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Really simple question but I cannot figure out the proof. If E1 is included in E2, how do I show that P(E1)<=P(E2)?

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    What's your level? How was probability introduced to you? – 2017-01-29
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    I am taking a basic probability course. We should be able to prove this using the axioms of probability. – 2017-01-29
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    So what are your axioms? – 2017-01-29
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    Hint: You should have around three (as Kolmogorov's probability theory does). What are the ones *you* use in your course? How do you think they might be relevant? – 2017-01-29
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    These are my axioms: i. P(S) = 1 ii. 0 ≤ P(E) ≤ 1 iii. If E1 and E2 are disjoint then P(E1 U E2) = P(E1) + P(E2) – 2017-01-29
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    Ok. so take $E_1$ and define $F=E_2-E_1$ (the complement of $E_1$ in $E_2$). What is $E_1\cap F$? What can you say about its probability? – 2017-01-29
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    You have $E_1 \subset E_2$. Draw a figure. Can you pick two disjoint sets out of the figure? Apply the third axiom to them and you have $P(E_2) = P(E_1) + P(\text{something})$, now use the second axiom. – 2017-01-29
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    Well then (i) we're not dealing with the (entire) sample space, (ii) is useful to show $P(E_1) \leq P(E_1)+\text{something positive} = P(E_2)$, (iii) would be useful if that something were the probability of ... – 2017-01-29

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Your third point says that for disjoint events, the probabilities are additive.

Then, we can do this : $E_1$ and $E_1^c \cap E_2$ are disjoint events. Hence, we can sum them up: $P(E_1 \cup (E_1^c \cap E_2)) = P(E_1) + P(E_1^c \cap E_2)$.

Now, note that $ P(E_1^c \cap E_2) \geq 0$, and also, $E_1 \cup (E_1^c \cap E_2) = E_2$, because $E_1 \subset E_2$. Hence, we get $P(E_2) \geq P(E_1)$.

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Let $$E_2 = E_1 \cup E_3$$ where $$E_3 = E_2-E_1.$$ Notice that $$E_1 \cap E_3 = \emptyset$$ According to first Kolmogorov axiom, we always have $P(E)\ge0$, and according to the third axiom, for two non-intersecting sets $A$ and $B$, $P(A \cup B) = P(A)+P(B)$. Therefore $$P(E_2) = P(E_3 \cup E_1) = P(E_3) + P(E_1) \ge P(E_1).$$

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First, you need to prove that your third axiom is true for any finite collection of pairwise disjoint sets. Then, note that the three sets $E_1\cap E_2$, $E_1\setminus E_2$ and $E_2\setminus E_1$ are disjoint. Lastly, use the fact that $E_1\subseteq E_2$ to simplify the descriptions of the above sets.