Consider a situation in which [tex]$P(X)=\frac{4}{5}$[/tex] and [tex]$P(Y)=\frac{1}{4}$[/tex]. If [tex]$P(X$[/tex] and [tex]$Y)$[/tex] is [tex]$=\frac{1}{5}$[/tex], which best describes the events?
A. They are independent because [tex]$P(X) \cdot P(Y)=P(X$[/tex] and [tex]$Y)$[/tex].
B. They are independent because [tex]$P( X )+P( Y )=P( X$[/tex] and Y [tex]$)$[/tex].
C. They are dependent because [tex]$P( X ) \cdot P( Y )=P( X$[/tex] and Y [tex]$)$[/tex].
D. They are dependent because [tex]$P(X)+P(Y)=P(X$[/tex] and [tex]$Y)$[/tex].
Answer (2)
The events X and Y are independent because the product of their individual probabilities equals the probability of them occurring together. Therefore, the correct answer is option A: They are independent because P(X) \cdot P(Y) = P(X and Y).
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Calculate P ( X ) × P ( Y ) = 5 4 × 4 1 = 5 1 .
Compare P ( X ) × P ( Y ) with P ( X and Y ) = 5 1 .
Since P ( X ) × P ( Y ) = P ( X and Y ) , events X and Y are independent.
The events are independent because P ( X ) × P ( Y ) = P ( X and Y ) .
Explanation
Analyze the given probabilities We are given the probabilities P ( X ) = 5 4 , P ( Y ) = 4 1 , and P ( X and Y ) = 5 1 . We need to determine if events X and Y are independent or dependent.
Check for independence To check for independence, we need to verify if P ( X ) ⋅ P ( Y ) = P ( X and Y ) . Let's calculate P ( X ) ⋅ P ( Y ) : P ( X ) ⋅ P ( Y ) = 5 4 ⋅ 4 1 = 20 4 = 5 1 Since P ( X ) ⋅ P ( Y ) = 5 1 and P ( X and Y ) = 5 1 , we have P ( X ) ⋅ P ( Y ) = P ( X and Y ) . This indicates that the events are independent.
Check the alternative condition Now, let's check if P ( X ) + P ( Y ) = P ( X and Y ) .
P ( X ) + P ( Y ) = 5 4 + 4 1 = 20 16 + 20 5 = 20 21 Since P ( X ) + P ( Y ) = 20 21 and P ( X and Y ) = 5 1 , we have P ( X ) + P ( Y ) = P ( X and Y ) .
Conclusion Since P ( X ) ⋅ P ( Y ) = P ( X and Y ) , the events X and Y are independent. The correct statement is: They are independent because P ( X ) ⋅ P ( Y ) = P ( X and Y ) .
Examples
Understanding independence and dependence of events is crucial in many real-world scenarios. For instance, consider a quality control process in a factory. If the probability of a product passing inspection A is independent of the probability of it passing inspection B, then the overall quality control process can be optimized by analyzing each inspection separately. This can save time and resources compared to situations where the inspections are dependent, requiring a more complex analysis.
