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JEE Advanced 2021 Paper 1

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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 .

Section: Mathematics

Question : 44 of 57

Marks: +1, -0

The value of

\;\frac{125}{4} p_{2}

Solution:

(24.50) Since

p_{2}=

probability that minimum of chosen numbers is at most 40

=1

- probability that minimum of chosen numbers is greater than or equal to 41

=1-\;\frac{60 \times 60 \times 60}{100 \times 100 \times 100}=1-\;\frac{27}{125}=\;\frac{98}{125}

\therefore\;\frac{125}{4} p_{2} = \frac{125}{4} \times \frac{98}{125} = 24.50

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