The value of $$\frac{125}{4} p_2$$ is .........

To find $$p_2$$, we interpret the event in words:
“Minimum of the three chosen numbers is at most 40.”
It is usually simpler to work with the complement of this event.

Step 1 - Define the complement
The complement is: “All three chosen numbers are > 40”, i.e. each number lies in $$\{41,42,\ldots ,100\}$$.

Step 2 - Probability for one draw
Count of favourable numbers for the complement in one draw = $$100-40=60$$.
Probability that a single draw gives a number > 40 is therefore$$\frac{60}{100}=\frac{3}{5}$$.

Step 3 - Independent draws with replacement
Because the three draws are independent (replacement), the probability that all three numbers exceed 40 equals$$\left(\frac{3}{5}\right)^{3}=\frac{27}{125}\;.\tag{-1}$$

Step 4 - Compute $$p_2$$ using the complement
$$p_2 = 1-\text{(probability of complement)} = 1-\frac{27}{125}=\frac{98}{125}\;.\tag{-2}$$

Step 5 - Evaluate $$\dfrac{125}{4}\,p_2$$
$$\frac{125}{4}\,p_2 = \frac{125}{4}\times\frac{98}{125}= \frac{98}{4}=24.5\;.\tag{-3}$$

Hence, the required value is 24.5, which lies in the given range 24.40 - 24.60.

Final answer: 24.5