Two numbers k_1 and k_2 are randomly chosen f | vedclass

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(C) The value of $i^k$ for any natural number $k$ can only be one of the four values: $\{i, -1, -i, 1\}$.Since $k_1$ and $k_2$ are chosen randomly,the

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$\frac{3}{4}$

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(C) The value of $i^k$ for any natural number $k$ can only be one of the four values: $\{i, -1, -i, 1\}$.Since $k_1$ and $k_2$ are chosen randomly,there are $4 	imes 4 = 16$ possible pairs of values for $(i^{k_1}, i^{k_2})$.We want to find the probability that $i^{k_1} + i^{k_2} 
eq 0$,which is equivalent to $1 - P(i^{k_1} + i^{k_2} = 0)$.The condition $i^{k_1} + i^{k_2} = 0$ implies $i^{k_1} = -i^{k_2}$.The possible pairs $(i^{k_1}, i^{k_2})$ that satisfy this are: $(i, -i), (-i, i), (1, -1), (-1, 1)$.There are $4$ such unfavorable cases.Thus,the number of favorable cases is $16 - 4 = 12$.The probability is $\frac{12}{16} = \frac{3}{4}$.

Two numbers $k_1$ and $k_2$ are randomly chosen from the set of natural numbers. Then,the probability that the value of $i^{k_1} + i^{k_2}$ (where $i = \sqrt{-1}$) is non-zero,equals:

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