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Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ..., 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40
Q. The value of 125/4p2 is _______.
Q. The value of 125/4p2 is _______.
Correct answer is '24.5'. Can you explain this answer?
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Most Upvoted Answer
Three numbers are chosen at random, one after another with replacement...
To solve this problem, we can break it down into two parts: finding the probability of the maximum of chosen numbers being at least 81 (p1), and finding the probability of the minimum of chosen numbers being at most 40 (p2).
Finding p1:
To find p1, we need to determine the number of ways in which the maximum of chosen numbers can be at least 81. Since the numbers are chosen with replacement, each number has an equal probability of being chosen at each step.
- The maximum number can be 81, 82, 83, ..., 100. There are 100 - 80 = 20 possible values for the maximum number.
- For each maximum number, there are 100 possible choices for each of the other two numbers. Therefore, there are 100^2 = 10,000 ways to choose the other two numbers.
- Since the total number of possible choices for three numbers is 100^3 = 1,000,000 (since there are 100 choices for each number), we have p1 = (20 * 10,000) / 1,000,000 = 200,000 / 1,000,000 = 1/5.
Finding p2:
To find p2, we need to determine the number of ways in which the minimum of chosen numbers can be at most 40. Again, since the numbers are chosen with replacement, each number has an equal probability of being chosen at each step.
- The minimum number can be 1, 2, 3, ..., 40. There are 40 possible values for the minimum number.
- For each minimum number, there are 100 possible choices for each of the other two numbers. Therefore, there are 100^2 = 10,000 ways to choose the other two numbers.
- Since the total number of possible choices for three numbers is 100^3 = 1,000,000, we have p2 = (40 * 10,000) / 1,000,000 = 400,000 / 1,000,000 = 2/5.
Calculating 125/4p2:
To calculate 125/4p2, we substitute the value of p2 into the expression: (125/4) * (2/5) = 250/20 = 12.5.
Therefore, the value of 125/4p2 is 12.5.
Finding p1:
To find p1, we need to determine the number of ways in which the maximum of chosen numbers can be at least 81. Since the numbers are chosen with replacement, each number has an equal probability of being chosen at each step.
- The maximum number can be 81, 82, 83, ..., 100. There are 100 - 80 = 20 possible values for the maximum number.
- For each maximum number, there are 100 possible choices for each of the other two numbers. Therefore, there are 100^2 = 10,000 ways to choose the other two numbers.
- Since the total number of possible choices for three numbers is 100^3 = 1,000,000 (since there are 100 choices for each number), we have p1 = (20 * 10,000) / 1,000,000 = 200,000 / 1,000,000 = 1/5.
Finding p2:
To find p2, we need to determine the number of ways in which the minimum of chosen numbers can be at most 40. Again, since the numbers are chosen with replacement, each number has an equal probability of being chosen at each step.
- The minimum number can be 1, 2, 3, ..., 40. There are 40 possible values for the minimum number.
- For each minimum number, there are 100 possible choices for each of the other two numbers. Therefore, there are 100^2 = 10,000 ways to choose the other two numbers.
- Since the total number of possible choices for three numbers is 100^3 = 1,000,000, we have p2 = (40 * 10,000) / 1,000,000 = 400,000 / 1,000,000 = 2/5.
Calculating 125/4p2:
To calculate 125/4p2, we substitute the value of p2 into the expression: (125/4) * (2/5) = 250/20 = 12.5.
Therefore, the value of 125/4p2 is 12.5.
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Community Answer
Three numbers are chosen at random, one after another with replacement...
p2 = 1 - (Probability that 3 chosen numbers are greater than 40)
So, 125/4 × p2 = 125/4 x 98/125
= 24.5
So, 125/4 × p2 = 125/4 x 98/125
= 24.5
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Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ..., 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40Q.The value of 125/4p2 is _______.Correct answer is '24.5'. Can you explain this answer?
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