An integer is chosen at random from the first 50 integers. What is the probability that the integer is neither divisible by 5 nor 9?

Total number of integers in the sample space, $n(S) = 50$.

Let $A$ be the event that the chosen integer is divisible by $5$, and $B$ be the event that it is divisible by $9$.

The number of integers divisible by $5$ is $n(A) = 10$.

The number of integers divisible by $9$ is $n(B) = 5$.

The number of integers divisible by both $5$ and $9$ (i.e., divisible by $45$) is $n(A \cap B) = 1$.

The number of integers divisible by either $5$ or $9$ is given by:

$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 10 + 5 - 1 = 14$

The number of integers neither divisible by $5$ nor $9$ is:

$n(A' \cap B') = n(S) - n(A \cup B) = 50 - 14 = 36$

The required probability is:

$P(A' \cap B') = \dfrac{36}{50} = \dfrac{18}{25}$

Answer: $18/25$