Conditional Probability of Five Digit Numbers Formed from Specific Digits

To find the value of $38p$, we need to calculate the conditional probability $p$ that a randomly chosen element from the set $X$ is a multiple of $20$, given that it is a multiple of $5$.

By definition of conditional probability: $p = P(\text{multiple of } 20 \mid \text{multiple of } 5) = \frac{n(\text{multiples of } 20 \text{ in } X)}{n(\text{multiples of } 5 \text{ in } X)}$

The available multiset of digits is $S = \{1, 2, 2, 2, 4, 4, 0\}$.

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Step 1: Find the number of elements in $X$ that are multiples of 5

Let $A$ be the set of five-digit numbers in $X$ that are multiples of $5$.

For a number to be a multiple of $5$, its units (last) digit must be $0$ or $5$. Since $5$ is not in $S$, the last digit must be $0$.

With $0$ fixed at the units place, the remaining four digits are chosen from $S' = \{1, 2, 2, 2, 4, 4\}$. Since none of the remaining digits is $0$, any four-digit arrangement automatically yields a valid five-digit number.

We count the number of permutations of 4 digits selected from $S'$ based on the frequency of the digit $2$:

1. Three 2s:

   • Remaining digit is $1$: $\{2, 2, 2, 1\} \implies \frac{4!}{3!} = 4$ numbers
   • Remaining digit is $4$: $\{2, 2, 2, 4\} \implies \frac{4!}{3!} = 4$ numbers

2. Two 2s:

   • Remaining digits are $1, 4$: $\{2, 2, 1, 4\} \implies \frac{4!}{2!} = 12$ numbers
   • Remaining digits are $4, 4$: $\{2, 2, 4, 4\} \implies \frac{4!}{2!2!} = 6$ numbers

3. One 2:

   • Remaining digits are $1, 4, 4$: $\{2, 1, 4, 4\} \implies \frac{4!}{2!} = 12$ numbers

4. Zero 2s:

   • Impossible, as there are only 3 non-$2$ digits available in $S'$.

Summing all the cases for $n(A)$: $n(A) = 4 + 4 + 12 + 6 + 12 = 38$

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Step 2: Find the number of elements in $X$ that are multiples of 20

Let $B$ be the set of five-digit numbers in $X$ that are multiples of $20$.

A number is a multiple of $20$ if its last digit is $0$ and its tens digit is even (i.e., $2$ or $4$). Thus, the numbers in $B$ must end in either $20$ or $40$.

Case 1: Numbers ending in $20$

The last two digits are fixed as $2, 0$. The remaining $3$ digits are chosen from $S \setminus \{2, 0\} = \{1, 2, 2, 4, 4\}$:

• Two 2s:
  • $\{2, 2, 1\} \implies \frac{3!}{2!} = 3$
  • $\{2, 2, 4\} \implies \frac{3!}{2!} = 3$
• One 2:
  • $\{2, 1, 4\} \implies 3! = 6$
  • $\{2, 4, 4\} \implies \frac{3!}{2!} = 3$
• Zero 2s:
  • $\{1, 4, 4\} \implies \frac{3!}{2!} = 3$

Total numbers ending in $20$: $3 + 3 + 6 + 3 + 3 = 18$

Case 2: Numbers ending in $40$

The last two digits are fixed as $4, 0$. The remaining $3$ digits are chosen from $S \setminus \{4, 0\} = \{1, 2, 2, 2, 4\}$:

• Three 2s:
  • $\{2, 2, 2\} \implies \frac{3!}{3!} = 1$
• Two 2s:
  • $\{2, 2, 1\} \implies \frac{3!}{2!} = 3$
  • $\{2, 2, 4\} \implies \frac{3!}{2!} = 3$
• One 2:
  • $\{2, 1, 4\} \implies 3! = 6$
• Zero 2s: Impossible (only two non-$2$ digits available).

Total numbers ending in $40$: $1 + 3 + 3 + 6 = 13$

Summing both cases for $n(B)$: $n(B) = 18 + 13 = 31$

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Step 3: Calculate $p$ and $38p$

The conditional probability $p$ is: $p = \frac{n(B)}{n(A)} = \frac{31}{38}$

Therefore: $38p = 38 \times \frac{31}{38} = 31$