Solve for [tex]\( x \)[/tex].

[tex]\[ 3x = 6x - 2 \][/tex]


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Init Test

Find [tex]\( P(C \mid Y) \)[/tex] from the information in the table.

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline & X & Y & Z & Total \\
\hline A & 32 & 10 & 28 & 70 \\
\hline B & 6 & 5 & 25 & 36 \\
\hline C & 18 & 15 & 7 & 40 \\
\hline Total & 56 & 30 & 60 & 146 \\
\hline
\end{tabular}
\][/tex]

To the nearest tenth, what is the value of [tex]\( P( C \mid Y ) \)[/tex]?

A. 0.4
B. 0.5
C. 0.7
D. 0.8



Answer :

To find [tex]\( P(C \mid Y) \)[/tex], we need to use the concept of conditional probability. The conditional probability formula is:

[tex]\[ P(C \mid Y) = \frac{P(C \cap Y)}{P(Y)} \][/tex]

From the table, we have the following relevant information:

1. The total number of events where [tex]\( Y \)[/tex] occurs, [tex]\( P(Y) \)[/tex]:
- [tex]\( Y \)[/tex] appears in the columns with totals: [tex]\( 10 + 5 + 15 = 30 \)[/tex].

2. The total number of events where both [tex]\( C \)[/tex] and [tex]\( Y \)[/tex] occur, [tex]\( P(C \cap Y) \)[/tex]:
- There are 15 events where both [tex]\( C \)[/tex] and [tex]\( Y \)[/tex] are true.

Using this information, we can calculate:

[tex]\[ P(C \mid Y) = \frac{P(C \cap Y)}{P(Y)} = \frac{15}{30} = 0.5 \][/tex]

Thus, the value of [tex]\( P(C \mid Y) \)[/tex], rounded to the nearest tenth, is [tex]\( 0.5 \)[/tex].

So the correct answer is:

[tex]\[ \boxed{0.5} \][/tex]