Out of the first 20 natural numbers {1, 2, 3, ..., 20}, one number is selected at random. What is the probability that it is either an even number or a prime number?

Introduction / Context:
This question combines basic number properties with probability. We choose one number at random from the first 20 natural numbers and want the probability that it is even or prime. Because some numbers are both even and prime, we must handle the overlap correctly using set operations.

Given Data / Assumptions:

• Sample space: numbers from 1 to 20 inclusive.
• Total numbers available = 20.
• Even numbers are multiples of 2.
• Prime numbers are positive integers greater than 1 with no positive divisors other than 1 and themselves.
• Each number from 1 to 20 is equally likely to be selected.

Concept / Approach:
We use the formula for the size of a union of two sets: |E ∪ P| = |E| + |P| - |E ∩ P|, where E is the set of even numbers and P is the set of primes. The required probability is |E ∪ P| divided by 20. The intersection must be considered because 2 is both even and prime.

Step-by-Step Solution:
Even numbers from 1 to 20: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 (10 numbers).Prime numbers from 1 to 20: 2, 3, 5, 7, 11, 13, 17, 19 (8 numbers).Intersection of even and prime sets: only the number 2.So |E| = 10, |P| = 8 and |E ∩ P| = 1.Size of E ∪ P = |E| + |P| - |E ∩ P| = 10 + 8 - 1 = 17.Total possible outcomes = 20.Hence probability that the selected number is even or prime = 17 / 20.

Verification / Alternative check:
We can list the union explicitly: {2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20}. There are 17 numbers in this set. Since there are 20 numbers in total from 1 to 20, the probability is 17/20, which matches our earlier calculation using the inclusion exclusion principle.

Why Other Options Are Wrong:
The value 1 would mean that every number from 1 to 20 is either even or prime, which is false because 9 and 15, for example, are neither even nor prime. The fraction 16/19 and the value 3/2 are impossible probabilities, and 3/2 is greater than 1. Only 17/20 is consistent with the correct counts of the union and sample space size.

Common Pitfalls:
Students sometimes double count 2 when they add the counts of even and prime numbers, forgetting to subtract the intersection. Others mistakenly include 1 as a prime number, which would make the prime set larger and the answer incorrect. Applying the inclusion exclusion formula carefully helps avoid these issues.

Final Answer:
The probability that the selected number is either even or prime is 17/20.