Sure, let's find [tex]\( P(A \cap B) \)[/tex] using the given probabilities.
We know:
- [tex]\( P(A) = \frac{2}{5} \)[/tex]
- [tex]\( P(B \mid A) = \frac{9}{20} \)[/tex]
To find [tex]\( P(A \cap B) \)[/tex], we use the definition of conditional probability, which is given by:
[tex]\[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \][/tex]
Rearranging this formula to solve for [tex]\( P(A \cap B) \)[/tex], we get:
[tex]\[ P(A \cap B) = P(B \mid A) \cdot P(A) \][/tex]
Substitute the given values into this formula:
[tex]\[ P(A \cap B) = \frac{9}{20} \cdot \frac{2}{5} \][/tex]
To multiply the fractions, multiply the numerators together and the denominators together:
[tex]\[ P(A \cap B) = \frac{9 \times 2}{20 \times 5} \][/tex]
[tex]\[ P(A \cap B) = \frac{18}{100} \][/tex]
Simplifying [tex]\( \frac{18}{100} \)[/tex] gives:
[tex]\[ P(A \cap B) = \frac{9}{50} \][/tex]
Thus, the missing probability [tex]\( P(A \cap B) \)[/tex] is [tex]\( \frac{9}{50} \)[/tex], which corresponds to option D.
