In a single toss of a fair die, the odds against the event that number 4 or 5 turns up is

The sample space for a fair die toss is $S = \{1, 2, 3, 4, 5, 6\}$ with $n(S) = 6$ total outcomes.

Let $E$ represent the event that a 4 or 5 appears, so $E = \{4, 5\}$ with $n(E) = 2$ favorable outcomes.
The probability is $P(E) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3}$.

The odds against $E$ are expressed as $P(E') : P(E)$, where $E'$ denotes the complement of $E$.
The complement consists of outcomes $\{1, 2, 3, 6\}$, so $P(E') = \frac{4}{6} = \frac{2}{3}$.

The odds ratio becomes $\frac{2}{3} : \frac{1}{3}$, which simplifies to $2 : 1$ after multiplying both terms by 3.

Odds against the event: $\boxed{\text{A}}$