Given the probabilities:
\[ P(A) = P(B) = \frac{5}{13}, \quad P(A \cap B) = \frac{2}{5}, \] we need to find \( P(A \cup B) \).
Using the formula for the union of two events, we have:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the given values: \[ P(A \cup B) = \frac{5}{13} + \frac{5}{13} - \frac{2}{5} \]
Now, simplify:
\[ P(A \cup B) = \frac{10}{13} - \frac{2}{5} \] To subtract these fractions, find a common denominator (the LCM of 13 and 5 is 65): \[ P(A \cup B) = \frac{50}{65} - \frac{26}{65} = \frac{24}{65} \]
\[ P(A \cup B) = \frac{24}{65} \]
Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability that neither of the lightbulbs selected will be defective?