A number is selected at random from the integers 1 to 30 inclusive. What is the probability that the selected number is divisible by 3 or by 7 (or by both)?

Introduction / Context:
This question uses basic number theory and probability. You must choose a random integer between 1 and 30 and find the probability that it is divisible by 3 or by 7. Because some numbers may be divisible by both 3 and 7, you should apply the inclusion exclusion principle for counting to avoid double counting overlap.

Given Data / Assumptions:

• The sample space is the set of integers from 1 to 30 inclusive.
• Total number of possible outcomes = 30.
• We are interested in numbers divisible by 3 or 7.
• A number divisible by both 3 and 7 is divisible by 21.

Concept / Approach:
Let A be the event that the chosen number is divisible by 3, and let B be the event that it is divisible by 7. We use: P(A or B) = P(A) + P(B) - P(A and B). We first count the numbers divisible by 3, then those divisible by 7, then subtract the numbers divisible by both 3 and 7 (that is, divisible by 21) to avoid double counting.

Step-by-Step Solution:
Step 1: Count numbers between 1 and 30 that are divisible by 3. They are 3, 6, 9, ..., 30. The largest multiple of 3 less than or equal to 30 is 30 = 3 * 10. So, there are 10 multiples of 3 in this range. Step 2: Count numbers between 1 and 30 that are divisible by 7. They are 7, 14, 21, 28. The largest multiple of 7 less than or equal to 30 is 28 = 7 * 4. So, there are 4 multiples of 7 in this range. Step 3: Count numbers divisible by both 3 and 7, that is, by 21. Between 1 and 30, the multiples of 21 are just 21 itself, so there is 1 such number. Step 4: Use the inclusion exclusion principle to count numbers divisible by 3 or 7. Count = 10 + 4 - 1 = 13. Step 5: Total numbers in the sample space = 30. Step 6: Probability that a randomly selected number is divisible by 3 or 7 = 13 / 30.

Verification / Alternative check:
You can verify by listing these numbers explicitly: 3, 6, 7, 9, 12, 14, 15, 18, 21, 24, 27, 28, 30. Counting them one by one gives exactly 13 numbers. Because the sample space has 30 equally likely outcomes, the probability 13/30 is consistent with this direct enumeration.

Why Other Options Are Wrong:
Options 11/30 and 7/30 are too small and result from undercounting the qualifying numbers or failing to use inclusion exclusion correctly. Option 2/15 equals 4/30, which is far lower than the correct value and suggests that only multiples of 7 were counted. Option 1/3 equals 10/30, which is the probability of divisibility by 3 alone and ignores the additional multiples of 7 that are not multiples of 3.

Common Pitfalls:
The most common error is to simply add the probabilities of divisibility by 3 and by 7 without subtracting the overlap, leading to 10/30 + 4/30 = 14/30. This overcounts the multiples of 21. Another mistake is to forget that the range is inclusive of both 1 and 30. Always check end points and remember to subtract the intersection when using the rule for P(A or B) in situations where the sets can overlap.

Final Answer:
The probability that the selected number is divisible by 3 or by 7 is 13/30.