If $e_1, e_2, e_3, e_4$ are the four elementary outcomes in a sample space and $P(e_1) = .1$, $P(e_2) = .5$, $P(e_3) = .1$, then the probability of $e_4$ is ______.

> [!TLDR] > The four probabilities must sum to 1 (by the probability axiom for a finite sample space). Given three probabilities, the fourth is found by subtraction: $P(e_4) = 1 - (0.1 + 0.5 + 0.1) = 0.3$. The core idea here is one of the most fundamental rules in probability: the **total probability of all elementary outcomes in a sample space is exactly 1**. This isn't arbitrary — it comes from the fact that when you perform an experiment, *something* must happen. The set of all elementary outcomes covers every possible result, so their probabilities must add up to certainty. Let's see why this applies directly. 1. **Identify the sample space.** The problem tells us there are exactly four elementary outcomes: $e_1, e_2, e_3, e_4$. These are mutually exclusive (no two can happen at the same time) and exhaustive (one of them must occur). That's the definition of a sample space for a discrete experiment. 2. **Recall the probability axiom for finite sample spaces.** For a finite set of elementary outcomes, the sum of their probabilities is 1. In symbols: $$P(e_1) + P(e_2) + P(e_3) + P(e_4) = 1$$ > [!FORMULA] > For any finite sample space $\{e_1, e_2, \dots, e_n\}$, > $$\sum_{i=1}^n P(e_i) = 1$$ 3. **Plug in the known values.** We are given: $$P(e_1) = 0.1, \quad P(e_2) = 0.5, \quad P(e_3) = 0.1$$ So the sum of the first three is: $$0.1 + 0.5 + 0.1 = 0.7$$ 4. **Solve for the missing probability.** Let $x = P(e_4)$. Then: $$0.7 + x = 1$$ $$x = 1 - 0.7 = 0.3$$ …

If e₁, e₂, e₃, e₄ are the four elementary outcomes in a sample space and P(e₁) = .1, P(e₂) = .5, P(e₃) = .1, t

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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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The four probabilities must sum to 1 (by the probability axiom for a finite sample space). Given three probabilities, the fourth is found by subtraction: $P(e_4) = 1 - (0.1 + 0.5 + 0.1) = 0.3$ .

The core idea here is one of the most fundamental rules in probability: the total probability of all elementary outcomes in a sample space is exactly 1. This isn't arbitrary — it comes from the fact that when you perform an experiment, something must happen. The set of all elementary outcomes covers every possible result, so their probabilities must add up to certainty.

Let's see why this applies directly.

Identify the sample space.

The problem tells us there are exactly four elementary outcomes: $e_1, e_2, e_3, e_4$ . These are mutually exclusive (no two can happen at the same time) and exhaustive (one of them must occur). That's the definition of a sample space for a discrete experiment.
Recall the probability axiom for finite sample spaces.

For a finite set of elementary outcomes, the sum of their probabilities is 1. In symbols:

P(e_1) + P(e_2) + P(e_3) + P(e_4) = 1

\{e_1, e_2, \dots, e_n\}

Plug in the known values. We are given:

P(e_1) = 0.1, \quad P(e_2) = 0.5, \quad P(e_3) = 0.1

So the sum of the first three is:

0.1 + 0.5 + 0.1 = 0.7

Solve for the missing probability. Let $x = P(e_4)$ . Then:

0.7 + x = 1

$x = 1 - 0.7 = 0.3$ …

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

AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$X$
›Reveal solutionSolution
Two linear equations (normalization and the mean) in the two unknowns $P(X{=}0)$ and $P(X{=}1)$ solve directly to give $P(X{=}0)=2/5$ .

Concept and Intuition

A discrete random variable's full probability distribution must sum to 1, and its mean is the weighted sum $\sum x_iP(X{=}x_i)$ . With two of the four probabilities related ( $P(X{=}3)=2P(X{=}1)$ ) and one given numerically ( $P(X{=}2)=0.3$ ), we get exactly two equations for the two remaining unknowns.

Step-by-Step Solution

Let $P(X{=}0)=p_0$ , $P(X{=}1)=p_1$ . Then $P(X{=}2)=0.3$ , $P(X{=}3)=2p_1$ .

Normalization: $p_0+p_1+0.3+2p_1=1\Rightarrow p_0+3p_1=0.7$ .

Mean: $0\cdot p_0+1\cdot p_1+2(0.3)+3(2p_1)=1.3\Rightarrow p_1+0.6+6p_1=1.3\Rightarrow7p_1=0.7\Rightarrow p_1=0.1$ . …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ

P(x \le 2) =

›Reveal solution

Solve for $k$ using $\sum P(X=x_i)=1$ , then sum the first three probabilities for $P(X\le2)$ . Answer: $\dfrac{83}{100}$ .

Concept and Intuition

A probability distribution's values must sum to exactly 1. This gives an equation in $k$ that we solve, keeping only the root that yields non-negative probabilities. Then $P(X\le2)$ is simply the sum of the probabilities at $x=0,1,2$ .

Step-by-Step Solution

Sum all probabilities and set equal to 1:

3k+5k+3k^2+(4k^2+k)+3k^2 = 1

10k^2 + 9k = 1 \implies 10k^2+9k-1=0

Solve the quadratic using the formula:

k = \frac{-9\pm\sqrt{81+40}}{20}=\frac{-9\pm11}{20}

So $k=\dfrac{2}{20}=\dfrac1{10}$ or $k=\dfrac{-20}{20}=-1$ .

3. Reject $k=-1$ (gives negative probabilities); take $k=0.1$ . …

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AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
$k =$
›Reveal solutionSolution
This tests the basic normalization condition for a discrete probability distribution — all probabilities must add to $1$ . Answer: $k=\dfrac{1}{10}$ .

Concept and Intuition

A probability distribution assigns a probability $P(X=x)$ to every possible outcome, and since the outcomes are exhaustive and mutually exclusive, the total probability must be exactly $1$ . This single normalization equation is usually enough to pin down an unknown constant like $k$ .

Step-by-Step Solution

List the probabilities: $P(X=1)=2k,\ P(X=2)=4k,\ P(X=3)=3k,\ P(X=4)=k$ .

Sum them and set equal to $1$ : $2k+4k+3k+k = 10k = 1$ .

Solve: $k = \dfrac{1}{10}$ . …
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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$\{0,1,2\}$
›Reveal solutionSolution
The three probabilities must add to 1; substituting each option, only $C=\frac13$ satisfies the cubic AND keeps every probability in $[0,1]$ .

Concept and Intuition

For any discrete random variable, the probabilities over its whole range must sum to exactly 1. Writing that sum-to-1 condition down turns the problem into solving a polynomial equation in $C$ — but since $C$ is a probability parameter, we must also sanity-check that the resulting individual probabilities are all non-negative and at most 1 (a root of the polynomial that fails this check would be spurious/extraneous for a genuine probability distribution).

Step-by-Step Solution

Sum-to-one condition: $P(X=0)+P(X=1)+P(X=2)=1$ $3C^3+(4C-10C^2)+(5C-1)=1$ .

Simplify: $3C^3-10C^2+9C-1=1 \Rightarrow 3C^3-10C^2+9C-2=0$ .

Try the given options as roots. $C=\frac13$ : $3\cdot\frac{1}{27}-10\cdot\frac19+9\cdot\frac13-2 = \frac{1}{9}-\frac{10}{9}+3-2=-1+1=0$ . This works. …
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AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
$\dfrac{4}{7},\dfrac{2}{7},\dfrac{1}{7}$
›Reveal solutionSolution
This is a simple ratio/probability partition problem: express all three probabilities as multiples of the smallest one and use that they sum to 1.

Concept and Intuition

When probabilities of mutually exclusive, exhaustive events are given in a chain of ratios, express everything as a multiple of the smallest and normalize using the fact that all probabilities must add to 1.

Step-by-Step Solution

Let $P(C) = x$ .

$P(B) = 2\,P(C) = 2x$ .

$P(A) = 2\,P(B) = 4x$ .

Since exactly one of A, B, C wins, $P(A)+P(B)+P(C) = 1 \Rightarrow 4x+2x+x = 7x = 1 \Rightarrow x = \dfrac{1}{7}$ . …
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AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
$\xi$
(i) $P(\phi)=0$
(ii) If $E^c$ is the complementary event of E, then $P(E^c) = 1-P(E)$
(iii) $0 \le P(E) \le 1,\ \forall E \subseteq S$
(iv) If $E_1 \subseteq E_2$ then $P(E_2) \le P(E_1)$ (A)
(iii) (B)
(iv) (C)
(ii) (D) (i)
›Reveal solutionSolution
Three of the four listed statements are genuine probability axioms/consequences; (iv) reverses the correct monotonicity direction, so it is the one that is NOT satisfied.

Concept and Intuition

Probability is monotonic with respect to subset inclusion: a smaller (more restrictive) event can never be more likely than a larger event that contains it. If $E_1\subseteq E_2$ , every outcome favouring $E_1$ also favours $E_2$ , so $P(E_1)\le P(E_2)$ .

Step-by-Step Solution

(i) $P(\phi)=0$ — a standard axiom, true.

(ii) $P(E^c)=1-P(E)$ — follows since $E$ and $E^c$ partition $S$ , true.

(iii) $0\le P(E)\le1$ for every $E\subseteq S$ — the basic probability axiom, true. …
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AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
$1/16$
›Reveal solutionSolution
This tests enumerating the number of ways to get each possible sum with two dice, restricted to prime sums, over the 36 equally-likely outcomes. Answer: $5/12$ .

Concept and Intuition

With two fair six-sided dice, there are $6\times6=36$ equally likely ordered outcomes. Each possible total from 2 to 12 has a known number of ways it can occur (a classic triangular-then-decreasing pattern: 1,2,3,4,5,6,5,4,3,2,1 for sums 2 through 12). We simply sum the counts corresponding to prime totals.

Step-by-Step Solution

Total outcomes when rolling two dice: $6\times6=36$ .

Number of ways to get each sum: sum=2→1 way, 3→2, 4→3, 5→4, 6→5, 7→6, 8→5, 9→4, 10→3, 11→2, 12→1.

Identify which sums from 2 to 12 are prime: 2, 3, 5, 7, 11 (4, 6, 8, 9, 10, 12 are composite; 1 is not a possible sum here anyway).

Sum the corresponding counts: for sum 2: 1 way; sum 3: 2 ways; sum 5: 4 ways; sum 7: 6 ways; sum 11: 2 ways. Total favorable $= 1+2+4+6+2 = 15$ . …
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AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
$X$
›Reveal solutionSolution
Requiring the probabilities to sum to 1 over $x=1,2,3,\dots$ forces $e^C=2$ , so $C=\ln2$ .

Concept and Intuition

Any valid probability mass function must have its probabilities sum to exactly 1 over its whole range. Recognizing $\sum_{x=0}^\infty \frac{C^x}{x!}=e^C$ as the Taylor series of $e^C$ lets us convert the normalization condition into a simple exponential equation.

Step-by-Step Solution

Since $X$ takes values $1,2,3,\dots$ , normalization requires $\displaystyle\sum_{x=1}^{\infty} P(X=x) = 1$ , i.e. $\displaystyle\sum_{x=1}^{\infty}\frac{C^x}{x!}=1$ .

Recall the Taylor series $e^C = \displaystyle\sum_{x=0}^{\infty}\frac{C^x}{x!} = 1 + \sum_{x=1}^{\infty}\frac{C^x}{x!}$ .

So $\displaystyle\sum_{x=1}^{\infty}\frac{C^x}{x!} = e^C - 1$ .

Set this equal to 1: $e^C - 1 = 1 \Rightarrow e^C = 2$ .

Taking natural log of both sides: $C = \log 2$ . …
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AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
$A, B, C$
›Reveal solutionSolution
Every probability expression must be a valid probability ( $0$ to $1$ ), and mutual exclusivity forces the sum of the three to be at most $1$ — intersecting all these constraints gives the answer.

Concept and Intuition

For probabilities to make sense, $0\le P(A),P(B),P(C)\le 1$ . Mutual exclusivity means $P(A\cup B\cup C)=P(A)+P(B)+P(C)\le 1$ (it could be less than 1 if there's a residual sample-space probability outside all three). All these inequalities on $x$ must hold simultaneously.

Step-by-Step Solution

$0\le\dfrac{3x+1}{3}\le1 \Rightarrow -1\le 3x\le 2 \Rightarrow -\dfrac13\le x\le \dfrac23$ .

$0\le\dfrac{1-x}{4}\le1 \Rightarrow -3\le x\le 1$ .

$0\le\dfrac{1-2x}{2}\le1 \Rightarrow -\dfrac12\le x\le \dfrac12$ .

Sum condition: $\dfrac{3x+1}{3}+\dfrac{1-x}{4}+\dfrac{1-2x}{2}\le 1$ . Over a common denominator 12: $\dfrac{4(3x+1)+3(1-x)+6(1-2x)}{12}\le1 \Rightarrow \dfrac{-3x+13}{12}\le1 \Rightarrow -3x\le-1 \Rightarrow x\ge\dfrac13$ . …
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AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$S$
›Reveal solutionSolution
Count $(b,c)$ pairs satisfying the real-roots discriminant condition out of all 36 equally likely pairs. Answer: (C).

Concept and Intuition

$x^2+bx+c=0$ has real roots exactly when the discriminant $b^2-4c\geq0$ , i.e., $c \le b^2/4$ . Since $b,c$ are each chosen uniformly from $\{1,2,3,4,5,6\}$ , there are $6\times6=36$ equally likely pairs total.

Step-by-Step Solution

For $b=1$ : need $c\le0.25$ — no valid $c$ (0 pairs).

For $b=2$ : need $c\le1$ — $c=1$ (1 pair).

For $b=3$ : need $c\le2.25$ — $c=1,2$ (2 pairs).

For $b=4$ : need $c\le4$ — $c=1,2,3,4$ (4 pairs).

For $b=5$ : need $c\le6.25$ — all $c=1,\ldots,6$ qualify (6 pairs). …
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←Previous questionQ30 — The probability that the home team will win an upcoming football game …→Next questionQ32 — Let S = \1, 2, 3, 4, 5, 6\ and E = \1, 3, 5\, then barE is _____.

$x_i$	0	1	2	3	4
$P(X=x_i)$	$3k$	$5k$	$3k^2$	$4k^2+k$	$3k^2$
(A) $\dfrac{14}{25}$
(B) $\dfrac{23}{32}$
(C) $\dfrac{41}{49}$
(D) $\dfrac{83}{100}$

X = x	1	2	3	4
P(X = x)	2k	4k	3k	k
(A) $\dfrac{1}{10}$
(B) $\dfrac{2}{10}$
(C) $\dfrac{3}{10}$
(D) $\dfrac{4}{10}$

Q.If e1,e2,e3,e4e_1, e_2, e_3, e_4e1​,e2​,e3​,e4​ are the four elementary outcomes in a sample space and P(e1)=.1P(e_1) = .1P(e1​)=.1, P(e2)=.5P(e_2) = .5P(e2​)=.5, P(e3)=.1P(e_3) = .1P(e3​)=.1, then the probability of e4e_4e4​ is ______.

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Q.If $e_1, e_2, e_3, e_4$ are the four elementary outcomes in a sample space and $P(e_1) = .1$ , $P(e_2) = .5$ , $P(e_3) = .1$ , then the probability of $e_4$ is ______.

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