JEE Main 2020 (9 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Probability: A random variable $X$ has the following probability distribution: X: 1, 2, 3, 4, 5; P(X): $K^{2}, 2 K, K, 2 K, 5 K^{2}$ . Then $P (X > 2)$ is equal to:

Q: Wait, why do we reject the negative root $K = -1$? It is a valid solution to the quadratic equation, right?

Mathematically, yes, $K = -1$ is a root of the quadratic equation. However, in the physical world of probability, we have a strict constraint: $P(X) \ge 0$ for all $X$. If you plug $K = -1$ into $P(2) = 2K$, you get $-2$, which is physically impossible. Always check your domain constraints!

Q: The question asks for $P(X > 2)$. Does that include 2?

Great catch! The inequality is strict: $X > 2$. This means we exclude 2 and only sum the probabilities for $X = 3, 4, 5$. If it were $X \ge 2$, we would include $P(2)$ as well. Always read the inequality sign carefully.

Select Answer:

Visualized Solution

Probability Distribution of $X$

Discrete random variable $X \in {1, 2, 3, 4, 5}$
Probabilities: $P (1) = K^{2}$ , $P (2) = 2 K$ , $P (3) = K$ , $P (4) = 2 K$ , $P (5) = 5 K^{2}$
Objective: Find $P (X > 2)$

The Total Probability Axiom

Fundamental Axiom: $\sum P (X_{i}) = 1$
The sum of all individual probabilities must exactly equal $1$ .

Summing the Probabilities

Summing all probabilities: $K^{2} + 2 K + K + 2 K + 5 K^{2} = 1$

Forming the Quadratic Equation

Group like terms: $(K^{2} + 5 K^{2}) + (2 K + K + 2 K) = 1$
Simplified Equation: $6 K^{2} + 5 K = 1$
Standard Form: $6 K^{2} + 5 K - 1 = 0$

Splitting the Middle Term

Product needed: $6 \times (- 1) = - 6$
Sum needed: $5$
Split $5 K$ into $6 K - K$
Equation: $6 K^{2} + 6 K - K - 1 = 0$

Factoring the Expression

Factor by grouping: $6 K (K + 1) - 1 (K + 1) = 0$
Common factor: $(6 K - 1) (K + 1) = 0$

Finding the Values of $K$

Case 1: $6 K - 1 = 0 \Rightarrow K = \frac{1}{6}$
Case 2: $K + 1 = 0 \Rightarrow K = - 1$

Validating the Constant $K$

Constraint: $P (X) \geq 0$ for all $X$
If $K = - 1$ , then $P (2) = 2 (- 1) = - 2$ (Impossible)
Therefore, $K = \frac{1}{6}$ is the only valid solution

Defining the Event $X > 2$

Event $X > 2$ includes $X \in {3, 4, 5}$
$P (X > 2) = P (3) + P (4) + P (5)$

Expressing $P (X > 2)$

Substituting expressions: $P (X > 2) = K + 2 K + 5 K^{2}$
Simplify expression: $3 K + 5 K^{2}$

Substituting the Value of $K$

Substitute $K = \frac{1}{6}$ : $3 (\frac{1}{6}) + 5 (\frac{1}{6})^{2}$
$= \frac{3}{6} + 5 (\frac{1}{36})$
$= \frac{18}{36} + \frac{5}{36}$

Final Calculation of $P (X > 2)$

Sum: $\frac{18}{36} + \frac{5}{36} = \frac{23}{36}$
Final Answer: $P (X > 2) = \frac{23}{36}$

00:00 / 00:00

The Sigma Insight: Random Variables and Probability Distributions

Solution Diagram

Analyzing the Probability Space

The system is defined by a random variable

X

taking values in the set

{1, 2, 3, 4, 5}

. The probabilities associated with these outcomes are given as:

P (1) = K^{2}

P (2) = 2 K

P (3) = K

P (4) = 2 K

, and

P (5) = 5 K^{2}

According to the fundamental axiom of probability, the sum of all probabilities in a sample space must equal exactly 1. We establish our anchor equation:

K^{2} + 2 K + K + 2 K + 5 K^{2} = 1

Solving for the Constant $K$

We group the like terms to simplify the expression. Combining the

K^{2}

terms and the

K

terms, we obtain:

6 K^{2} + 5 K - 1 = 0

To solve this quadratic equation, we look for two numbers that multiply to

- 6

and add to

5

. These numbers are

6

and

- 1

. We split the middle term and factor by grouping:

6 K^{2} + 6 K - K - 1 = 0

(6 K - 1) (K + 1) = 0

This yields two potential candidates for

K

K = 1/6

K = - 1

Applying Physical Constraints

We must perform a reality check on our solutions. Since a probability cannot be negative,

K = - 1

is rejected because it would result in negative probabilities (e.g.,

P (2) = 2 (- 1) = - 2

Therefore, we discard the negative root and accept the valid value: $K = 1/6$

Final Calculation of $P (X > 2)$

We are tasked with finding the probability

P (X > 2)

. This corresponds to the sum of the probabilities for

X = 3

X = 4

, and

X = 5

P (X > 2) = P (3) + P (4) + P (5) = K + 2 K + 5 K^{2} = 3 K + 5 K^{2}

Substituting

K = 1/6

into the expression:

P (X > 2) = 3 (\frac{1}{6}) + 5 (\frac{1}{6})^{2}

Simplifying the terms:

P (X > 2) = \frac{3}{6} + \frac{5}{36} = \frac{18}{36} + \frac{5}{36}

The final result is: $P (X > 2) = \frac{23}{36}$

Common Traps & FAQs

Wait, why do we reject the negative root

K = - 1

? It is a valid solution to the quadratic equation, right?

Mathematically, yes,

K = - 1

is a root of the quadratic equation. However, in the physical world of probability, we have a strict constraint:

P (X) \geq 0

for all

X

. If you plug

K = - 1

into

P (2) = 2 K

, you get

- 2

, which is physically impossible. Always check your domain constraints!

Is there a faster way to find

K

without solving the quadratic equation?

Not really. Since the probabilities are defined by a polynomial in

K

, the quadratic equation is the most direct path. Don't fear the algebra—it is just a tool to enforce the consistency of the probability space.

The question asks for

P (X > 2)

. Does that include 2?

Great catch! The inequality is strict:

X > 2

. This means we exclude 2 and only sum the probabilities for

X = 3, 4, 5

. If it were

X \geq 2

, we would include

P (2)

as well. Always read the inequality sign carefully.

Animated Solution for Mathematics - Probability: A random variable $X$ has the following probability distribution: X: 1, 2, 3, 4, 5; P(X): $K^{2}, 2 K, K, 2 K, 5 K^{2}$ . Then $P (X > 2)$ is equal to:

Select Answer:

Visualized Solution

Probability Distribution of $X$

The Total Probability Axiom

Summing the Probabilities

Forming the Quadratic Equation

Splitting the Middle Term

Factoring the Expression

Finding the Values of $K$

Validating the Constant $K$

Defining the Event $X > 2$

Expressing $P (X > 2)$

Substituting the Value of $K$

Final Calculation of $P (X > 2)$

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Probability Space

Solving for the Constant $K$

Applying Physical Constraints

Final Calculation of $P (X > 2)$

Similar Questions

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Comprehension Passage

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Select Answer:

Visualized Solution

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

The Total Probability Axiom

Summing the Probabilities

Forming the Quadratic Equation

Splitting the Middle Term

Factoring the Expression

Finding the Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

Validating the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

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Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X>2)

Substituting the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X>2)

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Probability Space

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Applying Physical Constraints

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X>2)

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Defining the Event $X > 2$

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Substituting the Value of $K$

Final Calculation of $P (X > 2)$

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Final Calculation of $P (X > 2)$

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If the probability that the random variable $X$ takes the value $x$ is given by $P (X = x) = k (x + 1) 3^{- x}$ , $x = 0, 1, 2, 3, \dots$ , where $k$ is a constant, then $P (X \geq 3)$ is equal to

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