Problem 9 We say that $E \subset F$ if e... [FREE SOLUTION]

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Introductory to Probability Models

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9 Edition

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

The set $F$ can be expressed as a union of two disjoint sets $E$ and $FE^c$, where $FE^c$ represents the set of points in $F$ but not in $E$. Using the additivity property of probability for disjoint sets, we obtain $P(F) = P(E) + P\left(F \cap E^c\right)$. Since the probability of any set is non-negative, we have $P\left(F \cap E^c\right) \geq 0$, which implies that $P(F) \geq P(E)$.

Step by step solution

Express $F$ as a union of two disjoint sets

Since $E\subset F$, the set $F$ can be written as a union of two disjoint sets, namely $E$ and $FE^c$, where $FE^c$ represents the set of points in $F$ but not in $E$. So we have: $$ F = E \cup \left( F \cap E^c \right) $$

Use the additivity property of probability

Using the additivity property of probability for disjoint sets, we can obtain the probability of $F$ as follows: $$ P(F) = P\left( E \cup \left( F \cap E^c \right) \right) = P(E) + P\left(F \cap E^c\right) $$

Show that $P(F) \geq P(E)$

Since the probability of any set is non-negative, we have $P\left(F \cap E^c\right) \geq 0$. Consequently: $$ P(F) = P(E) + P\left(F \cap E^c\right) \geq P(E) $$ This completes the proof that if $E \subset F$, then $P(F) \geq P(E)$.

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91影视

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

Step by step solution

Express \(F\) as a union of two disjoint sets

Use the additivity property of probability

Show that \(P(F) \geq P(E)\)

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