Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, .

Q: Why can't I just calculate $P(\text{max} \ge 81)$ directly? It seems straightforward.

Direct calculation requires summing the probabilities of the maximum being 81, 82, ..., 100. While possible, it is computationally expensive and prone to arithmetic errors. The complement rule, $P(A) = 1 - P(A^c)$, collapses these 20 cases into one simple calculation. In JEE Advanced, efficiency is just as important as accuracy.

Q: Why is the probability of the maximum being $\le 80$ equal to $\left(\frac{80}{100}\right)^3$?

Because we are sampling with replacement! Each of the three draws is an independent event. For the maximum to be $\le 80$, the first number must be $\le 80$ (prob $\frac{80}{100}$), the second must be $\le 80$ (prob $\frac{80}{100}$), and the third must be $\le 80$ (prob $\frac{80}{100}$). Since they are independent, we multiply them: $\frac{80}{100} \times \frac{80}{100} \times \frac{80}{100}$.

Q: I got confused with the range for $p_2$. Why is it $100 - 41 + 1$?

That is the standard formula for the number of integers in an inclusive range $[a, b]$, which is $b - a + 1$. Here, the range is $[41, 100]$. So, $100 - 41 + 1 = 60$. Always double-check your boundaries; it is a common place to lose marks!

Question

Passage:
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.

Question 1:
The value of $\frac{625}{4}p_1$ is

Question 2:
The value of $\frac{125}{4}p_2$ is

Sigma Percentile · Accepted Answer

### The Art of the Complement: A Probability Masterclass

Imagine you are standing before a vast urn containing 100 balls, numbered from 1 to 100. You are tasked with drawing three balls, one after another, with replacement.

This act of replacement ensures that every draw is an independent event. The total number of possible outcomes for these three draws is $100^3$.

### Phase 1: The $p_1$ Journey - Taming the Maximum

We seek $p_1$, the probability that the maximum of the three chosen numbers is at least 81. Calculating this directly would require summing cases from 81 to 100, which is inefficient.

Instead, we use the **Complement Rule**. The complement of "the maximum is at least 81" is "the maximum is 80 or less."

Mathematically, we express this as:

$$p_1 = 1 - P(	ext{max} \le 80)$$

For the maximum to be 80 or less, every chosen number must fall within the set $\{1, 2, \dots, 80\}$. The probability of picking a number $\le 80$ in a single draw is $\frac{80}{100} = \frac{4}{5}$.

Since we have three independent draws, we cube this probability:

$$P(	ext{max} \le 80) = \left( \frac{4}{5} ight)^3 = \frac{64}{125}$$

Thus, $p_1 = 1 - \frac{64}{125} = \frac{61}{125}$. The problem asks for the value of $\frac{625}{4}p_1$:

$$\frac{625}{4} 	imes \frac{61}{125} = \frac{5 	imes 61}{4} = \frac{305}{4} = \mathbf{76.25}$$

### Phase 2: The $p_2$ Journey - Mastering the Minimum

Now, we turn to $p_2$, the probability that the minimum of the chosen numbers is at most 40. The complement of "the minimum is at most 40" is "the minimum is strictly greater than 40," which implies the minimum is at least 41.

We define this as:

$$p_2 = 1 - P(	ext{min} \ge 41)$$

For the minimum to be 41 or higher, every number must be in the range $[41, 100]$. The number of integers in this range is $100 - 41 + 1 = 60$.

The probability of picking a number $\ge 41$ in a single draw is $\frac{60}{100} = \frac{3}{5}$. Cubing this for three independent draws:

$$P(	ext{min} \ge 41) = \left( \frac{3}{5} ight)^3 = \frac{27}{125}$$

This yields $p_2 = 1 - \frac{27}{125} = \frac{98}{125}$. Finally, we evaluate $\frac{125}{4}p_2$:

$$\frac{125}{4} 	imes \frac{98}{125} = \frac{98}{4} = \mathbf{24.5}$$

### Conclusion: The Elegance of Logic

We have successfully navigated these probability challenges by utilizing the complement rule. This approach transformed potentially grueling summations into simple arithmetic.

In competitive examinations, the most powerful tool in your arsenal is the ability to choose the path of least resistance. Master this logic, and you will find that even the most complex probability problems become manageable.

Animated Solution for Mathematics - Probability: Comprehension Passage

The value of $\frac{625}{4} p_{1}$ is

Enter Numerical Value:

The value of $\frac{125}{4} p_{2}$ is

Enter Numerical Value:

Visualized Solution

Sample Space Setup

Defining $p_{1}$ via Complement

Probability of $max \leq 80$

Evaluating $p_{1}$

Calculating $\frac{625}{4} p_{1}$

Defining $p_{2}$ via Complement

Probability of $min \geq 41$

Evaluating $p_{2}$

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

Final Conclusion

The Sigma Insight: Classical Definition of Probability

The Art of the Complement

Phase 1

Phase 2

Conclusion

Similar Questions

Let $E_{1}, E_{2}, E_{3}$ be three mutually exclusive events such that $P (E_{1}) = \frac{2 + 3 p}{6}$ , $P (E_{2}) = \frac{2 - p}{8}$ and $P (E_{3}) = \frac{1 - p}{2}$ . If the maximum and minimum values of $p$ are $p_{1}$ and $p_{2}$ , then $(p_{1} + p_{2})$ is equal to :

Two numbers are selected randomly from the set $S = {1, 2, 3, 4, 5, 6}$ without replacement one by one. The probability that minimum of the two numbers is less than 4 is

$A$ is a set containing $n$ elements. A subset $P$ of $A$ is chosen at random. The set $A$ is reconstructed by replacing the elements of $P$ . A subset $Q$ of $A$ is again chosen at random. Find the probability that $P$ and $Q$ have no common elements.

The probability that two randomly selected subsets of the set ${1, 2, 3, 4, 5}$ have exactly two elements in their intersection, is:

If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is

Two distinct numbers $a$ and $b$ are selected at random from ${1, 2, 3, \dots, 50}$ . The probability, that their product $ab$ is divisible by 3, is :

Two integers $x$ and $y$ are chosen with replacement from the set ${0, 1, 2, 3, \dots, 10}$ . Then the probability that $∣ x - y ∣ > 5$ is :

Three distinct numbers are selected randomly from the set ${1, 2, 3, \dots, 40}$ . If the probability, that the selected numbers are in an increasing G.P. is $\frac{m}{n}$ , $g cd (m, n) = 1$ , then $m + n$ is equal to _____.

An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown $n$ times and the list of $n$ numbers showing up is noted. What is the probability that, among the numbers 1, 2, 3, 4, 5, 6, only three numbers appear in this list?

If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $i^{t h}$ roll than the number obtained in the $(i - 1)^{t h}$ roll, $i = 2, 3$ , is equal to

Animated Solution for Mathematics - Probability: Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of 4625​p1​ is

Enter Numerical Value:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of 4125​p2​ is

Enter Numerical Value:

Visualized Solution

Sample Space Setup

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p1​ via Complement

Probability of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } max≤80

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4625​p1​

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p2​ via Complement

Probability of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } min≥41

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4125​p2​

Final Conclusion

The Sigma Insight: Classical Definition of Probability

The Art of the Complement

Phase 1

Phase 2

Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two numbers are selected randomly from the set S={1,2,3,4,5,6} without replacement one by one. The probability that minimum of the two numbers is less than 4 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability that two randomly selected subsets of the set {1,2,3,4,5} have exactly two elements in their intersection, is:

If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two distinct numbers a and b are selected at random from {1,2,3,…,50}. The probability, that their product ab is divisible by 3, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two integers x and y are chosen with replacement from the set {0,1,2,3,…,10}. Then the probability that ∣x−y∣>5 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If an unbiased dice is rolled thrice, then the probability of getting a greater number in the ith roll than the number obtained in the (i−1)th roll, i=2,3, is equal to

The value of $\frac{625}{4} p_{1}$ is

The value of $\frac{125}{4} p_{2}$ is

Defining $p_{1}$ via Complement

Probability of $max \leq 80$

Evaluating $p_{1}$

Calculating $\frac{625}{4} p_{1}$

Defining $p_{2}$ via Complement

Probability of $min \geq 41$

Evaluating $p_{2}$

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

Two numbers are selected randomly from the set $S = {1, 2, 3, 4, 5, 6}$ without replacement one by one. The probability that minimum of the two numbers is less than 4 is

The probability that two randomly selected subsets of the set ${1, 2, 3, 4, 5}$ have exactly two elements in their intersection, is:

Two distinct numbers $a$ and $b$ are selected at random from ${1, 2, 3, \dots, 50}$ . The probability, that their product $ab$ is divisible by 3, is :

Two integers $x$ and $y$ are chosen with replacement from the set ${0, 1, 2, 3, \dots, 10}$ . Then the probability that $∣ x - y ∣ > 5$ is :

If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $i^{t h}$ roll than the number obtained in the $(i - 1)^{t h}$ roll, $i = 2, 3$ , is equal to