Let's analyze the problem step-by-step:
Problem Restatement
You have tickets numbered from 1 to 1100. One ticket is drawn at random. You want to find the probability that the drawn ticket's number contains the digit 2 at least once.
Step 1: Total number of tickets
Total tickets = 1100 (from 1 to 1100 inclusive).
Step 2: Define the event
Event A = "the ticket number contains digit 2". We want P(A)=number of tickets containing digit 21100P(A)=\frac{\text{number of tickets containing digit 2}}{1100}P(A)=1100number of tickets containing digit 2.
Step 3: Use complementary counting
It's easier to count tickets not containing the digit 2, then subtract from total. Let AcA^cAc = "ticket number does NOT contain digit 2". Then:
P(A)=1−P(Ac)=1−number of tickets without digit 21100P(A)=1-P(A^c)=1-\frac{\text{number of tickets without digit 2}}{1100}P(A)=1−P(Ac)=1−1100number of tickets without digit 2
Step 4: Count numbers without digit 2 from 1 to 1100
Break into two parts:
- From 1 to 999
- From 1000 to 1100
Counting numbers without digit 2 from 1 to 999
Think of these as 3-digit numbers with leading zeros allowed:
- Range: 000 to 999 (1000 numbers)
- We want to count numbers with digits from {0,1,3,4,5,6,7,8,9} only (digits excluding 2)
- Number of allowed digits = 9 (0,1,3,4,5,6,7,8,9)
Number of 3-digit numbers (including leading zeros) without digit 2:
9×9×9=93=7299\times 9\times 9=9^3=7299×9×9=93=729
This counts numbers from 000 to 999 without digit 2. Since 000 corresponds to 0 (which is not in our ticket range), the count of numbers from 1 to 999 without digit 2 is:
729−1=728729-1=728729−1=728
Counting numbers without digit 2 from 1000 to 1100
Numbers: 1000, 1001, ..., 1100 (101 numbers) Check each number for digit 2:
- 1000: digits 1,0,0,0 → no 2
- 1001: digits 1,0,0,1 → no 2
- ...
- 1099: digits 1,0,9,9 → no 2
- 1100: digits 1,1,0,0 → no 2
None of these numbers have digit 2 because the hundreds digit is 0 or 1, tens and units digits are less than 10, and none are 2. So all 101 numbers from 1000 to 1100 do not contain digit 2.
Step 5: Total numbers without digit 2
728+101=829728+101=829728+101=829
Step 6: Calculate probability
P(A)=1−8291100=1100−8291100=2711100≈0.2464P(A)=1-\frac{829}{1100}=\frac{1100-829}{1100}=\frac{271}{1100}\approx 0.2464P(A)=1−1100829=11001100−829=1100271≈0.2464
Final answer:
The probability that a randomly drawn ticket from 1 to 1100 contains the digit 2 is approximately 0.2464, or 24.64%.
