Let X be the set of all five-digit numbers formed using 1, 2, 2, 2, 4, 4, 0. For example, 22240 is in X while 02244 and 44422 are not in X. Suppose that each element of X has an equal chance of being chosen. Let p be the conditional probability that an element chosen at random is a multiple of 20 given that it is a multiple of 5. Then the value of 38p is equal to:

To find the value of 38p, let us analyze the problem step-by-step.

Step 1: Understand the formation of set X
Set X consists of all five-digit numbers formed using the multiset of available digits: {0, 1, 2, 2, 2, 4, 4}.
A five-digit number cannot have 0 as its first (ten-thousands) digit.

Step 2: Define the events and the conditional probability p
Let event A be the event that a chosen number from X is a multiple of 20.
Let event B be the event that a chosen number from X is a multiple of 5.
We are given that each element of X has an equal chance of being chosen.
The conditional probability p is defined as:

$p=P\left(A\right|B)=\frac{n(A\cap B)}{n\left(B\right)}$

Since any number that is a multiple of 20 is automatically a multiple of 5, we have A ∩ B = A.
Thus, p = n(A) / n(B), where:

• n(B) is the number of elements in X that are multiples of 5.
• n(A) is the number of elements in X that are multiples of 20.

Step 3: Calculate n(B) - Count of 5-digit numbers that are multiples of 5
A number is a multiple of 5 if its units digit is 0 or 5. Since 5 is not in our set of available digits, any multiple of 5 in X must end in 0.
So, the units digit is fixed as 0.
The remaining 4 digits (ten-thousands, thousands, hundreds, tens) must be chosen from the multiset {1, 2, 2, 2, 4, 4}.

We analyze the possible combinations of 4 digits chosen from {1, 2, 2, 2, 4, 4}:

1. Three 2s and one other digit:
• Case 1a: Digits are {2, 2, 2, 1}. Number of arrangements = 4! / 3! = 4.
• Case 1b: Digits are {2, 2, 2, 4}. Number of arrangements = 4! / 3! = 4.

2. Two 2s and two 4s:
• Case 2: Digits are {2, 2, 4, 4}. Number of arrangements = 4! / (2! × 2!) = 6.

3. Two 2s, one 4, and one 1:
• Case 3: Digits are {2, 2, 4, 1}. Number of arrangements = 4! / 2! = 12.

4. One 2, two 4s, and one 1:
• Case 4: Digits are {2, 4, 4, 1}. Number of arrangements = 4! / 2! = 12.

Summing these cases gives total n(B):

$n\left(B\right)=4+4+6+12+12=38$

Step 4: Calculate n(A) - Count of 5-digit numbers that are multiples of 20
A number is a multiple of 20 if it ends in 00, 20, 40, 60, or 80.
Since our available digits are {0, 1, 2, 2, 2, 4, 4}, the last two digits (tens and units) of a multiple of 20 must be either 20 or 40 (00 is not possible as there is only one 0).

• Case A: Last two digits are 20
The units and tens digits are fixed as 20.
The first three digits (ten-thousands, thousands, hundreds) must be formed from the remaining digits in {1, 2, 2, 4, 4}.
- Subcase A1: Choose three digits as {2, 2, 1}. Arrangements = 3! / 2! = 3.
- Subcase A2: Choose three digits as {2, 2, 4}. Arrangements = 3! / 2! = 3.
- Subcase A3: Choose three digits as {4, 4, 1}. Arrangements = 3! / 2! = 3.
- Subcase A4: Choose three digits as {4, 4, 2}. Arrangements = 3! / 2! = 3.
- Subcase A5: Choose three digits as {2, 4, 1}. Arrangements = 3! = 6.
Total arrangements ending in 20 = 3 + 3 + 3 + 3 + 6 = 18.

• Case B: Last two digits are 40
The units and tens digits are fixed as 40.
The first three digits must be formed from the remaining digits in {1, 2, 2, 2, 4}.
- Subcase B1: Choose three digits as {2, 2, 2}. Arrangements = 3! / 3! = 1.
- Subcase B2: Choose three digits as {2, 2, 1}. Arrangements = 3! / 2! = 3.
- Subcase B3: Choose three digits as {2, 2, 4}. Arrangements = 3! / 2! = 3.
- Subcase B4: Choose three digits as {2, 4, 1}. Arrangements = 3! = 6.
Total arrangements ending in 40 = 1 + 3 + 3 + 6 = 13.

Summing both cases gives total n(A):

$n\left(A\right)=18+13=31$

Step 5: Compute conditional probability p and 38p
Using our results from Steps 3 and 4:

$p=\frac{n\left(A\right)}{n\left(B\right)}=\frac{31}{38}$

Therefore, the value of 38p is:

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

The correct answer is 31.