What is the probability of choosing a square between $2$ and $100$ (both numbers inclusive)?

\frac{1}{11}

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\frac{1}{10}

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\frac{8}{11}

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\frac{9}{100}

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Solution

The correct option is A $\frac{1}{11}$
Total Numbers: $99$
Total squared numbers: $9$ $\to {4, 9, 16, 25, 36, 49, 64, 81, 100}$
Required probability:
$P = \frac{9}{99} = \frac{1}{11}$

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Similar questions

(10)^{9} + 2 (11)^{1} (10)^{8} + 3 (11)^{2} (10)^{7} + \dots \dots + 10 (11)^{9} = k (10)^{9}

(10)^{9} + 2 (11)^{1} (10)^{8} + 3 (11)^{2} (10)^{7} + . . . + 10 (11)^{9} = k (10)^{9},

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