A person A was issued a $10$ digit mobile phone number having his $6$ digit birth date (DDMMYY format) in it. A person $'A'$ was issued a $10$ digit mobile phone number having his $6$ digit birth date (DDMMYY format) in it. Let $E$ be the event that his birth was in first $9$ days of a month, then the probability of the occurrence of the event $E$ is
My Approach: There are $5$ ways to choose $6$ places for birth date in (DDMMYY) format.
$n(E)=5\cdot9\cdot 12\cdot 100 \cdot 10^{4}$
Note: There are $9$  ways to select DD, $12$ ways to select MM and $100$ ways to select YY remaining $4$ places can be filled in $10^{4}$ ways.
Sample Space:
$n(S)=5\cdot30\cdot12\cdot100\cdot 10^{4}$
Note: There are $30$  ways to select DD considering $30$ days in a month. $12$ ways to select MM and $100$ ways to select YY remaining $4$ places can be filled in $10^{4}$ ways.
So, $P(E)=\dfrac{n(E)}{n(s)}=\dfrac{9}{30}=0.3$ but given answer is $(.72)$.
Where am I going Wrong?
Similar Question What are the odds are getting your 6-digit birth date in your 10-digit phone number?
 A: The supposed answer of $0.72$ probability seems far too high.
Assuming that there are no restrictions on what constitutes a $10$ digit mobile number, there are $5$ possible starting spots for $DDMMYY$,
but only one in $10^6$  chance of getting all the $6$ digits right.
(the other four digits can be anything, they don't matter),
Thus $Pr = \Large\frac 5{10^6}$
PS:
No restrictions are mentioned in the question, which means that numbers can range from $0000000000$ to $9999999999$
This can't be right, but then whatever  restrictions exist need to be clearly spelt out in the question (or be a matter of common knowledge in your country)
Even with some restrictions, a probability of $0.72$ seems  unlikely.

New interpretation after OP's comment
The link you had given computed the probability of having your entire DDMMYY embedded in your phone number, but apparently the question is to find the Pr that your birthdate (DD) is in the phone in the format DDMMYY !
It has further assumed that a phone number can't start with $0$
Then there are only $4$ place where $0D$ can start, and only $1-9$ possibilities for $D$,(as birth was in the first $9$ days of a month), then
$n(E)=4\cdot  9 \cdot 10^{3},\;\;n(S)=5\cdot 10 \cdot 10^3$, so $P(E) = 0.72$
