The question in the image asks:

"You pick 2 digits (0-9) at random without replacement and write them in the order picked. What is the probability that you have written the first 2 digits of your phone number? Assume there are no repeats of digits in your phone number."

Let's break this down step by step:

Step 1: Total possible outcomes

You are selecting 2 digits from a total of 10 digits (0 through 9) without replacement.

• For the first digit, you have 10 possible choices.
• For the second digit, since you're not replacing the first one, you have 9 remaining choices.

Thus, the total number of possible outcomes is: $10 \times 9 = 90$

Step 2: Favorable outcomes

Assume the first two digits of your phone number are given. Let's call them $d_1$ and $d_2$. Since you're only considering one specific combination of digits in that particular order (matching the first two digits of your phone number), there is only 1 favorable outcome.

Step 3: Probability

The probability is the ratio of favorable outcomes to total outcomes: $\text{Probability} = \frac{1}{90}$

Thus, the probability that you have written the first 2 digits of your phone number is $\frac{1}{90}$.

Would you like any further clarification or additional details?

Here are 5 related questions for further exploration:

1. How would the probability change if you were to pick 3 digits instead of 2?
2. What if the digits were picked with replacement? How would the probability differ?
3. How would the probability be affected if the first digit couldn't be zero?
4. How does the concept of permutations apply to this problem?
5. Can you generalize this problem to a phone number with more than two digits?

Tip: Always pay attention to whether the problem involves replacement or not, as this significantly impacts the total number of possible outcomes.