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Sagot :
To find the conditional probability [tex]\( P(C \mid Y) \)[/tex], we need to understand the concept of conditional probability. Conditional probability, [tex]\( P(A \mid B) \)[/tex], is the probability of event [tex]\( A \)[/tex] occurring given that event [tex]\( B \)[/tex] has occurred. It is given by the formula:
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]
In this problem, [tex]\( C \)[/tex] represents the event of falling into category [tex]\( C \)[/tex], and [tex]\( Y \)[/tex] represents the event of fall into category [tex]\( Y \)[/tex].
Looking at the provided table:
\begin{tabular}{|c|c|c|c|c|}
\hline & [tex]$X$[/tex] & [tex]$Y$[/tex] & [tex]$Z$[/tex] & 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}
We are interested in [tex]\( P(C \mid Y) \)[/tex]:
1. Identify [tex]\( P(C \cap Y) \)[/tex] (The Probability of both [tex]\( C \)[/tex] and [tex]\( Y \)[/tex] occurring):
The number of occurrences of [tex]\( C \)[/tex] and [tex]\( Y \)[/tex] together is given in the table as 15.
2. Identify [tex]\( P(Y) \)[/tex] (The Probability of [tex]\( Y \)[/tex] occurring):
The total number of occurrences of [tex]\( Y \)[/tex] is given in the table as 30.
So, to calculate [tex]\( P(C \mid Y) \)[/tex]:
[tex]\[ P(C \mid Y) = \frac{P(C \cap Y)}{P(Y)} = \frac{\text{Number of occurrences of both } C \text{ and } Y}{\text{Total number of occurrences of } Y} \][/tex]
[tex]\[ P(C \mid Y) = \frac{15}{30} = 0.5 \][/tex]
To the nearest tenth, the value of [tex]\( P(C \mid Y) \)[/tex] is [tex]\( 0.5 \)[/tex].
Therefore, the correct answer is:
[tex]\[ 0.5 \][/tex]
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]
In this problem, [tex]\( C \)[/tex] represents the event of falling into category [tex]\( C \)[/tex], and [tex]\( Y \)[/tex] represents the event of fall into category [tex]\( Y \)[/tex].
Looking at the provided table:
\begin{tabular}{|c|c|c|c|c|}
\hline & [tex]$X$[/tex] & [tex]$Y$[/tex] & [tex]$Z$[/tex] & 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}
We are interested in [tex]\( P(C \mid Y) \)[/tex]:
1. Identify [tex]\( P(C \cap Y) \)[/tex] (The Probability of both [tex]\( C \)[/tex] and [tex]\( Y \)[/tex] occurring):
The number of occurrences of [tex]\( C \)[/tex] and [tex]\( Y \)[/tex] together is given in the table as 15.
2. Identify [tex]\( P(Y) \)[/tex] (The Probability of [tex]\( Y \)[/tex] occurring):
The total number of occurrences of [tex]\( Y \)[/tex] is given in the table as 30.
So, to calculate [tex]\( P(C \mid Y) \)[/tex]:
[tex]\[ P(C \mid Y) = \frac{P(C \cap Y)}{P(Y)} = \frac{\text{Number of occurrences of both } C \text{ and } Y}{\text{Total number of occurrences of } Y} \][/tex]
[tex]\[ P(C \mid Y) = \frac{15}{30} = 0.5 \][/tex]
To the nearest tenth, the value of [tex]\( P(C \mid Y) \)[/tex] is [tex]\( 0.5 \)[/tex].
Therefore, the correct answer is:
[tex]\[ 0.5 \][/tex]
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