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This week in school, there is a 75% probability of having a fire drill, a 50% probability of a tornado drill, and a 25% probability of having both drills. Let event [tex]\( F \)[/tex] be a fire drill and event [tex]\( T \)[/tex] be a tornado drill. Are the two events independent?

A. No, because [tex]\( P(F \cap T) \neq P(F) \cdot P(T) \)[/tex].
B. No, because [tex]\( P(F \cap T) \neq P(F) + P(T) \)[/tex].
C. Yes, because [tex]\( P(F \cap T) = P(F) \cdot P(T) \)[/tex].
D. Yes, because [tex]\( P(F) = P(T) + P(F \cap T) \)[/tex].



Answer :

GinnyAnswer
To determine if the events [tex]\( F \)[/tex] (fire drill) and [tex]\( T \)[/tex] (tornado drill) are independent, we need to check if the probability of both events occurring simultaneously, [tex]\( P(F \cap T) \)[/tex], equals the product of the individual probabilities of each event, [tex]\( P(F) \cdot P(T) \)[/tex].

Here are the given probabilities:
- [tex]\( P(F) = 0.75 \)[/tex]
- [tex]\( P(T) = 0.5 \)[/tex]
- [tex]\( P(F \cap T) = 0.25 \)[/tex]

First, let's calculate the product of [tex]\( P(F) \)[/tex] and [tex]\( P(T) \)[/tex]:
[tex]\[ P(F) \cdot P(T) = 0.75 \cdot 0.5 = 0.375 \][/tex]

Next, we will compare [tex]\( P(F \cap T) \)[/tex] with [tex]\( P(F) \cdot P(T) \)[/tex]:
[tex]\[ P(F \cap T) = 0.25 \][/tex]
[tex]\[ P(F) \cdot P(T) = 0.375 \][/tex]

Since [tex]\( P(F \cap T) \neq P(F) \cdot P(T) \)[/tex], the two events are not independent. Therefore, the correct answer is:

No, because [tex]\( P(F \cap T) \neq P(F) \cdot P(T) \)[/tex].

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