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 \(\frac{2}{15}\)
  • C \(\frac{1}{2}\)
  • D \(\frac{1}{4}\)

The correct answer is C. \(\frac{1}{2}\)

There are a total of 6 ways to remove two numbers at random from the numbers 1, 2, 3, and 4.

These are (1,2), (1,3), (1,4), (2,3), (2,4), and (3,4). 

Out of these 6 possibilities, 3 of them result in an even sum: (1,3), (2,4), and (3,4). 

Therefore, the probability that the sum of the numbers removed is even is 3/6 = 1/2. 

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