Question

We say that \(E \subset F\) if every point in \(E\) is also in \(F\). Show that if \(E \subset F\), then $$ P(F)=P(E)+P\left(F E^{c}\right) \geqslant P(E) $$

Short answer

We have shown that if E ⊆ F, then P(F) = P(E) + P(F ∩ Eᶜ) and P(F) ≥ P(E) by dividing F into disjoint parts, using the property of probabilities for disjoint sets, and noting that probabilities are non-negative.

Step-by-step solution

Understanding the notation

Recall that E ⊆ F means that every point in E is also in F. When we talk about probabilities defined on sets, we can write the probability of a set A as P(A). In our case, we have: - Probability of set E: P(E) - Probability of set F: P(F) The complement of E, denoted as Eᶜ, is the set of all points that are NOT in E. Finally, we have F ∩ Eᶜ, which represents the set of points belonging to both F and E's complement, i.e., the points in F that are not in E. Our goal is to prove that P(F) = P(E) + P(F ∩ Eᶜ) and P(F) ≥ P(E).

The probability of F, E, and their complements

Since E ⊆ F, we can divide the set F into two disjoint parts: the part that belongs to E and the part that does not belong to E. This can be represented as: F = E ∪ (F ∩ Eᶜ) Now, we want to express P(F) as a sum of probabilities of E and F ∩ Eᶜ.

Using the property of probabilities for disjoint sets

Since E and F ∩ Eᶜ are disjoint, we can use the property of probabilities that states that the probability of a union of disjoint sets is the sum of their probabilities: P(F) = P(E ∪ (F ∩ Eᶜ)) = P(E) + P(F ∩ Eᶜ) We have now shown the first part of the given expression.

Showing P(F) ≥ P(E)

In order to show that P(F) ≥ P(E), we need to observe that the probability of F ∩ Eᶜ is a non-negative value since probabilities are always non-negative. Therefore, we can write: P(F ∩ Eᶜ) ≥ 0 Adding P(E) to both sides of the inequality, we get: P(E) + P(F ∩ Eᶜ) ≥ P(E) Since we have already found that P(F) = P(E) + P(F ∩ Eᶜ), we can now show the second part of the given expression: P(F) ≥ P(E) Thus, we have successfully shown that if E ⊆ F, then P(F) = P(E) + P(F ∩ Eᶜ) and P(F) ≥ P(E).