Answer :
To determine if two events [tex]\( A \)[/tex] (rolling a number greater than 4) and [tex]\( B \)[/tex] (rolling an even number) are independent, we need to check if the probability of their intersection [tex]\( P(A \cap B) \)[/tex] equals the product of their individual probabilities [tex]\( P(A) \times P(B) \)[/tex].
First, let’s define the probabilities:
1. [tex]\( P(A) \)[/tex] is the probability of rolling a number greater than 4. On a six-sided die, the possible outcomes greater than 4 are 5 and 6. Therefore:
[tex]\[ P(A) = \frac{2}{6} \][/tex]
2. [tex]\( P(B) \)[/tex] is the probability of rolling an even number. The even numbers on a six-sided die are 2, 4, and 6. Therefore:
[tex]\[ P(B) = \frac{3}{6} \][/tex]
3. [tex]\( P(A \cap B) \)[/tex] is the probability of both events happening (rolling a number that is both greater than 4 and even). The only number that fits this criterion is 6. Therefore:
[tex]\[ P(A \cap B) = \frac{1}{6} \][/tex]
For events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] to be independent, the following condition must hold:
[tex]\[ P(A \cap B) = P(A) \times P(B) \][/tex]
First, let’s compute [tex]\( P(A) \times P(B) \)[/tex]:
[tex]\[ P(A) \times P(B) = \frac{2}{6} \times \frac{3}{6} = \frac{6}{36} = \frac{1}{6} \][/tex]
Therefore, [tex]\( P(A \cap B) = P(A) \times P(B) \)[/tex] holds true.
Now, let’s double-check with the given question by examining each option for the necessary calculations:
1. [tex]\(\frac{\frac{2}{6}}{\frac{3}{6}} \neq \frac{3}{6}\)[/tex]
[tex]\[ \frac{\frac{2}{6}}{\frac{3}{6}} = \frac{2}{3} \quad \text{and} \quad \frac{3}{6} = 0.5 \quad \Rightarrow \quad \frac{2}{3} \neq 0.5 \][/tex]
2. [tex]\(\frac{\frac{3}{6}}{\frac{1}{6}} \neq \frac{2}{6}\)[/tex]
[tex]\[ \frac{\frac{3}{6}}{\frac{1}{6}} = 3 \quad \text{and} \quad \frac{2}{6} = 0.333... \quad \Rightarrow \quad 3 \neq 0.333... \][/tex]
3. [tex]\(\frac{\frac{3}{6}}{\frac{2}{6}} \neq \frac{2}{6}\)[/tex]
[tex]\[ \frac{\frac{3}{6}}{\frac{2}{6}} = \frac{3}{2} = 1.5 \quad \text{and} \quad \frac{2}{6} = 0.333... \quad \Rightarrow \quad 1.5 \neq 0.333... \][/tex]
4. [tex]\(\frac{\frac{1}{6}}{\frac{2}{6}} \neq \frac{3}{6}\)[/tex]
[tex]\[ \frac{\frac{1}{6}}{\frac{2}{6}} = \frac{1}{2} = 0.5 \quad \text{and} \quad\frac{3}{6} = 0.5 \quad \Rightarrow \quad 0.5 = 0.5 \][/tex]
Based on the comparisons, option [tex]\( \frac{\frac{1}{6}}{\frac{2}{6}} \neq \frac{3}{6} \)[/tex] correctly represents the calculation used to determine the independence condition that fails in this given context.
Therefore, the correct answer is:
[tex]\[ \frac{\frac{1}{6}}{\frac{2}{6}} \neq \frac{3}{6} \][/tex]
First, let’s define the probabilities:
1. [tex]\( P(A) \)[/tex] is the probability of rolling a number greater than 4. On a six-sided die, the possible outcomes greater than 4 are 5 and 6. Therefore:
[tex]\[ P(A) = \frac{2}{6} \][/tex]
2. [tex]\( P(B) \)[/tex] is the probability of rolling an even number. The even numbers on a six-sided die are 2, 4, and 6. Therefore:
[tex]\[ P(B) = \frac{3}{6} \][/tex]
3. [tex]\( P(A \cap B) \)[/tex] is the probability of both events happening (rolling a number that is both greater than 4 and even). The only number that fits this criterion is 6. Therefore:
[tex]\[ P(A \cap B) = \frac{1}{6} \][/tex]
For events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] to be independent, the following condition must hold:
[tex]\[ P(A \cap B) = P(A) \times P(B) \][/tex]
First, let’s compute [tex]\( P(A) \times P(B) \)[/tex]:
[tex]\[ P(A) \times P(B) = \frac{2}{6} \times \frac{3}{6} = \frac{6}{36} = \frac{1}{6} \][/tex]
Therefore, [tex]\( P(A \cap B) = P(A) \times P(B) \)[/tex] holds true.
Now, let’s double-check with the given question by examining each option for the necessary calculations:
1. [tex]\(\frac{\frac{2}{6}}{\frac{3}{6}} \neq \frac{3}{6}\)[/tex]
[tex]\[ \frac{\frac{2}{6}}{\frac{3}{6}} = \frac{2}{3} \quad \text{and} \quad \frac{3}{6} = 0.5 \quad \Rightarrow \quad \frac{2}{3} \neq 0.5 \][/tex]
2. [tex]\(\frac{\frac{3}{6}}{\frac{1}{6}} \neq \frac{2}{6}\)[/tex]
[tex]\[ \frac{\frac{3}{6}}{\frac{1}{6}} = 3 \quad \text{and} \quad \frac{2}{6} = 0.333... \quad \Rightarrow \quad 3 \neq 0.333... \][/tex]
3. [tex]\(\frac{\frac{3}{6}}{\frac{2}{6}} \neq \frac{2}{6}\)[/tex]
[tex]\[ \frac{\frac{3}{6}}{\frac{2}{6}} = \frac{3}{2} = 1.5 \quad \text{and} \quad \frac{2}{6} = 0.333... \quad \Rightarrow \quad 1.5 \neq 0.333... \][/tex]
4. [tex]\(\frac{\frac{1}{6}}{\frac{2}{6}} \neq \frac{3}{6}\)[/tex]
[tex]\[ \frac{\frac{1}{6}}{\frac{2}{6}} = \frac{1}{2} = 0.5 \quad \text{and} \quad\frac{3}{6} = 0.5 \quad \Rightarrow \quad 0.5 = 0.5 \][/tex]
Based on the comparisons, option [tex]\( \frac{\frac{1}{6}}{\frac{2}{6}} \neq \frac{3}{6} \)[/tex] correctly represents the calculation used to determine the independence condition that fails in this given context.
Therefore, the correct answer is:
[tex]\[ \frac{\frac{1}{6}}{\frac{2}{6}} \neq \frac{3}{6} \][/tex]
