Answer:
Explanation:
Certainly! To show that \( P(G \cap S)' > P(G \cup S') \), you can use set theory and the properties of probability.
Let's break down the notation:
- \( G \cap S \) represents the intersection of events G and S.
- \( G \cup S' \) represents the union of event G and the complement of event S.
The complement of an event is denoted by adding a prime (') to its symbol.
Now, let's express the probabilities using set notation:
\[ P(G \cap S)' \]
This is the probability of the complement of the intersection of G and S.
\[ P(G \cap S)' = 1 - P(G \cap S) \]
Similarly,
\[ P(G \cup S') \]
This is the probability of the union of G and the complement of S.
Now, to show that \( P(G \cap S)' > P(G \cup S') \), you need to prove:
\[ 1 - P(G \cap S) > P(G \cup S') \]
Here's a step-by-step explanation:
1. Start with the Inequality:
\[ 1 - P(G \cap S) > P(G \cup S') \]
2. Use Probability Properties:
\[ 1 - P(G \cap S) = P((G \cap S)') \]
So now, the inequality becomes:
\[ P((G \cap S)') > P(G \cup S') \]
3. Express in Terms of Intersections and Unions:
\[ P((G \cap S)') = P(G' \cup S') \]
Now, the inequality is:
\[ P(G' \cup S') > P(G \cup S') \]
4. Use Probability Properties Again:
\[ P(G' \cup S') = 1 - P(G \cap S') \]
Now, the inequality is:
\[ 1 - P(G \cap S') > P(G \cup S') \]
5. Simplify:
\[ P(G \cup S') < 1 \]
This inequality is true because the probability of the union of two events is always less than or equal to 1. Therefore, you've shown that \( P(G \cap S)' > P(G \cup S') \).