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Problem of the Week

Problem D and Solution

Tickets

Problem

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.
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?

Ticket numbered 0826

Solution

First we must determine the total number of possible four-digit ticket numbers. There are 10 choices for the first digit. For each of these choices there are 10 choices for the second digit. Therefore, there are \(10\times 10=100\) choices 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 choices 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 must determine the number of ticket numbers with a digit sum of 34 or higher. We will consider cases.

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 more by the number of possible ticket numbers. The probability of getting a ticket with a digit sum of 34 or higher is \(\frac{15}{10\,000}=\frac{3}{2000}\). Another way of looking at this result is out of every \(2000\) tickets you could expect, on average, to find 3 with a digit sum of 34 or higher.