The probability that a certain kind of component will survive a given test is $\frac{2}{3}$. The probability that at most 2 components out of 4 tested, will survive is

Let $X$ be the number of components surviving the test, following a binomial distribution with $n = 4$ and success probability $p = \frac{2}{3}$.

To find $P(X \leq 2)$, compute $P(X = 0) + P(X = 1) + P(X = 2)$ using the binomial PMF $P(X = k) = \binom{4}{k} \left(\frac{2}{3}\right)^k \left(\frac{1}{3}\right)^{4 - k}$.

Evaluate each term:
$P(X = 0) = \binom{4}{0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^4 = \frac{1}{3^4}$
$P(X = 1) = \binom{4}{1} \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^3 = \frac{8}{3^4}$
$P(X = 2) = \binom{4}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^2 = \frac{24}{3^4}$

The sum is $P(X \leq 2) = \frac{1 + 8 + 24}{3^4} = \frac{33}{3^4}$.

This matches option C.