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 $\frac{125}{4} p_{2}$ is

Question

TestHub · Accepted Answer

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, $p_{2} = \frac{100^{3} - 60^{3}}{100^{3}}$

$= \frac{20^{3} (5^{3} - 3^{3})}{20^{3} \times 5^{3}} = \frac{98}{125}$

Hence, $\frac{125}{4} p_{2} = \frac{125}{4} \times \frac{98}{125} = 24.50$

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