Q.If an event is impossible, then what will be its probability?
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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: 61+61=31.
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)≥0for 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.
If A and B are mutually exclusive (they cannot happen together, i.e., A∩B=∅), then:
P(A∪B)=P(A)+P(B)
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 61+61.
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:
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Complement rule: P(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(not A)=P(S)=1.
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Probability of an impossible event: P(∅)=0. Since S and ∅ are mutually exclusive and S∪∅=S, we get P(S)+P(∅)=P(S), so P(∅)=0. …
An impossible event corresponds to the empty set, which has probability 0 by definition. …
By definition, an impossible event is the empty set ∅, and P(∅)=0.
An impossible event is one that can never occur under the given random experiment; it corresponds to the empty subset ∅ of the sample space.
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Showing the 12 most recent of 14 on this concept.
- CBSE 2026Set ANNUAL1 markQ.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 …
- CBSE 2025Set sz1 markMCQQ.For an event A if P(A)=53, then P(A′) is equal to :(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)=53: …
- CBSE 2025Set ANNUAL1 markMCQQ.Which of the following cannot be the probability of occurrence of an event?(a) 0(b) 4−3(c) 43(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 …
- CBSE 2024Set ANNUAL1 markMCQQ.If 113 is probability of an event A, what is the probability of the event 'not A'?(a) 118(b) 119(c) 111(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)=113, the complement rule states:
P(A′)=1−P(A)
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- CBSE 2024Set ANNUAL1 markQ.If an event is impossible, then what will be its probability?
›Reveal solutionSolution
By definition, an impossible event is the empty set ∅, and P(∅)=0.
An impossible event is one that can never occur under the given random experiment; it corresponds to the empty subset ∅ of the sample space.
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- CBSE 2024Set ANNUAL1 markQ.Fill in the blank: P(A)+P(A)= ______.
›Reveal solutionSolution
An event and its complement are exhaustive and mutually exclusive, so their probabilities sum to 1.
Step 1. A and 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 markMCQQ.If E is an impossible event, then P(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 ϕ — an event that contains no outcome …
- CBSE 2023Set annual1 markQ.If P(A)=52, find P(not A).
›Reveal solutionSolution
P(not A)=1−P(A)=53.
For any event A, the event 'not A' (its complement) together with A covers the whole sample space, so
P(A)+P(not A)=1.
Given P(A)=52, …
- CBSE 2022Set ANNUAL1 markMCQQ.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∩B⊆A, we always have P(A∩B)≤P(A) — the reverse inequality in (c) is false in general.
For any two events A,B: A∩B⊆A, A∩B⊆B, and A∩B⊆A∪B.
- (a) P(A∩B)≤P(A) — true, since A∩B⊆A.
- (b) P(A∩B)≤P(A∪B) — true, since A∩B⊆A∪B. …
- CBSE 2022Set ANNUAL1 markMCQQ.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≤P(E)≤1; a certain event has probability 1 and an impossible event 0. So probability can never be greater than …
- CBSE 2021Set ANNUAL1 markMCQQ.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∩B⊆A, we always have P(A∩B)≤P(A) — the reverse (option c) is the false one.
For any two events, A∩B is a subset of both A and B, and also a subset of A∪B. From basic set-subset monotonicity of probability (X⊆Y⇒P(X)≤P(Y)):
- (a) A∩B⊆A∪B⇒P(A∩B)≤P(A∪B) — true
- (b) A∩B⊆A⇒P(A∩B)≤P(A) — true
- (d) B⊆A∪B⇒P(B)≤P(A∪B) — true …
- CBSE 2021Set ANNUAL1 markMCQQ.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(not A)=1−P(A)=9/11.
For any event A, P(not A)=P(A′)=1−P(A).
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