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 $\dfrac{625}{4}{p}_{1}$ 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{625}{4}{p}_{1}

is

Quantrex Academy · Accepted Answer

Correct answer: 0

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, ${p}_{1} = \dfrac{{100}^{3} - {80}^{3}}{{100}^{3}}$

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

Hence, $\dfrac{625}{4}{p}_{1} = \dfrac{625}{4} \times \dfrac{61}{125} = 76 . 25$

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