Question:

**Step 1: Analyze the conditions** A number in $X$ is a 5-digit number (meaning the first digit cannot be 0). We are looking for $p = P(\text{multiple of } 20 \mid \text{multiple of } 5) = \frac{n(\text{multiple of } 20)}{n(\text{multiple of } 5)}$.\nA number is a multiple of 5 if it ends in 0 or 5. Since we only have 0, it must end in 0. So $n(\text{multiple of } 5)$ is the number of 5-digit numbers ending in 0.\nA number is a multiple of 20 if it ends in 00, 20, 40, 60, or 80. Since we only have one 0, it cannot end in 00. It can end in 20 or 40. So $n(\text{multiple of } 20)$ is the number of 5-digit numbers ending in 20 or 40. **Step 2: Count multiples of 5 (ending in 0)** The last digit is 0. The remaining 4 digits are chosen from $\{1, 2, 2, 2, 4, 4\}$. We have several cases based on the multiset of 4 digits chosen:\nCase 1: Three 2s (2, 2, 2). The fourth digit can be 1 or 4. \n- {2, 2, 2, 1}: $4!/3! = 4$ arrangements.\n- {2, 2, 2, 4}: $4!/3! = 4$ arrangements.\nCase 2: Two 4s (4, 4). The remaining two digits can be {1, 2} or {2, 2}.\n- {4, 4, 1, 2}: $4!/2! = 12$ arrangements.\n- {4, 4, 2, 2}: $4!/(2!2!) = 6$ arrangements.\nCase 3: Two 2s, one 4, one 1 (2, 2, 4, 1): $4!/2! = 12$ arrangements.\nTotal number of ways to form the first 4 digits = $4 + 4 + 12 + 6 + 12 = 38$.\nSo $n(\text{multiple of } 5) = 38$. **Step 3: Count multiples of 20** These numbers end in 20 or 40. We can count the number ending in 10 (which are NOT multiples of 20) and subtract from 38, as the tens digit must be from $\{1, 2, 4\}$.\nFor ending in 10: The last two digits are 10. The remaining 3 digits are chosen from {2, 2, 2, 4, 4}.\nSubsets of size 3:\n- {2, 2, 2} $\implies 3!/3! = 1$\n- {2, 2, 4} $\implies 3!/2! = 3$\n- {2, 4, 4} $\implies 3!/2! = 3$\nTotal ending in 10 = $1 + 3 + 3 = 7$.\nSo multiples of 20 are $38 - 7 = 31$. **Step 4: Calculate probability** The conditional probability is $p = \frac{31}{38}$.\nThen $38p = 31$.