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

$\text{Digits available: } 1,\ 2,\ 2,\ 2,\ 4,\ 4,\ 0$

$X = \{\text{all five-digit numbers using these digits}\}$

$\text{Each element equally likely}$

$p = P(\text{multiple of } 20 \mid \text{multiple of } 5)$

$\text{To find: } 38p$

$p = \dfrac{|A \cap B|}{|A|},\quad A = \{\text{mult of } 5\},\ B = \{\text{mult of } 20\}$

$B \subseteq A \Rightarrow A \cap B = B$

$\text{So we need } |A| \text{ and } |B|$

$\_\ \_\ \_\ \_\ 0 \quad (\text{multiple of } 5)$

$\text{first four digits from } \{1, 2, 2, 2, 4, 4\}$

hline #1 #2 #4 ways hline 0 2 2 6 0 3 1 4 1 1 2 12 1 2 1 12 1 3 0 4 hline

$|A| = 6 + 4 + 12 + 12 + 4 = 38$

$\text{multiple of } 20 \Rightarrow \text{last two digits } 20 \text{ or } 40$

$\text{ends in } 20:\ \text{first three from } \{1, 2, 2, 4, 4\}$

$3 + 3 + 3 + 6 + 3 = 18$

$\text{ends in } 40:\ \text{first three from } \{1, 2, 2, 2, 4\}$

$1 + 3 + 3 + 6 = 13$

$|B| = 18 + 13 = 31$

$p = \dfrac{|B|}{|A|} = \dfrac{31}{38}$

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