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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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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