POTW Secondary School is putting on a play. Tickets for the play are numbered, and each ticket consists of exactly four digits chosen from the digits \(0\) to \(9\) (digits may be repeated).
Every possible ticket is printed exactly once. If tickets are handed out in a random order, what is the probability that first ticket handed out has digits whose sum is \(34\) or higher?
First, we determine the total number of possible four-digit ticket numbers. There are \(10\) choices for the first digit. For each of these possibilities, there are \(10\) choices for the second digit. Therefore, there are \(10\times 10=100\) possibilities for the first two digits. For each of these possibilities, there are \(10\) choices for the third digit. Therefore, there are \(100\times 10=1000\) possibilities for the first three digits. And finally, for each of these \(1000\) possibilities for the first three digits, there are \(10\) choices for the fourth digit. Therefore, there are \(1000\times 10= 10\,000\) possible four-digit ticket numbers.
Now, we determine the number of ticket numbers with a digit sum of \(34\) or higher. We consider cases.
Case 1: The ticket number contains four \(9\)s.
If the digits are all \(9\)s, the sum
is \(36\), which is \(34\) or higher. There is one ticket with
all \(9\)s for digits.
Case 2: The ticket number contains three \(9\)s and one other digit.
The three \(9\)s sum to \(27\). In order to get a sum of \(34\) or higher, the fourth digit must be
\(7\) or \(8\). Thus, there are two choices for the
fourth digit. Once the digit is chosen, there are four places to put the
digit. Once the digit is placed, the digits in the remaining spots must
be \(9\)s. Therefore, there are \(2\times 4 =8\) ticket numbers containing
three \(9\)s. (It is possible to list
them: \(7999\), \(8999\), \(9799\), \(9899\), \(9979\), \(9989\), \(9997\), \(9998\).)
Case 3: The ticket number contains two \(9\)s.
The two \(9\)s sum to \(18\). In order to get a sum of \(34\) or higher, we need a sum of \(34-18=16\) or higher from the two remaining
digits. The only way to do this, since we cannot use more \(9\)s, is to use two \(8\)s. The two \(8\)s can be placed in six ways and then the
\(9\)s must go in the remaining spots.
Therefore, there are six ticket numbers containing two \(9\)s. (It is possible to list them: \(8899\), \(8989\), \(8998\), \(9889\), \(9898\), \(9988\).)
Case 4: The ticket number contains one \(9\).
In order to get a sum of \(34\) or
higher, we need a sum of \(34-9=25\) or
higher from the remaining three digits. But this sum would be made from
three digits chosen from the digits \(0\) to \(8\). The maximum possible sum would be
\(24\) if three \(8\)s were used. We need the sum to be \(25\) or higher. Therefore, no ticket number
containing only one \(9\) will produce
a sum of \(34\) or higher. It should be
noted that a ticket number with zero nines would not produce a sum of
\(34\) or higher either.
Therefore, the number of ticket numbers with a digit sum of \(34\) or higher is \(1+8+6=15\). To calculate the probability we divide the number of tickets with a digit sum of \(34\) or higher by the number of possible ticket numbers. Therefore, the probability that the first ticket handed out has digits whose sum is \(34\) or higher is \(\frac{15}{10\,000}=\frac{3}{2000}\).
Another way of looking at this result is that out of every \(2000\) tickets you could expect, on average, \(3\) tickets with a digit sum of \(34\) or higher.