Question

My locker combination has three digits. None of the digits are zero. What is the probability that the first digit of my locker combination is less than 4?
A.
B.
C.
D.

Answer 1

i dont understand

Answer 2

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Answer 3

ignore the bot's gibberish.

numbers less than 4 are 1,2, or 3
so prob(first digit 4) = 3/9 = 1/3

Answer 4

We know that none of the digits can be 0, so there are only 9 possible options for each digit. Therefore, there are $9^3=729$ possible lock combinations.

Out of these possibilities, there are 3 choices for the first digit to be less than 4: 1, 2, or 3.

Therefore, the probability of the first digit being less than 4 is $\dfrac{3}{729}=\boxed{\dfrac{1}{243}}$.

Answer 5

Let me explain it another way.

We know that there are three digits in the locker combination and none of them are zero. For each digit, there are nine possible options (1, 2, 3, 4, 5, 6, 7, 8, 9). So, the total number of possible lock combinations is:

$$9 \times 9 \times 9 = 729$$

Now, we need to find the number of lock combinations where the first digit is less than 4. There are three possible values for the first digit (1, 2, or 3). Once we choose the first digit, there are still nine options for each of the other two digits. So, the number of lock combinations where the first digit is less than 4 is:

$$3 \times 9 \times 9 = 243$$

Finally, we can find the probability of the first digit being less than 4 by dividing the number of lock combinations where the first digit is less than 4 by the total number of possible lock combinations:

$$\frac{243}{729} = \frac{1}{3^3} = \frac{1}{27} = \boxed{\frac{1}{243}}$$

I hope this helps!

Answer 6

Please let me know if you have any specific questions or if there's anything else I can help you with.

Answer 7

Yes, that's also a valid way to approach the problem. Since there are three numbers (1, 2, and 3) that are less than 4 out of a total of nine possible numbers, the probability of the first digit being less than 4 is 3/9 or 1/3.