1. Define the event

Let success mean:

In one throw of 4 fair dice, at least 2 dice show either 3 or 5.

For a single die:

• Favorable outcomes for “3 or 5” are $\{3,5\}$, so $p = \frac{2}{6} = \frac{1}{3}$
• Hence, $q = 1-p = \frac{2}{3}$

So for 4 dice, the number of dice showing 3 or 5 follows a binomial distribution: $X \sim \text{Binomial}(4,\tfrac13)$

We need $P(X \ge 2) = P(X=2)+P(X=3)+P(X=4)$

2. Compute the required probability

$P(X=2)=\binom{4}{2}\left(\frac13\right)^2\left(\frac23\right)^2$ $=6\cdot \frac19\cdot \frac49=\frac{24}{81}=\frac{8}{27}$

$P(X=3)=\binom{4}{3}\left(\frac13\right)^3\left(\frac23\right)$ $=4\cdot \frac{1}{27}\cdot \frac23=\frac{8}{81}$

$P(X=4)=\binom{4}{4}\left(\frac13\right)^4=\frac{1}{81}$

Therefore, $P(X\ge 2)=\frac{24}{81}+\frac{8}{81}+\frac{1}{81}=\frac{33}{81}=\frac{11}{27}$

3. Expected number in 27 independent trials

If this experiment is repeated 27 times, and each time success probability is $\frac{11}{27}$, then the expected number of successes is

$E = n p = 27 \cdot \frac{11}{27} = 11$

4. Final answer

The expected number of times is $\boxed{11}$