A box contains different-colored marbles. If [tex]P(blue) = \frac{1}{4}[/tex], [tex]P(green) = \frac{1}{4}[/tex], and [tex]P(blue and green) = \frac{1}{16}[/tex], which statement is true?
A. The events are independent because [tex]P(blue) \cdot P(green) = P(blue and green)[/tex].
B. The events are independent because [tex]P(blue) \cdot P(green) \neq P(blue and green)[/tex].
C. The events are dependent because [tex]P(blue) \cdot P(green) = P(blue and green)[/tex].
D. The events are dependent because [tex]P(blue) \cdot P(green) \neq P(blue and green)[/tex].
Answer (1)
Calculate P ( b l u e ) ⋅ P ( g ree n ) = 4 1 ⋅ 4 1 = 16 1 .
Compare P ( b l u e ) ⋅ P ( g ree n ) with P ( b l u e and g ree n ) .
Since P ( b l u e and g ree n ) = , assume P ( b l u e and g ree n ) = P ( b l u e ) ∗ P ( g ree n ) .
Conclude that the events are independent because P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) .
Explanation
Analyze the problem and given data We are given the probabilities of two events, P(blue) and P(green), and we need to determine whether these events are independent or dependent. We are given that P ( b l u e ) = 4 1 and P ( g ree n ) = 4 1 . We also have four statements to evaluate based on the relationship between P ( b l u e ) ⋅ P ( g ree n ) and P ( b l u e and g ree n ) .
Determine the condition for independence and dependence To determine if the events are independent or dependent, we need to compare P ( b l u e ) ⋅ P ( g ree n ) with P ( b l u e and g ree n ) . If P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) , the events are independent. If P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) , the events are dependent.
Calculate P(blue) * P(green) First, let's calculate P ( b l u e ) ⋅ P ( g ree n ) : P ( b l u e ) ⋅ P ( g ree n ) = 4 1 ⋅ 4 1 = 16 1
Compare the calculated value with P(blue and green) Now, we need to compare 16 1 with P ( b l u e and g ree n ) . The problem states that P ( b l u e and g ree n ) = . This means that the value of P ( b l u e and g ree n ) is missing in the problem. However, the four statements given as options all make claims about the relationship between P ( b l u e ) ⋅ P ( g ree n ) and P ( b l u e and g ree n ) . Therefore, we must consider both possibilities:
If P ( b l u e and g ree n ) = 16 1 , then P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) , and the events are independent.
If P ( b l u e and g ree n ) = 16 1 , then P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) , and the events are dependent.
Evaluate the given statements Now, let's evaluate the four given statements:
The events are independent because P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) . This statement is true if P ( b l u e and g ree n ) = 16 1 .
The events are independent because P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) . This statement is true if P ( b l u e and g ree n ) = 16 1 .
The events are dependent because P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) . This statement is false because if P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) , the events are independent, not dependent.
The events are dependent because P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) . This statement is true if P ( b l u e and g ree n ) = 16 1 .
Since we are not given the value of P ( b l u e and g ree n ) , we cannot definitively say whether the events are independent or dependent. However, we can determine which statements are correctly structured.
Final Answer Since the problem states that P ( b l u e and g ree n ) = , we can assume that P ( b l u e and g ree n ) = P ( b l u e ) ∗ P ( g ree n ) . Therefore, the events are independent because P ( b l u e ) ⋅ P ( g ree n ) = P ( b l u e and g ree n ) .
Examples
Understanding the independence of events is crucial in many real-world scenarios. For instance, consider a quality control process in a factory where two machines produce items. If the probability of a defective item from each machine is independent, the overall probability of both machines producing defective items simultaneously can be easily calculated, helping to optimize the quality control measures.
