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?
- \(\frac{2}{3}\)
- \(\frac{2}{15}\)
-
\(\frac{1}{2}\)
- \(\frac{1}{4}\)
The correct answer is: C
Explanation
\(\begin{array}{c|c} 1 & 2 & 3 & 4\\\hline 1(1, 1) & (1, 2) & (1, 3) & (1, 4)\\ \hline 2(2, 1) & (2 , 2) & (2, 3) & (2, 4) \\ \hline 3(3, 1) & (3, 2) & (3, 3) & (3, 4)\\ \hline 4(4, 1) & (4, 2) & (4, 3) & (4, 4)\end{array}\)
sample space = 16
sum of nos. removed are (2), 3, (4), 5
3, (4), 5, (6)
(4), 5, (6), 7
(5), 6, 7, (8)
Even nos. = 8 of them
Pr(even sum) = \(\frac{8}{16}\)
= \(\frac{1}{2}\)