Explain why [tex][tex]$P( A \mid D )$[/tex][/tex] and [tex][tex]$P( D \mid A )$[/tex][/tex] from the table below are not equal.

\begin{tabular}{|c|c|c|c|}
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & C & D & Total \\
\hline
A & 6 & 2 & 8 \\
\hline
B & 1 & 8 & 9 \\
\hline
Total & 7 & 10 & 17 \\
\hline
\end{tabular}



Answer :

Let's delve into the concepts of conditional probability and analyze why [tex]\( P(A \mid D) \)[/tex] and [tex]\( P(D \mid A) \)[/tex] are not equal using the provided table:

[tex]\[ \begin{tabular}{|c|c|c|c|} \cline { 2 - 4 } \multicolumn{1}{c|}{} & C & D & Total \\ \hline A & 6 & 2 & 8 \\ \hline B & 1 & 8 & 9 \\ \hline Total & 7 & 10 & 17 \\ \hline \end{tabular} \][/tex]

### Step-by-Step Solution:

1. Understanding the Table:
- Rows represent events A and B.
- Columns represent events C and D.
- Cell Values represent the joint occurrences of the events.

2. Calculating [tex]\( P(A \mid D) \)[/tex]:
- [tex]\( P(A \mid D) \)[/tex] is the probability of event A occurring given that event D has occurred.
- Using the conditional probability formula: [tex]\( P(A \mid D) = \frac{P(A \cap D)}{P(D)} \)[/tex].

From the table:
- [tex]\( P(A \cap D) \)[/tex] is the number of instances where both A and D occur, which is the value at the intersection of row A and column D (2). Thus, [tex]\( P(A \cap D) = 2 \)[/tex].
- [tex]\( P(D) \)[/tex] is the total number of instances where D occurs, which is the sum of the values in column D (10). Thus, [tex]\( P(D) = 10 \)[/tex].

Now, we can calculate:
[tex]\[ P(A \mid D) = \frac{2}{10} = 0.2 \][/tex]

3. Calculating [tex]\( P(D \mid A) \)[/tex]:
- [tex]\( P(D \mid A) \)[/tex] is the probability of event D occurring given that event A has occurred.
- Using the conditional probability formula: [tex]\( P(D \mid A) = \frac{P(D \cap A)}{P(A)} \)[/tex].

From the table:
- [tex]\( P(D \cap A) \)[/tex] is the number of instances where both D and A occur, which is again the value at the intersection of row A and column D (2). Thus, [tex]\( P(D \cap A) = 2 \)[/tex].
- [tex]\( P(A) \)[/tex] is the total number of instances where A occurs, which is the sum of the values in row A (8). Thus, [tex]\( P(A) = 8 \)[/tex].

Now, we can calculate:
[tex]\[ P(D \mid A) = \frac{2}{8} = 0.25 \][/tex]

4. Comparison and Explanation:
- When we compare the two probabilities [tex]\( P(A \mid D) \)[/tex] and [tex]\( P(D \mid A) \)[/tex]:
[tex]\[ P(A \mid D) = 0.2 \quad \text{and} \quad P(D \mid A) = 0.25 \][/tex]
- These probabilities are not equal because they refer to different conditional scenarios. Specifically:
- [tex]\( P(A \mid D) \)[/tex] considers the fraction of A within the context of D's occurrences.
- [tex]\( P(D \mid A) \)[/tex] considers the fraction of D within the context of A's occurrences.

The denominators in these calculations reflect different contexts: total occurrences of D for [tex]\( P(A \mid D) \)[/tex] and total occurrences of A for [tex]\( P(D \mid A) \)[/tex]. Thus, the probabilities are computed with different bases, resulting in different values.

### Conclusion:
The probabilities [tex]\( P(A \mid D) \)[/tex] and [tex]\( P(D \mid A) \)[/tex] are not equal because they refer to different conditional probabilities and are calculated with different denominators. This difference in context and base of calculation leads to different results.