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 $\dfrac{125}{4}{p}_{2}$ is

Question

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

\dfrac{125}{4}{p}_{2}

is

Quantrex Academy · Accepted Answer

Correct answer: 0

Minimum of the chosen numbers is at most $40$

It means we have to choose all the numbers from $1$ to $40$ only
from $41$ 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 $41$ to $100$ $= 60 \times 60 \times 60 = {60}^{3}$

Total number of favourable cases $= {100}^{3} - {60}^{3}$

So, ${p}_{2} = \dfrac{{100}^{3} - {60}^{3}}{{100}^{3}}$

$= \dfrac{{20}^{3}\left({5}^{3} - {3}^{3}\right)}{{20}^{3} \times {5}^{3}} = \dfrac{98}{125}$

Hence, $\dfrac{125}{4}{p}_{2} = \dfrac{125}{4} \times \dfrac{98}{125} = 24 . 50$

Three numbers are chosen at random, one after another with...