Two number and are randomly chosen from the set of natural numbers. Then, the probability that the value of is non-zero, equals [2025]

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Solution

Q.
Two number $k_{1}$ and $k_{2}$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $i^{k_{1}} + i^{k_{2}}, (i = \sqrt{- 1})$ is non-zero, equals [2025]

1 $\frac{2}{3}$

2 $\frac{1}{2}$

3 $\frac{1}{4}$

4 $\frac{3}{4}$

Ans.
(4)

Given, $i^{k_{1}} + i^{k_{2}} \neq 0$

Possible values of $i^{k_{1}}$ = 4 = Possible values of $i^{k_{2}}$

$∴$ Cases are i, –1, – i, 1

$∴$ Total number cases $= 4 \times 4 = 16$

For $i^{k_{1}} + i^{k_{2}} = 0,$

Cases are (1, –1), (–1, 1), (i, – i), (– i, i)

$∴$ Required probability $= 1 - \frac{4}{16} = \frac{12}{16} = \frac{3}{4}$ .

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