Let $a = i^{k_{1}} + i^{k_{2}} + i^{k_{3}} + i^{k_{4}}, (i = \sqrt{- 1})$ where each $k_{n}$ is randomly chosen from the set ${1, 2, 3, 4}$ . The probability that $a = 0$ , is

\frac{7}{64}

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\frac{9}{64}

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\frac{37}{256}

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\frac{39}{256}

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Solution

The correct option is B $\frac{9}{64}$
The power of $i$ contain two sets of numbers that are additive inverse of each other, namely $(1, - 1)$ and $(i, - i)$ . Thus the only sets of four numbers that will satisfy $a = 0$ are permutations of either $(1, 1, - 1, - 1)$ and $(i, - i, 1, - 1)$ . The first two have ${^{4} C}_{2} = 6$ distinct arrangements each, while the last has $4! = 24$ total arrangements, giving $2 (6) + 24) = 36$ overall. There are $4^{4} = 256$ possibilities, giving a probability of $\frac{36}{256} = \frac{9}{64}$

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