We are tasked with finding the probability that a number chosen from the digits 1 through 30 is either an even number or a number divisible by 3.

Step 1: Total number of outcomes

The total number of digits from 1 to 30 is 30, so the total number of possible outcomes is: $\text{Total outcomes} = 30$

Step 2: Counting favorable outcomes

We now need to count how many numbers are either:

1. Even numbers.
2. Divisible by 3.

Even numbers:

The even numbers between 1 and 30 are: $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30$ There are 15 even numbers.

Numbers divisible by 3:

The numbers divisible by 3 between 1 and 30 are: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$ There are 10 numbers divisible by 3.

Step 3: Applying the inclusion-exclusion principle

Some numbers are counted twice because they are both even and divisible by 3. These numbers are: $6, 12, 18, 24, 30$ There are 5 such numbers.

Using the inclusion-exclusion principle, the number of favorable outcomes is: $\text{Favorable outcomes} = (\text{Even numbers}) + (\text{Divisible by 3}) - (\text{Both even and divisible by 3})$ $\text{Favorable outcomes} = 15 + 10 - 5 = 20$

Step 4: Calculating the probability

The probability is the ratio of favorable outcomes to the total number of outcomes: $P = \frac{20}{30} = \frac{2}{3}$

Final Answer:

The probability that the chosen number is either an even number or divisible by 3 is $\frac{2}{3}$.

Would you like further clarification or details?

Here are some related questions to expand your understanding:

1. What is the probability of selecting a number divisible by both 2 and 3?
2. How many numbers between 1 and 30 are divisible by 5?
3. What is the probability of selecting an odd number from 1 to 30?
4. How many prime numbers exist between 1 and 30?
5. What is the probability that a number chosen from 1 to 30 is divisible by 4?

Tip: When dealing with probabilities of unions (like "or" conditions), using the inclusion-exclusion principle helps avoid overcounting.