A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $$\dfrac{k}{3^{11}}$$, then $$k$$ is equal to

For throwing a pair of dice, the ways to get a total of 5 are: (1,4), (2,3), (3,2), (4,1) = 4 outcomes out of 36.

Probability of success: $$p = \frac{4}{36} = \frac{1}{9}$$

Probability of failure: $$q = 1 - \frac{1}{9} = \frac{8}{9}$$

The dice are thrown 5 times. We need P(at least 4 successes) = P(4) + P(5).

$$ P(5) = \binom{5}{5}\left(\frac{1}{9}\right)^5 = \frac{1}{9^5} $$

$$ P(4) = \binom{5}{4}\left(\frac{1}{9}\right)^4\left(\frac{8}{9}\right) = \frac{5 \times 8}{9^5} = \frac{40}{9^5} $$

$$ P(\geq 4) = \frac{41}{9^5} = \frac{41}{59049} $$

Converting to the form $$\frac{k}{3^{11}}$$: Since $$9^5 = 3^{10}$$, we have:

$$ \frac{41}{3^{10}} = \frac{41 \times 3}{3^{11}} = \frac{123}{3^{11}} $$

Therefore $$k = \mathbf{123}$$.