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Test the pair of events B and C for independence based on the following table. Events [tex]$A$[/tex], [tex]$B$[/tex], and [tex]$C$[/tex] are mutually exclusive. Events [tex]$D$[/tex] and [tex]$E$[/tex] are mutually exclusive.

\begin{tabular}{|c|c|c|c|c|}
\hline & A & B & C & Totals \\
\hline D & 0.16 & 0.12 & 0.12 & 0.40 \\
\hline E & 0.24 & 0.18 & 0.18 & 0.60 \\
\hline Totals & 0.40 & 0.30 & 0.30 & 1.00 \\
\hline
\end{tabular}

Are the events [tex]$B$[/tex] and [tex]$C$[/tex] independent? Select the correct answer below and fill in the answer boxes to complete your choice.

A. No, they are not independent because [tex]$P(B \cap C) \neq P(B) P(C)$[/tex]. [tex]$P(B \cap C)=$[/tex] [tex]$\square$[/tex] and [tex]$P(B) P(C)=$[/tex] [tex]$\square$[/tex]

B. Yes, they are independent because [tex]$P(B \cap C)=P(B) P(C)$[/tex]. [tex]$P(B \cap C)=$[/tex] [tex]$\square$[/tex] and [tex]$P(B) P(C)=$[/tex] [tex]$\square$[/tex]

C. Yes, they are independent because [tex]$P(B \cap C)=P(B) P(C)$[/tex]. [tex]$P(B \cap C)=$[/tex] [tex]$\square$[/tex] and [tex]$P(B) P(C)=$[/tex] [tex]$\square$[/tex]

D. No, they are not independent because [tex]$P(B \cap C) \neq P(B) P(C)$[/tex]. [tex]$P(B \cap C)=$[/tex] [tex]$\square$[/tex] and [tex]$P(B) P(C)=$[/tex] [tex]$\square$[/tex]

Sagot :

GinnyAnswer
To test whether events [tex]\( B \)[/tex] and [tex]\( C \)[/tex] are independent, we need to compare [tex]\( P(B \cap C) \)[/tex] with [tex]\( P(B) \cdot P(C) \)[/tex].

### Step-by-Step Solution:

1. Identify the probabilities from the table:
- [tex]\( P(B) = 0.30 \)[/tex]
- [tex]\( P(C) = 0.30 \)[/tex]

2. Calculate the joint probability [tex]\( P(B \cap C) \)[/tex]:
Since [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] are mutually exclusive, there is no overlap between these events. Hence,
[tex]\[ P(B \cap C) = 0.0 \][/tex]

3. Calculate the product [tex]\( P(B) \cdot P(C) \)[/tex]:
[tex]\[ P(B) \cdot P(C) = 0.30 \times 0.30 = 0.09 \][/tex]

4. Compare [tex]\( P(B \cap C) \)[/tex] with [tex]\( P(B) \cdot P(C) \)[/tex]:
[tex]\[ P(B \cap C) = 0.0 \quad \text{and} \quad P(B) \cdot P(C) = 0.09 \][/tex]
Since [tex]\( P(B \cap C) \neq P(B) \cdot P(C) \)[/tex], the events [tex]\( B \)[/tex] and [tex]\( C \)[/tex] are not independent.

### Answer:

A. No, they are not independent because [tex]\( P(B \cap C) \neq P(B) \cdot P(C) \)[/tex]. [tex]\( P(B \cap C) = 0.0 \)[/tex] and [tex]\( P(B) \cdot P(C) = 0.09 \)[/tex].

So, the filled in answer choice is:
```markdown
A. No, they are not independent because [tex]\( P(B \cap C) \neq P(B) \cdot P(C) \)[/tex]. [tex]\( P(B \cap C) = 0.0 \)[/tex] and [tex]\( P(B) \cdot P(C) = 0.09 \)[/tex].
```
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