A random variable X has the following probability distribution: X: 1 2 3 4 5 P(x): K², 2k, K, 2K, 5K

Q: Wait, why do we reject the negative root $K = -1$?

Great question! In probability theory, the probability of any event $P(x)$ must satisfy $0 \le P(x) \le 1$. If we chose $K = -1$, then $P(2) = 2K = -2$, which is physically impossible. Probability is a measure of likelihood, and it cannot be negative.

Q: Can I use $P(X > 2) = 1 - P(X \le 2)$ instead?

Absolutely! That is a brilliant alternative approach. You would calculate $1 - (P(1) + P(2))$, which is $1 - (K^2 + 2K)$. Both methods will lead you to the same correct answer. It is always good to have a backup strategy!

Q: Why is the sum of probabilities always 1?

This is the fundamental axiom of probability. It represents the fact that the random variable $X$ must take one of the values in its sample space. The sum of all possible outcomes covers the entire 'certainty' of the experiment, which is defined as 1.

Question

A random variable $X$ has the following probability distribution:
$X: 1, 2, 3, 4, 5$
$P(x): K^2, 2k, K, 2K, 5K^2$
Then $P(x > 2)$ is equal to :

Options:

(A) 23/36

(B) 7/12

(C) 1/36

(D) 1/6

Sigma Percentile · Accepted Answer

### The Beauty of Discrete Probability

Welcome, fellow traveler on the JEE journey. Today, we are going to unravel a problem that might look like simple algebra, but it is actually a beautiful dance of logic and constraints.

We are dealing with a discrete random variable $X$ that takes values in the set $\{1, 2, 3, 4, 5\}$. Each of these outcomes has a probability attached to it, defined by an unknown constant $K$. Our mission is to find the probability that $X$ is strictly greater than 2.

### Phase 1: The Law of Total Probability

Before we dive into the calculations, let's ground ourselves in the most fundamental rule of probability: the Law of Total Probability. Imagine you are looking at a set of bars on a graph, where each bar represents the probability of an outcome.

If these bars represent all possible outcomes, their total height must be exactly 1. Mathematically, we express this as:

$$\sum P(x_i) = 1$$

This is our master key. Without this, we are lost in a sea of variables. We sum up our given probabilities:

$$K^2 + 2K + K + 2K + 5K^2 = 1$$

### Phase 2: The Quadratic Challenge

Now, let's clean up this equation. We group the like terms together. We have $K^2$ terms and $K$ terms.

Combining them, we get $6K^2 + 5K = 1$. Moving the 1 to the left side, we arrive at the standard quadratic form:

$$6K^2 + 5K - 1 = 0$$

This is where many students rush and make a mistake. We need to factor this carefully. We are looking for two numbers that multiply to $-6$ and add to $5$. Those numbers are $6$ and $-1$.

So, we split the middle term:

$$6K^2 + 6K - K - 1 = 0$$

Factoring this, we get $(6K - 1)(K + 1) = 0$. This gives us two potential values for $K$:

$$K = \frac{1}{6} \quad 	ext{and} \quad K = -1$$

### Phase 3: The Physical Reality

Here is where the JEE examiner tests your conceptual depth. We have two mathematical solutions, but do they both make physical sense?

We must check our constraint: $P(x) \ge 0$ for all $x$. If we plug $K = -1$ back into our probability distribution, we get negative probabilities, which is a physical impossibility.

Thus, we must reject $K = -1$ and accept **$K = \frac{1}{6}$** as our only valid constant.

### Phase 4: The Final Calculation

With $K = \frac{1}{6}$ in our toolkit, we can now tackle the target: $P(X > 2)$. This condition includes the outcomes $X = 3$, $X = 4$, and $X = 5$.

We sum their probabilities:

$$P(3) + P(4) + P(5) = K + 2K + 5K^2 = 3K + 5K^2$$

Now, we substitute $K = \frac{1}{6}$ into this expression:

$$3\left(\frac{1}{6}ight) + 5\left(\frac{1}{6}ight)^2$$

This simplifies to:

$$\frac{3}{6} + \frac{5}{36}$$

To add these, we convert $\frac{3}{6}$ to $\frac{18}{36}$. Finally:

$$\frac{18}{36} + \frac{5}{36} = \frac{23}{36}$$

### Conclusion

And there you have it! The probability is **$\frac{23}{36}$**.

The journey from the abstract sum of probabilities to the final fraction is a testament to the power of systematic thinking. Always remember: in JEE, the math is just the language; the logic is the story. Keep practicing, stay curious, and never stop questioning the 'why' behind the equations.

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

Understanding the Probability Distribution

The Fundamental Law of Total Probability

Setting up the Equation

Combining Like Terms

Forming the Quadratic Equation

Factoring the Quadratic

Solving for $K$

Constraint Check: $K > 0$

Identifying the Target: $P (X > 2)$

Summing Target Probabilities

Substitution of $K = \frac{1}{6}$

Fraction Addition

Final Conclusion

The Sigma Insight: Random Variables and Probability Distributions

The Beauty of Discrete Probability

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

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:

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

If a random variable $x$ has the probability distribution $x P (x) 00 1 2 k 2 k 3 3 k 4 2 k^{2} 5 2 k 6 k^{2} + k 7 7 k^{2}$ then $P (3 < x \leq 6)$ is equal to :

A random variable $X$ has Poisson distribution with mean 2. Then $P (X > 1.5)$ equals

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

A fair die is tossed until six is obtained on it. Let $X$ be the number of required tosses, then the conditional probability $P (X \geq 5∣ X > 2)$ is :

The probability distribution of random variable $X$ is given by: $X P (X) 1 K 2 2 K 3 2 K 4 3 K 5 K$ Let $p = P (1 < X < 4∣ X < 3)$ . If $5 p = λ K$ , then $λ$ equal to .

Comprehension Passage

The probability that $X = 3$ equals

The probability that $X \geq 3$ equals

The conditional probability that $X \geq 6$ given $X > 3$ equals

Animated Solution for Mathematics - Probability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A random variable X has the following probability distribution: X:1,2,3,4,5 P(x):K2,2k,K,2K,5K2 Then P(x>2) is equal to :