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

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

Are the events [tex]\( B \)[/tex] and [tex]\( A \)[/tex] independent? Select the correct answer below and fill in the answer boxes to complete your choice.

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

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

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

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

Sagot :

GinnyAnswer
To determine whether events B and A are independent, we must check if the probability of their intersection [tex]\( P(B \cap A) \)[/tex] is equal to the product of their individual probabilities [tex]\( P(B) \times P(A) \)[/tex].

From the given table, we can extract the following probabilities:
- [tex]\( P(A) = 0.30 \)[/tex]
- [tex]\( P(B) = 0.40 \)[/tex]
- [tex]\( P(B \cap A) = 0.16 \)[/tex] (this value corresponds to the cell in the table where both B and A occur)

Now, let's calculate [tex]\( P(B) \times P(A) \)[/tex]:
[tex]\[ P(B) \times P(A) = 0.40 \times 0.30 = 0.12 \][/tex]

We then compare [tex]\( P(B \cap A) \)[/tex] and [tex]\( P(B) \times P(A) \)[/tex]:
- [tex]\( P(B \cap A) = 0.16 \)[/tex]
- [tex]\( P(B) \times P(A) = 0.12 \)[/tex]

Since [tex]\( P(B \cap A) \neq P(B) \times P(A) \)[/tex] (specifically, [tex]\( 0.16 \neq 0.12 \)[/tex]), events B and A are not independent.

Therefore, the correct answer is:
B. No, they are not independent because [tex]\( P(B \cap A) \neq P(B) \times P(A) \)[/tex]. [tex]\( P(B \cap A) = 0.16 \)[/tex] and [tex]\( P(B) \times P(A) = 0.12 \)[/tex].
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