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Hi, can someone help with this ven diagram question?

Show that P(G n S)’ > P(G u S’)

n is meant to be the symbol for the intersection

u is meant to be the symbol for the union

Hi, can someone help with this ven diagram question? Show that P(G n S)’ > P(G-example-1
User Liviucmg
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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 \( P(G \cap S)' > P(G \cup S') \).

User Ioleo
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