This week in school, there is a 75 percent probabiity of having a fire drill, a 50 percent probability of a tornado drill, and a 25 percent probability of having both drils. Let event $$ F $$ be a fire drill and event T be a tornado drill. Are the two events independent? No, because $$ P(F \cap T)
eq P(F)=P(T) $$. No, because $$ P(F \cap T)
eq P(F)+P(T) $$. Yes, because $$ P(F \cap T)=P(F)=P(T) $$. Yes, because $$ P(F)=P(T)+P(F \cap T) $$.

Question

This week in school, there is a 75 percent probabiity of having a fire drill, a 50 percent probability of a tornado drill, and a 25 percent probability of having both drils. Let event $$ F $$ be a fire drill and event T be a tornado drill. Are the two events independent? No, because $$ P(F \cap T) 
eq P(F)=P(T) $$. No, because $$ P(F \cap T) 
eq P(F)+P(T) $$. Yes, because $$ P(F \cap T)=P(F)=P(T) $$. Yes, because $$ P(F)=P(T)+P(F \cap T) $$.

UpStudy AI · Accepted Answer

To determine whether the events $$ F $$ (fire drill) and $$ T $$ (tornado drill) are independent, we need to check if the probability of both events occurring together equals the product of their individual probabilities.

**Given:**
- $$ P(F) = 75\% = 0.75 $$
- $$ P(T) = 50\% = 0.50 $$
- $$ P(F \cap T) = 25\% = 0.25 $$

**Independence Condition:**
Two events $$ F $$ and $$ T $$ are independent if:
$$
P(F \cap T) = P(F) \times P(T)
$$
Calculating the product of the individual probabilities:
$$
P(F) \times P(T) = 0.75 \times 0.50 = 0.375
$$
Comparing this with the given joint probability:
$$
P(F \cap T) = 0.25 \neq 0.375 = P(F) \times P(T)
$$
Since $$ P(F \cap T) \neq P(F) \times P(T) $$, the events $$ F $$ and $$ T $$ are **not independent**.

**Correct Answer:**
No, because $$ P(F \cap T) \neq P(F)P(T) $$.
The events are not independent because the probability of both drills occurring together ($$ P(F \cap T) = 25\% $$) does not equal the product of their individual probabilities ($$ P(F) \times P(T) = 37.5\% $$).

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