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?
A
\(\frac{2}{3}\)
B
\(\frac{1}{3}\)
C
\(\frac{1}{2}\)
D
\(\frac{1}{4}\)
Verified Correct Answer: B.
Step-by-Step Mathematical Solution
Total pairs: C(4,2) = 6.
Favorable (even sum): both odd (1 way: {1,3}) or both even (1 way: {2,4}) → 2 ways.
P = \(\frac{2}{6}\) = \(\frac{1}{3}\).
Source: WAEC / JAMB / NECO Classroom Archive (Record #31489)