My locker combination has three digits. None of the digits are 0. What is the probability that the first digit of my locker combination is less than 3?
A. 1/3
B. 1/2
C. 2/9
D. 3/10
There are 9 digits to choose from for the first digit (since 0 is not allowed). Out of these 9 digits, only 3 are less than 3 (namely 1 and 2). Therefore, the probability that the first digit is less than 3 is 3/9, which simplifies to 1/3. So the answer is A.
1 and 2 are two numbers, not 3.
2/9 which is C
You are correct, and I apologize for my mistake. You are absolutely right that there are 2 digits less than 3 (namely 1 and 2), out of a total of 9 digits (excluding 0). Therefore, the probability that the first digit is less than 3 is 2/9, and the answer is indeed C. Thank you for catching my error!
To determine the probability that the first digit of your locker combination is less than 3, we need to first count the number of possible combinations and then count the number of combinations where the first digit is less than 3.
Since none of the digits can be 0, there are 9 choices for each of the three digits (1-9). Therefore, the total number of possible combinations is 9 * 9 * 9 = 729.
To count how many combinations have the first digit less than 3, we need to consider two cases: when the first digit is 1 and when it is 2.
Case 1: When the first digit is 1, there are 9 choices for the second digit (1-9), and 9 choices for the third digit (also 1-9). So, there are 9 * 9 = 81 combinations.
Case 2: When the first digit is 2, there are also 9 choices for the second digit and third digit. So, there are 9 * 9 = 81 combinations.
Combining both cases, there are 81 + 81 = 162 combinations where the first digit is less than 3.
Now, we can calculate the probability by dividing the number of desired outcomes (162) by the total number of possible outcomes (729): 162 / 729 = 2 / 9.
Therefore, the correct answer is C. 2/9.