Numbers are selected at random, one at a time from the two-digit numbers $00,01,02$, -------, 99 with replacement. An event E occurs only if the product of the two digits of a selected number is 24 . If four numbers are selected, then probability, that the event E occurs at least 3 times, is

Two-digit numbers from $00$ to $99$ comprise $100$ distinct outcomes. We seek the probability that the product of the digits equals $24$.

Let the number be represented as $10a + b$ where $a, b \in \{0, 1, ..., 9\}$. The product condition $a \cdot b = 24$ holds when $(a, b)$ is $(3,8)$, $(4,6)$, $(6,4)$, or $(8,3)$, corresponding to numbers $38$, $46$, $64$, and $83$.

The probability of selecting one such number is $p = \frac{4}{100} = \frac{1}{25}$, so $q = 1 - p = \frac{24}{25}$.

The scenario involves $n=4$ independent trials. The probability of at least $3$ successes is given by the binomial probabilities for $k=3$ and $k=4$.

$P(X=3) = \binom{4}{3} p^3 q = 4 \left(\frac{1}{25}\right)^3 \cdot \frac{24}{25} = \frac{96}{25^4}$

$P(X=4) = \binom{4}{4} p^4 = \frac{1}{25^4}$

$P(X \ge 3) = \frac{96}{25^4} + \frac{1}{25^4} = \frac{97}{25^4}$, which corresponds to option C.