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Two events, [tex]E_1[/tex] and [tex]E_2[/tex], are defined for a random experiment. What is the probability that at least one of the two events occurs in any trial of the experiment?

A. [tex]P(E_1) + P(E_2) - 2P(E_1 \cap E_2)[/tex]

B. [tex]P(E_1) + P(E_2) + P(E_1 \cap E_2)[/tex]

C. [tex]P(E_1) + P(E_2) - P(E_1 \cap E_2)[/tex]

D. [tex]P(E_1) - P(E_2) - P(E_1 \cap E_2)[/tex]

Sagot :

GinnyAnswer
To determine the probability that at least one of two events, [tex]\( E_1 \)[/tex] and [tex]\( E_2 \)[/tex], occurs, we use the principle of inclusion and exclusion in probability theory.

The probability that at least one of [tex]\( E_1 \)[/tex] or [tex]\( E_2 \)[/tex] occurs is represented by [tex]\( P(E_1 \cup E_2) \)[/tex].

According to the principle of inclusion and exclusion:
[tex]\[ P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \][/tex]

Here’s how this formula is derived:
1. [tex]\( P(E_1) \)[/tex] is the probability that event [tex]\( E_1 \)[/tex] occurs.
2. [tex]\( P(E_2) \)[/tex] is the probability that event [tex]\( E_2 \)[/tex] occurs.
3. Adding [tex]\( P(E_1) + P(E_2) \)[/tex] initially considers all outcomes of [tex]\( E_1 \)[/tex] and [tex]\( E_2 \)[/tex], but it counts the overlap (where both [tex]\( E_1 \)[/tex] and [tex]\( E_2 \)[/tex] occur) twice.
4. To correct for this double-counting, we subtract [tex]\( P(E_1 \cap E_2) \)[/tex], the probability that both events occur.

Thus, the correct probability that at least one of the two events, [tex]\( E_1 \)[/tex] or [tex]\( E_2 \)[/tex], occurs is:
[tex]\[ P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \][/tex]

So, the correct answer is:
C. [tex]\( P(E_1) + P(E_2) - P(E_1 \cap E_2) \)[/tex]
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