The probability distribution of random variable $$X$$ is given by

Let $$p = P(1 < X < 4 | X < 3)$$. If $$5p = \lambda K$$, then $$\lambda$$ is equal to _________.

We have a discrete random variable $$X$$ that can take the five values $$1, 2, 3, 4, 5$$ with probabilities $$K, 2K, 2K, 3K, K$$ respectively.

Because these five probabilities must add up to $$1$$, we first write the normalisation condition:

$$K + 2K + 2K + 3K + K = 1.$$

Combining like terms on the left-hand side gives

$$9K = 1.$$

Dividing both sides by $$9$$, we obtain

$$K = \frac{1}{9}.$$

Now we define two events that appear in the required conditional probability.

Event $$A : (1 < X < 4).$$ In words, $$X$$ must be strictly greater than $$1$$ and strictly less than $$4$$, so the possible values are $$X = 2$$ or $$X = 3$$.

Event $$B : (X < 3).$$ This means $$X$$ can be $$1$$ or $$2$$.

The conditional probability we want is

$$p = P\!\bigl(1 < X < 4 \,\bigl|\, X < 3\bigr) = P(A\,|\,B).$$

The definition of conditional probability says

$$P(A\,|\,B) = \frac{P(A \cap B)}{P(B)}.$$

So, we need both $$P(A \cap B)$$ and $$P(B).$$

The intersection $$A \cap B$$ consists of those $$X$$ that satisfy both conditions simultaneously. Event $$A$$ allows $$X = 2, 3,$$ while event $$B$$ allows $$X = 1, 2.$$ The only common value is $$X = 2.$$ Therefore

$$P(A \cap B) = P(X = 2) = 2K.$$

Next, for the denominator, we compute the probability of $$B$$:

$$P(B) = P(X = 1) + P(X = 2) = K + 2K = 3K.$$

Substituting these two results into the conditional-probability formula, we have

$$p = \frac{P(A \cap B)}{P(B)} = \frac{2K}{3K}.$$

The factor $$K$$ cancels, giving

$$p = \frac{2}{3}.$$

Now the problem states that $$5p = \lambda K$$. Substituting $$p = \dfrac{2}{3}$$ on the left and $$K = \dfrac{1}{9}$$ on the right, we get

$$5 \times \frac{2}{3} = \lambda \times \frac{1}{9}.$$

This simplifies to

$$\frac{10}{3} = \frac{\lambda}{9}.$$

Multiplying both sides by $$9$$ to isolate $$\lambda$$, we obtain

$$\lambda = 9 \times \frac{10}{3} = 3 \times 10 = 30.$$

So, the answer is $$30$$.