To determine whether events [tex]\( A \)[/tex] (home sharing) and [tex]\( B \)[/tex] (fly) are independent, we need to compare [tex]\( P(A) \)[/tex] (the probability of home sharing) with [tex]\( P(A \mid B) \)[/tex] (the probability of home sharing given that the traveler flies).
First, let's calculate [tex]\( P(A) \)[/tex]:
[tex]\( A \)[/tex] represents travelers who prefer home sharing. According to the table, there are 20 travelers who prefer home sharing out of a total of 100 travelers.
[tex]\[ P(A) = \frac{\text{Number of travelers who prefer home sharing}}{\text{Total number of travelers}} = \frac{20}{100} = 0.2 \][/tex]
Next, let's calculate [tex]\( P(B) \)[/tex]:
[tex]\( B \)[/tex] represents travelers who fly. According to the table, there are 20 travelers who fly out of a total of 100 travelers.
[tex]\[ P(B) = \frac{\text{Number of travelers who fly}}{\text{Total number of travelers}} = \frac{20}{100} = 0.2 \][/tex]
Now, let's calculate [tex]\( P(A \cap B) \)[/tex], the probability that a traveler both prefers home sharing and flies. According to the table, there are 4 travelers who both prefer home sharing and fly.
[tex]\[ P(A \cap B) = \frac{\text{Number of travelers who prefer home sharing and fly}}{\text{Total number of travelers}} = \frac{4}{100} = 0.04 \][/tex]
Now, we can calculate [tex]\( P(A \mid B) \)[/tex], the conditional probability that a traveler prefers home sharing given that they fly:
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.04}{0.2} = 0.2 \][/tex]
Finally, we compare [tex]\( P(A) \)[/tex] with [tex]\( P(A \mid B) \)[/tex]:
[tex]\[ P(A) = 0.2 \][/tex]
[tex]\[ P(A \mid B) = 0.2 \][/tex]
Since [tex]\( P(A) = P(A \mid B) \)[/tex], the events [tex]\( A \)[/tex] (home sharing) and [tex]\( B \)[/tex] (fly) are independent.
Therefore, the correct statement is:
Yes, [tex]\( P(A) = P(A \mid B) \)[/tex].
