A number is chosen at random from the integers 1 to 30. What is the probability that the chosen number is divisible by 3 or by 7?

Introduction:
This question is about basic probability using divisibility and counting. You must find how many numbers between 1 and 30 are divisible by 3 or by 7, and then divide by the total number of possibilities. It also illustrates the principle of inclusion and exclusion when dealing with "or" conditions in counting problems.

Given Data / Assumptions:

• We consider integers from 1 to 30 inclusive.
• Each integer is equally likely to be chosen.
• We want the probability that the chosen number is divisible by 3 or 7 (or both).
• Total number of possible outcomes = 30.

Concept / Approach:
Let A be the event "number divisible by 3" and B be the event "number divisible by 7". We want P(A ∪ B). Use inclusion–exclusion:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).In counting form, we first count multiples of 3, then multiples of 7, then subtract multiples common to both (i.e., multiples of lcm(3, 7) = 21), and finally divide by 30.

Step-by-Step Solution:
Step 1: Count numbers divisible by 3.Multiples of 3 up to 30: 3, 6, 9, ..., 30.Number of such multiples = 30 ÷ 3 = 10.Step 2: Count numbers divisible by 7.Multiples of 7 up to 30: 7, 14, 21, 28.Number of such multiples = 4.Step 3: Count numbers divisible by both 3 and 7.Numbers divisible by both 3 and 7 are multiples of lcm(3, 7) = 21.Multiples of 21 up to 30: only 21.So there is 1 such number.Step 4: Apply inclusion–exclusion to count numbers divisible by 3 or 7.Count(A ∪ B) = Count(A) + Count(B) − Count(A ∩ B).= 10 + 4 − 1 = 13.Step 5: Convert to probability.Total possibilities = 30 numbers.P(divisible by 3 or 7) = 13 / 30.

Verification / Alternative check:
We can explicitly list the 13 numbers: divisible by 3 → 3, 6, 9, 12, 15, 18, 21, 24, 27, 30; divisible by 7 → 7, 14, 21, 28. The union is {3, 6, 7, 9, 12, 14, 15, 18, 21, 24, 27, 28, 30}, which clearly has 13 elements, matching our count.

Why Other Options Are Wrong:
7/15 = 14/30, 2/5 = 12/30, 11/30 and 3/10 = 9/30 all correspond to incorrect counts of numbers satisfying the condition. Only 13/30 captures the correct 13 favourable outcomes out of 30 possible choices.

Common Pitfalls:
Common errors include double-counting multiples of both 3 and 7 or forgetting to subtract the overlap for 21. Some students only consider one of the sets or add the probabilities directly without adjusting for intersection. Always remember to use inclusion–exclusion when working with "or" events that can overlap.

Final Answer:
The probability that a randomly chosen number from 1 to 30 is divisible by 3 or by 7 is 13/30.