explain why p(a|d) and p(d|a) from the table below are not equal.\n| | c | d | total |\n|----|----|----|----|…

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. So, $P(A|D)=\frac{P(A\cap D)}{P(D)}$ and $P(D|A)=\frac{P(A\cap D)}{P(A)}$.

Step2: Calculate $P(A\cap D)$

From the table, $P(A\cap D)=\frac{2}{17}$ since there are 2 elements in the intersection of A and D out of a total of 17 elements.

Step3: Calculate $P(D)$

$P(D)=\frac{10}{17}$ as there are 10 elements in D out of 17 total elements. Then $P(A|D)=\frac{P(A\cap D)}{P(D)}=\frac{\frac{2}{17}}{\frac{10}{17}}=\frac{2}{10}=\frac{1}{5}$.

Step4: Calculate $P(A)$

$P(A)=\frac{8}{17}$ as there are 8 elements in A out of 17 total elements. Then $P(D|A)=\frac{P(A\cap D)}{P(A)}=\frac{\frac{2}{17}}{\frac{8}{17}}=\frac{2}{8}=\frac{1}{4}$.

Answer:

Since $P(A|D)=\frac{1}{5}$ and $P(D|A)=\frac{1}{4}$, $P(A|D)$ and $P(D|A)$ are not equal because the denominators in their conditional - probability calculations (i.e., $P(D)$ and $P(A)$) are different, which changes the value of the fractions even though the numerator ($P(A\cap D)$) is the same for both.