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Chapter 1: Problem 9

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

Expert verified
We have shown that if E 鈯� F, then P(F) = P(E) + P(F 鈭� E岫�) and P(F) 鈮� 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

01

Understanding the notation

Recall that E 鈯� 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 E岫�, is the set of all points that are NOT in E. Finally, we have F 鈭� E岫�, 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 鈭� E岫�) and P(F) 鈮� P(E).
02

The probability of F, E, and their complements

Since E 鈯� 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 鈭� (F 鈭� E岫�) Now, we want to express P(F) as a sum of probabilities of E and F 鈭� E岫�.
03

Using the property of probabilities for disjoint sets

Since E and F 鈭� E岫� 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 鈭� (F 鈭� E岫�)) = P(E) + P(F 鈭� E岫�) We have now shown the first part of the given expression.
04

Showing P(F) 鈮� P(E)

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

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