Part a-1: Number of possible outcomes when digits can be repeated

We are given a three-digit PIN where:

• The first digit cannot be zero.
• The digits can be repeated.

Let's break it down:

• The first digit can be any number from 1 to 9 (since zero is excluded), so there are 9 possible choices for the first digit.
• The second digit can be any number from 0 to 9 (since repetition is allowed), so there are 10 possible choices for the second digit.
• The third digit can also be any number from 0 to 9, so there are 10 possible choices for the third digit.

Therefore, the total number of possible outcomes when digits can be repeated is:

$9 \times 10 \times 10 = 900$

Part a-2: Number of possible outcomes when digits cannot be repeated

Now, let's calculate the number of possible outcomes when digits cannot be repeated:

• The first digit can still be any number from 1 to 9, so there are 9 choices.
• The second digit can be any number from 0 to 9, except the number already chosen for the first digit. So, there are 9 choices for the second digit.
• The third digit can be any number from 0 to 9, except the two numbers already chosen for the first and second digits. So, there are 8 choices for the third digit.

Therefore, the total number of possible outcomes when digits cannot be repeated is:

$9 \times 9 \times 8 = 648$

Part b: Probability that Mr. Jones and Mrs. Smith select the same PIN

To find the probability that both Mr. Jones and Mrs. Smith select the same PIN:

• The total number of possible PINs is 900 (from part a-1).
• If Mr. Jones selects a PIN, Mrs. Smith must select exactly the same PIN for them to match. So, the number of successful outcomes is 1.

The probability is the ratio of successful outcomes to the total outcomes. Therefore, the probability is:

$P(\text{same PIN}) = \frac{1}{900} \approx 0.001111$

Rounded to six decimal places, the probability is 0.001111.

Concept of probability used:

The concept of probability used here is Classical Probability. This is based on the assumption that each possible outcome is equally likely, which is the case in this problem where each PIN has an equal chance of being selected.

Summary:

1. a-1: The number of possible outcomes when digits can be repeated: 900.
2. a-2: The number of possible outcomes when digits cannot be repeated: 648.
3. b: The probability that Mr. Jones and Mrs. Smith select the same PIN: 0.001111.
4. The concept of probability used: Classical Probability.

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Would you like any further explanation or have any questions?

Here are some related questions to expand on this:

1. What is the probability of selecting a PIN where all digits are even?
2. How would the probability change if the first digit could be zero?
3. How many possible PINs are there if the digits are allowed to be letters as well as numbers?
4. What is the expected number of attempts it would take to match a specific PIN, assuming random selection?
5. What is the probability that Mr. Jones and Mrs. Smith select different PINs?

Tip: When calculating probabilities, always ensure that the outcomes are equally likely, as that forms the basis for classical probability calculations.