If a number $n$ is chosen at random from the set $\{11,12,13, \ldots \ldots, 30\}$. Then, the probability that $n$ is neither divisible by 3 nor divisible by 5, is

The set of numbers is $S = \{11, 12, 13, \ldots, 30\}$. The total number of elements in $S$ is $30 - 11 + 1 = 20$.

Let $A$ be the set of numbers in $S$ divisible by 3. These are $\{12, 15, 18, 21, 24, 27, 30\}$. The number of elements in $A$ is $n(A) = 7$.

Let $B$ be the set of numbers in $S$ divisible by 5. These are $\{15, 20, 25, 30\}$. The number of elements in $B$ is $n(B) = 4$.

The set of numbers divisible by both 3 and 5 (i.e., divisible by 15) is $A \cap B = \{15, 30\}$. The number of elements in $A \cap B$ is $n(A \cap B) = 2$.

The number of elements divisible by 3 or 5 is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 7 + 4 - 2 = 9$.

The number of elements neither divisible by 3 nor by 5 is $n(S) - n(A \cup B) = 20 - 9 = 11$.

The probability is $\dfrac{11}{20}$.

Answer: $\dfrac{11}{20}$