Three numbers are chosen at random, one after another with replacement, from the set $S = {1, 2, 3, \dots, 100}$ . 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 $\frac{625}{4} p_{1}$ is

Question

TestHub · Accepted Answer

Maximum of the chosen numbers is at least $81$

It means we have to choose at least one number from $81$ to $100$

Total number of possible selections $= 100 \times 100 \times 100 = 100^{3}$

Favourable cases $=$ Total $-$ unfavourable cases

Unfavourable cases are those in which we have selected all the three numbers form $1$ to $80$ $= 80 \times 80 \times 80 = 80^{3}$

Total number of favourable cases $= 100^{3} - 80^{3}$

So, $p_{1} = \frac{100^{3} - 80^{3}}{100^{3}}$

$= \frac{20^{3} (5^{3} - 4^{3})}{20^{3} \times 5^{3}} = \frac{125 - 64}{125} = \frac{61}{125}$

Hence, $\frac{625}{4} p_{1} = \frac{625}{4} \times \frac{61}{125} = 76.25$

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