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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 : 7 of 31

Marks: +1, -0

The value of

\;\frac{625}{4} p_{1}

Solution:

Since

p_{1}=

probability that maximum of chosen numbers is at least 81

p_{1}=1

- probability that maximum of chosen number is less than or equal to 80

p_{1}=1-\;\frac{80 \times 80 \times 80}{100 \times 100 \times 100}=1-\;\frac{64}{125}

p_{1}=\;\frac{61}{125}

\therefore \;\; \;\frac{625 p_{1}}{4} = \;\frac{625}{4} \times \;\frac{61}{125} = \;\frac{305}{4} = 76.25

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