Two numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?

1. Analyze the question

We are given a set of four numbers: $\{1, 2, 3, 4\}$. We need to select two numbers at random and find the probability that their sum is an even number.

2. Explain the concept

• Total Outcomes: We need to find the number of ways to choose 2 numbers out of 4. This is a combination problem: $^nC_r = \frac{n!}{r!(n-r)!}$.
• Condition for an Even Sum: The sum of two numbers is even if: • Both numbers are even (Even + Even = Even) • Both numbers are odd (Odd + Odd = Even)
• Probability Formula: $P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

3. Show the solution

Step 1: Calculate the total number of ways to choose 2 numbers from 4.

The possible pairs are: (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)

Step 2: Identify the pairs that result in an even sum.

• (1, 2) $\rightarrow$ Sum = 3 (Odd)
• (1, 3) $\rightarrow$ Sum = 4 (Even)
• (1, 4) $\rightarrow$ Sum = 5 (Odd)
• (2, 3) $\rightarrow$ Sum = 5 (Odd)
• (2, 4) $\rightarrow$ Sum = 6 (Even)
• (3, 4) $\rightarrow$ Sum = 7 (Odd)

There are 2 favorable outcomes: (1, 3) and (2, 4).

Step 3: Calculate the probability.

Simplifying the fraction:

4. Identify the answer

Comparing our result with the provided options, the correct choice is C.

5. Exam Tip

When dealing with "sum is even/odd" problems, remember the parity rules:

• Odd + Odd = Even
• Even + Even = Even
• Odd + Even = Odd Instead of listing all pairs, you can quickly count how many odd and even numbers are in the set and use combinations to find the favorable pairs. This saves time in the JAMB exam hall!