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The three number are chosen at random, one after another with replacement, from the set $S = {1, 2, 3, ....10} .$ Let $P_{1}$ be the probability that the maximum of chosen number is at least 81. Then $\frac{625}{4} P_{1} =__$

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Detailed Solution

$P_{1} = 1 -$ Probability that all the drawn numbers are less than or equal to 80.
$= 1 - \frac{80}{100} . \frac{80}{100} . \frac{80}{100} = 1 - \frac{64}{125} = \frac{61}{125}$

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