1. Tardigrade
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  4. Three numbers are chosen at random, one after another with replacement, from the set S = 1,2,3, ldots . ., 100 . Let p 1 be the probability that the maximum of chosen numbers is at least 81 and p 2 be the probability that the minimum of chosen numbers is at most 40 . The value of (125/4) p2 is .

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Q. Three numbers are chosen at random, one after another with replacement, from the set $S =\{1,2,3, \ldots . ., 100\}$. Let $p _{1}$ be the probability that the maximum of chosen numbers is at least $81$ and $p _{2}$ be the probability that the minimum of chosen numbers is at most $40$ .
The value of $\frac{125}{4} p_{2}$ is ______.

JEE AdvancedJEE Advanced 2021

Solution:

$P_{2}=1$-(Probability that 3 chosen numbers are greater than 40 )
$=1-\left(\frac{60}{100}\right)^{3}=1-\left(\frac{3}{5}\right)^{3}=\frac{98}{125}$
So, $\frac{125}{4}$
$P_{2}=\frac{125}{4} \times \frac{98}{125}=24.5$