Happy Chicken Year!

From the question, there are $6$ possible outcomes from this laying (R = red; B = blue; G = green) :

(R, R), (B, B), (G, G), (R, B), (B, G), (G, R)

Also, we know that for each outcome, such probability can be written as $\dfrac{1}{n}$ for some natural number $n$ .

By applying Venn's diagram, the probability for each event can be calculated similarly to the set operation:

$P(R \cup B \cup G) = 1 = P(R) + P(B) + P(G) - P(R \cap B) - P(B \cap G) - P(G \cap R) + P(R \cap B \cap G)$

However, in a day, the hen only lays $2$ eggs, so there is no way that there will be $3$ colored eggs. Hence, $P(R \cap B \cap G) = 0$ .

Then we can substitute in the values of $P(R)$ ; $P(B)$ ; $P(G)$ as given:

$P(R \cup B \cup G) = 1 = \dfrac{2}{3} + \dfrac{2}{4} + \dfrac{2}{5} - P(R \cap B) - P(B \cap G) - P(G \cap R)$

Then $P(R \cap B) + P(B \cap G) + P(G \cap R) = \dfrac{40+30+24-60}{60} = \dfrac{17}{30}$ .

Since we know that the probability of each color combination is the reciprocal of some natural number, $\dfrac{17}{30} = \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$ for positive integers $a, b, c$ .

Thus, its applicable summation can be:

$\dfrac{17}{30} = \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{15} = \dfrac{1}{6} + \dfrac{1}{5} + \dfrac{1}{5}\ = \dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{15}$ .

However, only the first summation can result in probability of form $\dfrac{1}{n}$ for three sole color outcomes. Therefore, $\dfrac{17}{30} = \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{15}$

Now that we know the probabilities of the color combination, we just need to match it to the right combination. This is important as, for example, if we let $P(R \cap B) = \dfrac{1}{3}$ ; $P(B \cap G) = \dfrac{1}{6}$ ; $P(G \cap R) = \dfrac{1}{15}$ , then the probability of getting only blue eggs in a day = $P(B) - (P(B \cap G) - P(R \cap B) = \dfrac{2}{4} - \dfrac{1}{6} - \dfrac{1}{3} = 0$ , which is contradicted.

By orienting the values to the optimal point, we will obtain:

$P(R \cap B) = \dfrac{1}{3}$

$P(B \cap G) = \dfrac{1}{15}$

$P(G \cap R) = \dfrac{1}{6}$

As a result, the probability of getting only blue eggs in a day = $P(B) - (P(B \cap G) - P(R \cap B) = \dfrac{2}{4} - \dfrac{1}{15} - \dfrac{1}{3} = \dfrac{1}{10}$ .

Therefore, the mentioned $x = \boxed{10}$ .