — Probability Axioms

Probability Axioms: From Intuition to Precision

Imagine you're rolling a fair six-sided die. Before you throw it, you know a few things for certain: the result will be one of the numbers 1 through 6. You also know that some outcomes are equally likely — each face has a 1-in-6 chance. And you know that the chance of getting either a 1 or a 2 is simply the sum of their individual chances: $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$ .

These three ideas — that probabilities are numbers between 0 and 1, that something must happen (total probability = 1), and that probabilities of non-overlapping events add — are the bedrock of all probability theory. They are so fundamental that we call them axioms: self-evident truths from which everything else is derived.

The Three Axioms (Kolmogorov's Axioms)

Let’s make this precise. We have a sample space $S$ — the set of all possible outcomes. An event $A$ is any subset of $S$ (like "rolling an even number" = $\{2,4,6\}$ ). The probability of an event $A$ is written $P(A)$ .

P(A) \geq 0 \quad \text{for every event } A

Axiom 1 (Non-negativity): A probability can never be negative. This matches our intuition: you can't have a "less than zero" chance of something happening. The smallest possible probability is 0 (an impossible event).

P(S) = 1

Axiom 2 (Normalization): The probability that some outcome in the sample space occurs is exactly 1. Something must happen. This is why we say "the die will show 1,2,3,4,5, or 6" with certainty.

A

Axiom 3 (Additivity): For events that don't overlap, the probability of "A or B" is just the sum of their individual probabilities. This is why the chance of rolling a 1 or a 2 is $\frac{1}{6} + \frac{1}{6}$ .

Watch out

This additivity only works for mutually exclusive events. If events can happen together (like "rolling an even number" and "rolling a number greater than 3"), you cannot simply add their probabilities — you'd double-count the overlap.

Why These Three Are Enough

From these three simple rules, we can derive everything else in probability. For example:

Complement rule: $P(\text{not } A) = 1 - P(A)$ . Why? Because $A$ and "not $A$ " are mutually exclusive and together cover the whole sample space. By Axiom 3: $P(A) + P(\text{not } A) = P(S) = 1$ .
Probability of an impossible event: $P(\emptyset) = 0$ . Since $S$ and $\emptyset$ are mutually exclusive and $S \cup \emptyset = S$ , we get $P(S) + P(\emptyset) = P(S)$ , so $P(\emptyset) = 0$ . …

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$P(\text{not }A) = 1 - P(A) = 9/11$ .

For any event A, $P(\text{not } A) = P(A') = 1 - P(A)$ .

…

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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 14 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. The sum of the probabilities of all the elementary events of an experiment is ______.$
›Reveal solutionSolution

The sum equals $1$ .

In any random experiment, the elementary events form the sample space $S$ , and they are mutually exclusive and exhaustive (exactly one of them must occur). Since $P(S)=1$ and $S$ is the union of all elementary events, the sum of their individual pro …

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CBSE 2025Set sz1 markMCQ
$P(A) = \dfrac{3}{5}$
(a) \dfrac{7}{8}
(b) \dfrac{5}{3}
(c) \dfrac{2}{3}
(d) \dfrac{2}{5}
›Reveal solutionSolution

Complementary events satisfy $P(A) + P(A') = 1$ , so with $P(A)=3/5$ , $P(A')=2/5$ .

For any event $A$ and its complement $A'$ (the event that $A$ does not occur), the probabilities always satisfy:

$P(A) + P(A') = 1.$

Given $P(A) = \dfrac{3}{5}$ : …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following cannot be the probability of occurrence of an event?$
(a) 0
(b) $\dfrac{-3}{4}$
(c) $\dfrac{3}{4}$
(d) 1
›Reveal solutionSolution

Probability values must always lie in the closed range $[0,1]$ .

$0$ and $1$ are the extreme allowed values, and $3/4$ lies between them — all valid. $-3/4$ is negativ …

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CBSE 2024Set ANNUAL1 markMCQ
$\dfrac{3}{11}$
(a) $\dfrac{8}{11}$
(b) $\dfrac{9}{11}$
(c) $\dfrac{1}{11}$
(d) None of these
›Reveal solutionSolution

The probability of an event NOT happening is $1$ minus the probability that it does happen.

Given $P(A) = \dfrac{3}{11}$ , the complement rule states:

$P(A') = 1 - P(A)$

…

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CBSE 2024Set ANNUAL1 mark
$Q. If an event is impossible, then what will be its probability?$
›Reveal solutionSolution

By definition, an impossible event is the empty set $\varnothing$ , and $P(\varnothing)=0$ .

An impossible event is one that can never occur under the given random experiment; it corresponds to the empty subset $\varnothing$ of the sample space.

…

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CBSE 2024Set ANNUAL1 mark
$P(A) + P(\overline{A}) =$
›Reveal solutionSolution

An event and its complement are exhaustive and mutually exclusive, so their probabilities sum to 1.

Step 1. $A$ and $\overline{A}$ are mutually exclusive (they cannot both occur) and exhaustive (together they cover the whole sample space $S$ ).

…

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CBSE 2023Set ANNUAL1 markMCQ
$E$
(a) 0
(b) 4
(c) 16
(d) 2
›Reveal solutionSolution

An impossible event has probability exactly 0, by definition.

In probability theory, the sample space $S$ has $P(S)=1$ . The impossible event is the empty set $\phi$ — an event that contains no outcome …

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CBSE 2023Set annual1 mark
$P(A) = \dfrac{2}{5}$
›Reveal solutionSolution

$P(\text{not }A) = 1 - P(A) = \dfrac{3}{5}$ .

For any event $A$ , the event 'not $A$ ' (its complement) together with $A$ covers the whole sample space, so

$P(A) + P(\text{not }A) = 1.$

Given $P(A) = \dfrac{2}{5}$ , …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. If A and B are any two events, then which of the following is not true?$
(a) P(A ∩ B) ≤ P(A)
(b) P(A ∩ B) ≤ P(A ∪ B)
(c) P(A) ≤ P(A ∩ B)
(d) P(B) ≤ P(A ∪ B)
›Reveal solutionSolution
Since $A \cap B \subseteq A$ , we always have $P(A \cap B) \le P(A)$ — the reverse inequality in (c) is false in general.

For any two events $A, B$ : $A \cap B \subseteq A$ , $A \cap B \subseteq B$ , and $A \cap B \subseteq A \cup B$ .

(a) $P(A \cap B) \le P(A)$ — true, since $A \cap B \subseteq A$ .

(b) $P(A \cap B) \le P(A \cup B)$ — true, since $A \cap B \subseteq A \cup B$ . …
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CBSE 2022Set ANNUAL1 markMCQ
$Q. The probability of an event cannot be greater than.........$
(a) one
(b) two
(c) zero
(d) three
›Reveal solutionSolution

Probability cannot exceed $1$ .

For any event $E$ , $0\le P(E)\le 1$ ; a certain event has probability $1$ and an impossible event $0$ . So probability can never be greater than …

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CBSE 2021Set ANNUAL1 markMCQ
$Q. If A and B are any two events, then which of the following is not true?$
(a) P(A∩B) ≤ P(A∪B)
(b) P(A∩B) ≤ P(A)
(c) P(A) ≤ P(A∩B)
(d) P(B) ≤ P(A∪B)
›Reveal solutionSolution
Since $A \cap B \subseteq A$ , we always have $P(A\cap B) \le P(A)$ — the reverse (option c) is the false one.

For any two events, $A \cap B$ is a subset of both A and B, and also a subset of $A \cup B$ . From basic set-subset monotonicity of probability ( $X \subseteq Y \Rightarrow P(X) \le P(Y)$ ):

(a) $A\cap B \subseteq A\cup B \Rightarrow P(A\cap B)\le P(A\cup B)$ — true

(b) $A\cap B \subseteq A \Rightarrow P(A\cap B)\le P(A)$ — true

(d) $B \subseteq A\cup B \Rightarrow P(B)\le P(A\cup B)$ — true …
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CBSE 2021Set ANNUAL1 markMCQ
$Q. If 2/11 is the probability of an event then the probability of the event "not A" is:$
(a) 0
(b) 2/11
(c) 9/11
(d) -2/11
›Reveal solutionSolution

$P(\text{not }A) = 1 - P(A) = 9/11$ .

For any event A, $P(\text{not } A) = P(A') = 1 - P(A)$ .

…

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Q.If 2/11 is the probability of an event then the probability of the event "not A" is:

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