Determine the probability that a randomly chosen integer is…

Determine the probability that a randomly chosen integer is not divisible $1^{\circ}$ by 2 nor by 3 (by neither of them); $2^{\circ}$ by 2 or by 3 (i.e. is not divisible by 2, or is not divisible by 3).

Step 1: The model

"A randomly chosen integer" means: its remainder when divided by $6$ is equally likely to be $0,1,\dots,5$ (take the integer uniformly from a long range). Then

\[ P(A)=\tfrac12,\quad P(B)=\tfrac13,\quad P(AB)=P(6\mid N)=\tfrac16=P(A)P(B). \]

Step 2: $1^{\circ}$ — divisible by neither 2 nor 3

By independence of $\overline A$ and $\overline B$:

\[ \boxed{P(\overline A\,\overline B)=\left(1-\tfrac12\right)\left(1-\tfrac13\right)=\tfrac13.} \]

Check: among $0,\dots,5$ only $1$ and $5$ qualify, $\frac26=\frac13$ ✓.

Step 3: $2^{\circ}$ — not divisible by 2, or not divisible by 3

By De Morgan, $\overline A+\overline B=\overline{AB}$ = "not divisible by $6$":

\[ \boxed{P(\overline A+\overline B)=1-P(AB)=1-\tfrac16=\tfrac56.} \]

Step 4: Table by remainder mod 6

Step 5: Another reading of part $2^{\circ}$

If part $2^{\circ}$ is read as "divisible by 2 or by 3" (without "not"), the answer is $1-\frac13=\frac23$. The two parts of the problem are meant to contrast "neither … nor" ($\frac13$) with "not … or not" ($\frac56$).