Question

In a certain show, a lottery ticket is numbered consecutively from 100 through 999 (both inclusive). What is the probability that a randomly selected ticket will have a number with a ten’s digit as “3”?

• A. \( \frac{1}{5} \)
• B. \( \frac{90}{899} \)
• C. \( \frac{1}{10} \)
• D. \( \frac{1}{11} \)
• E. \( \frac{10}{111} \)

Hint:

When calculating probability, always consider both the total possible outcomes and the favorable outcomes.

Answer 1

The Correct Option is C

Step 1: Total number of tickets. 
The tickets are numbered from 100 to 999, so there are a total of: \[ 999 - 100 + 1 = 900 \] tickets in total. 
Step 2: Favorable tickets. 
We want to find the tickets where the ten’s digit is “3”. These tickets are of the form \( \text{X}3\text{Y} \), where X and Y are any digits from 0 to 9. Thus, for each hundred’s place digit (from 1 to 9), there are 10 possible tickets with a ten’s digit of 3 (e.g., 130, 131, ..., 139 for the hundred’s place 1). So, there are: \[ 9 \times 10 = 90 \] favorable tickets. 
Step 3: Calculate probability. 
The probability is the ratio of favorable tickets to total tickets: \[ \frac{90}{900} = \frac{1}{10} \] Step 4: Conclusion. 
The correct answer is (C).