Answer :
To understand why [tex]\( P(A \mid D) \)[/tex] and [tex]\( P(D \mid A) \)[/tex] are not equal, let's break down these conditional probabilities step-by-step using the provided table.
### The Table:
[tex]\[ \begin{array}{|c|c|c|c|} \cline { 2 - 4 } \multicolumn{1}{c|}{} & C & D & Total \\ \hline A & 6 & 2 & 8 \\ \hline B & 1 & 8 & 9 \\ \hline Total & 7 & 10 & 17 \\ \hline \end{array} \][/tex]
### Total Probabilities
First, we note the total number of outcomes, which is 17.
### Calculating [tex]\( P(A \mid D) \)[/tex]
[tex]\( P(A \mid D) \)[/tex] is the probability of [tex]\( A \)[/tex] occurring given that [tex]\( D \)[/tex] has occurred. This is given by the formula:
[tex]\[ P(A \mid D) = \frac{P(A \cap D)}{P(D)} \][/tex]
#### Finding [tex]\( P(A \cap D) \)[/tex]
[tex]\( P(A \cap D) \)[/tex] is the probability that both [tex]\( A \)[/tex] and [tex]\( D \)[/tex] occur, which corresponds to the entry where both [tex]\( A \)[/tex] and [tex]\( D \)[/tex] intersect in the table. From the table, we see that [tex]\( A \)[/tex] and [tex]\( D \)[/tex] occur together 2 times out of the 17 total outcomes.
Thus,
[tex]\[ P(A \cap D) = \frac{2}{17} \][/tex]
#### Finding [tex]\( P(D) \)[/tex]
[tex]\( P(D) \)[/tex] is the probability of [tex]\( D \)[/tex] occurring. From the table, we see that [tex]\( D \)[/tex] occurs in 10 out of the 17 total outcomes.
Thus,
[tex]\[ P(D) = \frac{10}{17} \][/tex]
#### Calculating [tex]\( P(A \mid D) \)[/tex]
Now, we can calculate [tex]\( P(A \mid D) \)[/tex]:
[tex]\[ P(A \mid D) = \frac{P(A \cap D)}{P(D)} = \frac{\frac{2}{17}}{\frac{10}{17}} = \frac{2}{10} = 0.2 \][/tex]
### Calculating [tex]\( P(D \mid A) \)[/tex]
[tex]\( P(D \mid A) \)[/tex] is the probability of [tex]\( D \)[/tex] occurring given that [tex]\( A \)[/tex] has occurred. This is given by the formula:
[tex]\[ P(D \mid A) = \frac{P(D \cap A)}{P(A)} \][/tex]
#### Finding [tex]\( P(A) \)[/tex]
[tex]\( P(A) \)[/tex] is the probability of [tex]\( A \)[/tex] occurring. From the table, we see that [tex]\( A \)[/tex] occurs in 8 out of the 17 total outcomes.
Thus,
[tex]\[ P(A) = \frac{8}{17} \][/tex]
#### Finding [tex]\( P(D \cap A) \)[/tex]
We already know from the previous calculation that:
[tex]\[ P(D \cap A) = \frac{2}{17} \][/tex]
#### Calculating [tex]\( P(D \mid A) \)[/tex]
Now, we can calculate [tex]\( P(D \mid A) \)[/tex]:
[tex]\[ P(D \mid A) = \frac{P(D \cap A)}{P(A)} = \frac{\frac{2}{17}}{\frac{8}{17}} = \frac{2}{8} = 0.25 \][/tex]
### Conclusion
- [tex]\( P(A \mid D) = 0.2 \)[/tex]
- [tex]\( P(D \mid A) = 0.25 \)[/tex]
The reason [tex]\( P(A \mid D) \)[/tex] and [tex]\( P(D \mid A) \)[/tex] are not equal is because they measure different aspects of the data.
- [tex]\( P(A \mid D) \)[/tex] measures the probability of [tex]\( A \)[/tex] given that [tex]\( D \)[/tex] has already occurred.
- [tex]\( P(D \mid A) \)[/tex] measures the probability of [tex]\( D \)[/tex] given that [tex]\( A \)[/tex] has already occurred.
These are different conditional probabilities that depend on how the conditions (given events) influence the total probability. Therefore, it is expected for them to not be equal in most cases, and thus they represent different perspectives on the relationship between events [tex]\( A \)[/tex] and [tex]\( D \)[/tex].
### The Table:
[tex]\[ \begin{array}{|c|c|c|c|} \cline { 2 - 4 } \multicolumn{1}{c|}{} & C & D & Total \\ \hline A & 6 & 2 & 8 \\ \hline B & 1 & 8 & 9 \\ \hline Total & 7 & 10 & 17 \\ \hline \end{array} \][/tex]
### Total Probabilities
First, we note the total number of outcomes, which is 17.
### Calculating [tex]\( P(A \mid D) \)[/tex]
[tex]\( P(A \mid D) \)[/tex] is the probability of [tex]\( A \)[/tex] occurring given that [tex]\( D \)[/tex] has occurred. This is given by the formula:
[tex]\[ P(A \mid D) = \frac{P(A \cap D)}{P(D)} \][/tex]
#### Finding [tex]\( P(A \cap D) \)[/tex]
[tex]\( P(A \cap D) \)[/tex] is the probability that both [tex]\( A \)[/tex] and [tex]\( D \)[/tex] occur, which corresponds to the entry where both [tex]\( A \)[/tex] and [tex]\( D \)[/tex] intersect in the table. From the table, we see that [tex]\( A \)[/tex] and [tex]\( D \)[/tex] occur together 2 times out of the 17 total outcomes.
Thus,
[tex]\[ P(A \cap D) = \frac{2}{17} \][/tex]
#### Finding [tex]\( P(D) \)[/tex]
[tex]\( P(D) \)[/tex] is the probability of [tex]\( D \)[/tex] occurring. From the table, we see that [tex]\( D \)[/tex] occurs in 10 out of the 17 total outcomes.
Thus,
[tex]\[ P(D) = \frac{10}{17} \][/tex]
#### Calculating [tex]\( P(A \mid D) \)[/tex]
Now, we can calculate [tex]\( P(A \mid D) \)[/tex]:
[tex]\[ P(A \mid D) = \frac{P(A \cap D)}{P(D)} = \frac{\frac{2}{17}}{\frac{10}{17}} = \frac{2}{10} = 0.2 \][/tex]
### Calculating [tex]\( P(D \mid A) \)[/tex]
[tex]\( P(D \mid A) \)[/tex] is the probability of [tex]\( D \)[/tex] occurring given that [tex]\( A \)[/tex] has occurred. This is given by the formula:
[tex]\[ P(D \mid A) = \frac{P(D \cap A)}{P(A)} \][/tex]
#### Finding [tex]\( P(A) \)[/tex]
[tex]\( P(A) \)[/tex] is the probability of [tex]\( A \)[/tex] occurring. From the table, we see that [tex]\( A \)[/tex] occurs in 8 out of the 17 total outcomes.
Thus,
[tex]\[ P(A) = \frac{8}{17} \][/tex]
#### Finding [tex]\( P(D \cap A) \)[/tex]
We already know from the previous calculation that:
[tex]\[ P(D \cap A) = \frac{2}{17} \][/tex]
#### Calculating [tex]\( P(D \mid A) \)[/tex]
Now, we can calculate [tex]\( P(D \mid A) \)[/tex]:
[tex]\[ P(D \mid A) = \frac{P(D \cap A)}{P(A)} = \frac{\frac{2}{17}}{\frac{8}{17}} = \frac{2}{8} = 0.25 \][/tex]
### Conclusion
- [tex]\( P(A \mid D) = 0.2 \)[/tex]
- [tex]\( P(D \mid A) = 0.25 \)[/tex]
The reason [tex]\( P(A \mid D) \)[/tex] and [tex]\( P(D \mid A) \)[/tex] are not equal is because they measure different aspects of the data.
- [tex]\( P(A \mid D) \)[/tex] measures the probability of [tex]\( A \)[/tex] given that [tex]\( D \)[/tex] has already occurred.
- [tex]\( P(D \mid A) \)[/tex] measures the probability of [tex]\( D \)[/tex] given that [tex]\( A \)[/tex] has already occurred.
These are different conditional probabilities that depend on how the conditions (given events) influence the total probability. Therefore, it is expected for them to not be equal in most cases, and thus they represent different perspectives on the relationship between events [tex]\( A \)[/tex] and [tex]\( D \)[/tex].
