Question

The value of (\(\frac{625}{4}\))p₁ is _____ .

Answer 1

Correct Answer: 76.25

Step 1: Understanding the Problem
We are selecting three numbers, one after another, from the set \( S = \{1, 2, 3, \dots, 100\} \), with replacement. We need to find the value of \( \frac{625}{4} p_1 \), where \( p_1 \) is the probability that the maximum of the chosen numbers is at least 81. This can be interpreted as finding the probability that, among the three chosen numbers, at least one of them is greater than or equal to 81.

Step 2: Total Possible Outcomes
Since we are choosing three numbers from the set \( S \) with replacement, the total number of possible outcomes is: \[ 100 \times 100 \times 100 = 100^3 \]

Step 3: Probability that the Maximum is Less Than 81
Now, we calculate the complementary event where the maximum of the three chosen numbers is less than 81. This means that each of the three numbers must be chosen from the subset \( \{1, 2, \dots, 80\} \). There are 80 choices for each number, so the total number of outcomes where the maximum is less than 81 is: \[ 80 \times 80 \times 80 = 80^3 \] Thus, the probability that the maximum of the three chosen numbers is less than 81 is: \[ \frac{80^3}{100^3} \]

Step 4: Probability that the Maximum is at Least 81
The probability that the maximum is at least 81 is the complement of the probability that the maximum is less than 81. Therefore: \[ p_1 = 1 - \frac{80^3}{100^3} \] Substitute the values: \[ 80^3 = 512000 \quad \text{and} \quad 100^3 = 1000000 \] \[ p_1 = 1 - \frac{512000}{1000000} = 1 - 0.512 = 0.488 \]

Step 5: Calculating \( \frac{625}{4} p_1 \)
We are asked to find \( \frac{625}{4} p_1 \). Substituting the value of \( p_1 = 0.488 \): \[ \frac{625}{4} p_1 = \frac{625}{4} \times 0.488 = 625 \times 0.122 = 76.25 \] Thus, the value of \( \frac{625}{4} p_1 \) is: 76.25

Answer 2

Step 1: Understanding the Problem
We are given a set \( S = \{ 1, 2, 3, \dots, 100 \} \) and we randomly select three numbers from this set with replacement. We need to find the value of \( \left( \frac{125}{4} \right) p_2 \), where \( p_1 \) is the probability that the maximum of the chosen numbers is at least 81 and \( p_2 \) is the probability that the minimum of the chosen numbers is at most 40.

Step 2: Probability \( p_1 \) - Maximum of Chosen Numbers is at Least 81
We are asked to find the probability that the maximum of the three chosen numbers is at least 81. The total number of possible outcomes is: \[ \text{Total outcomes} = 100 \times 100 \times 100 = 100^3 \] Now, consider the complement, which is the case when all three chosen numbers are less than 81. The numbers that are less than 81 are from the set \( \{ 1, 2, 3, \dots, 80 \} \). The number of outcomes where all three numbers are from this set is: \[ \text{Outcomes where all numbers are less than 81} = 80 \times 80 \times 80 = 80^3 \] Thus, the probability that the maximum of the three numbers is at least 81 is: \[ p_1 = 1 - \frac{80^3}{100^3} \] We can compute this expression for \( p_1 \), but it is not necessary to continue as we are mainly concerned with \( p_2 \).

Step 3: Probability \( p_2 \) - Minimum of Chosen Numbers is at Most 40
Next, we are asked to find the probability that the minimum of the three chosen numbers is at most 40. Again, consider the complement, which is the case when all three chosen numbers are greater than 40. The numbers greater than 40 are from the set \( \{ 41, 42, \dots, 100 \} \). The number of outcomes where all three numbers are from this set is: \[ \text{Outcomes where all numbers are greater than 40} = 60 \times 60 \times 60 = 60^3 \] Thus, the probability that the minimum of the three numbers is at most 40 is: \[ p_2 = 1 - \frac{60^3}{100^3} \] Now, we will compute this value: \[ p_2 = 1 - \frac{60^3}{100^3} = 1 - \frac{216000}{1000000} = 1 - 0.216 = 0.784 \]

Step 4: Calculate \( \left( \frac{125}{4} \right) p_2 \)
We are asked to find the value of \( \left( \frac{125}{4} \right) p_2 \). Substituting \( p_2 = 0.784 \) into this expression: \[ \left( \frac{125}{4} \right) p_2 = \frac{125}{4} \times 0.784 = \frac{125 \times 0.784}{4} = \frac{98}{4} = 24.5 \] Thus, the value of \( \left( \frac{125}{4} \right) p_2 \) is \( \boxed{24.5} \).