Step-by-step Solution:

1. Understanding the Problem and Defining Events We are given a set of digits {1, 2, 2, 2, 4, 4, 0}. The set $X$ consists of all 5-digit numbers formed using these digits. We need to find the conditional probability $p$ that a randomly chosen number from $X$ is a multiple of 20, given that it is a multiple of 5. Finally, we need to compute the value of $38p$.

   Let's define the events:

   • $A$: The chosen number is a multiple of 5.
   • $B$: The chosen number is a multiple of 20.

   The required conditional probability is $p = P(B|A) = \frac{n(A \cap B)}{n(A)}$.

2. Analyzing the Condition for the Events

   • For a number to be a multiple of 5, its last digit must be 0 or 5. Given the available digits, the last digit must be 0.
   • For a number to be a multiple of 20, it must be a multiple of both 4 and 5. This means its last digit must be 0, and the number formed by the last two digits must be divisible by 4. The possible endings are ...20, ...40, ...60, ...80, ...00. From the available digits, the number must end in 20 or 40.
   • Notice that if a number is a multiple of 20 (event B), it is automatically a multiple of 5 (event A). Therefore, the intersection of events A and B, $A \cap B$, is simply event B.
   • So, the probability becomes $p = \frac{n(B)}{n(A)}$.

3. Calculating n(A): Number of 5-digit numbers that are multiples of 5 A number in $X$ is a multiple of 5 if its last digit is 0. We fix the last digit (5th position) as 0. We need to form a 4-digit number for the first four positions using 4 digits from the remaining set {1, 2, 2, 2, 4, 4}. Since 0 is at the end, the first digit cannot be 0, which is satisfied.

   We find the number of ways to arrange 4 digits chosen from {1, 2, 2, 2, 4, 4}:

   • Case 1: Digits chosen are {2, 2, 2, 4}. Number of permutations = $\frac{4!}{3!} = 4$.
   • Case 2: Digits chosen are {2, 2, 4, 4}. Number of permutations = $\frac{4!}{2!2!} = \frac{24}{4} = 6$.
   • Case 3: Digits chosen are {1, 2, 2, 2}. Number of permutations = $\frac{4!}{3!} = 4$.
   • Case 4: Digits chosen are {1, 2, 2, 4}. Number of permutations = $\frac{4!}{2!} = 12$.
   • Case 5: Digits chosen are {1, 2, 4, 4}. Number of permutations = $\frac{4!}{2!} = 12$.

   The total number of multiples of 5 is $n(A) = 4 + 6 + 4 + 12 + 12 = 38$.

4. Calculating n(B): Number of 5-digit numbers that are multiples of 20 A number in $X$ is a multiple of 20 if it ends in 20 or 40.

   • Subcase 1: The number ends in 20. The last two digits are fixed as 20. We have used one 2 and one 0. The remaining digits are {1, 2, 2, 4, 4}. We need to fill the first three positions by choosing 3 digits from this set and arranging them.

     • Digits chosen {1, 2, 2}: Permutations = $\frac{3!}{2!} = 3$.
     • Digits chosen {1, 2, 4}: Permutations = $3! = 6$.
     • Digits chosen {1, 4, 4}: Permutations = $\frac{3!}{2!} = 3$.
     • Digits chosen {2, 2, 4}: Permutations = $\frac{3!}{2!} = 3$.
     • Digits chosen {2, 4, 4}: Permutations = $\frac{3!}{2!} = 3$. Number of multiples of 20 ending in 20 = $3 + 6 + 3 + 3 + 3 = 18$.

   • Subcase 2: The number ends in 40. The last two digits are fixed as 40. We have used one 4 and one 0. The remaining digits are {1, 2, 2, 2, 4}. We need to fill the first three positions by choosing 3 digits from this set and arranging them.

     • Digits chosen {1, 2, 2}: Permutations = $\frac{3!}{2!} = 3$.
     • Digits chosen {1, 2, 4}: Permutations = $3! = 6$.
     • Digits chosen {2, 2, 2}: Permutations = $\frac{3!}{3!} = 1$.
     • Digits chosen {2, 2, 4}: Permutations = $\frac{3!}{2!} = 3$. Number of multiples of 20 ending in 40 = $3 + 6 + 1 + 3 = 13$.

   The total number of multiples of 20 is $n(B) = 18 + 13 = 31$.

5. Calculating the final value The conditional probability is $p = \frac{n(B)}{n(A)} = \frac{31}{38}$.

   The question asks for the value of $38p$. $38p = 38 \times \frac{31}{38} = 31.$

The final answer is 31.