A random variable X has the probability distribution: X: 1, 2, 3, 4, 5, 6, 7, 8; P(X): 0.15, 0.23, 0

Q: Wait, is 1 a prime number? I always get confused about this in exams.

This is a classic trap! By definition, a prime number must have exactly two distinct positive divisors: $1$ and itself. Since $1$ only has one divisor (itself), it is neither prime nor composite. So, when defining set $E$, exclude $1$ immediately. Keeping these basic definitions crystal clear is what separates the top rankers from the rest.

Q: Can I just use the formula $P(E \cup F) = P(E) + P(F)$? Why do I have to list the elements?

Great question! You can use the formula $P(E \cup F) = P(E) + P(F) - P(E \cap F)$, but notice that $E$ and $F$ are not mutually exclusive here. They share the elements $2$ and $3$. If you simply add $P(E)$ and $P(F)$, you will count the probabilities of $2$ and $3$ twice! Listing the elements is a foolproof way to avoid this double-counting error.

Q: What if the events were mutually exclusive? Would the method change?

If the events were mutually exclusive, meaning $E \cap F = \emptyset$, then yes, the intersection probability would be zero, and the formula would simplify to $P(E) + P(F)$. However, in probability, always check for overlap first. When in doubt, listing the unique elements of the union is the safest, most reliable strategy.

Question

A random variable $X$ has the probability distribution:

| $X$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $P(X)$ | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |

For the events $E = \{X 	ext{ is a prime number}\}$ and $F = \{X < 4\}$, the $P(E \cup F)$ is

Options:

(A) 0.50

(B) 0.77

(C) 0.35

(D) 0.87

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are not just solving a probability problem; we are learning to navigate the landscape of uncertainty. In the JEE Advanced arena, probability is not about guessing; it is about the rigorous, systematic accounting of possibilities.

We are given a discrete random variable $X$ with a distribution table where $X \in \{1, 2, 3, 4, 5, 6, 7, 8\}$. The first step is to verify the validity of the distribution by ensuring the sum of all probabilities equals unity:

$$\sum P(X) = 0.15 + 0.23 + 0.12 + 0.10 + 0.20 + 0.08 + 0.07 + 0.05 = 1.00$$

The system is closed, consistent, and ready for analysis.

### Defining the Players

We have two events, $E$ and $F$. Let us define them with absolute precision.

Event $E$ is the set of prime numbers in our sample space. Since $1$ is not prime, the primes in our set are $2, 3, 5,$ and $7$. Thus, we define our set as:

$$E = \{2, 3, 5, 7\}$$

Event $F$ is defined by the condition $X < 4$. The values that satisfy this inequality are $1, 2,$ and $3$. Thus:

$$F = \{1, 2, 3\}$$

Notice the overlap; both $2$ and $3$ appear in both sets. This is the crux of the problem, as we must ensure we do not count these probabilities twice.

### The Union of Possibilities

We need to find $P(E \cup F)$. The union $E \cup F$ represents the set of all outcomes that belong to either $E$ or $F$.

Combining our sets while ensuring each unique element is listed only once, we get:

$$E \cup F = \{1, 2, 3, 5, 7\}$$

Any outcome in this set makes the event $E \cup F$ occur.

### The Final Calculation

Since these are discrete, mutually exclusive outcomes, the probability of the union is the sum of the individual probabilities:

$$P(E \cup F) = P(1) + P(2) + P(3) + P(5) + P(7)$$

Extracting these values from our distribution table:

- $P(1) = 0.15$
- $P(2) = 0.23$
- $P(3) = 0.12$
- $P(5) = 0.20$
- $P(7) = 0.07$

Performing the summation:

$$P(E \cup F) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07$$

Calculating step-by-step:

$$0.15 + 0.23 = 0.38$$
$$0.38 + 0.12 = 0.50$$
$$0.50 + 0.20 = 0.70$$
$$0.70 + 0.07 = 0.77$$

### The Takeaway

The final result is **$0.77$**.

By systematically defining your sets and summing the probabilities, you eliminate the room for error. You have mastered the logic of the union; keep this discipline, and you will find that even the most complex probability problems become simple, elegant, and solvable.

Select Answer:

Visualized Solution

Analyze the Probability Distribution

Define Event $E$ : Prime Numbers

Define Event $F$ : $X < 4$

Find the Union $E \cup F$

Set Up the Probability Sum

Substitute the Values

Execute the Addition

Final Conclusion

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

Defining the Players

The Union of Possibilities

The Final Calculation

The Takeaway

Similar Questions

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

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:

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 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 six faced die is biased such that $3 \times P (a prime number) = 6 \times P (a composite number) = 2 \times P (1)$ . Let $X$ be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of $X$ is :

Let $X$ be a random variable such that the probability function of a distribution is given by $P (X = 0) = 1/2, P (X = j) = 1/ 3^{j} (j = 1, 2, 3, \dots, \infty)$ . Then the mean of the distribution and $P (X is positive and even)$ respectively are:

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 .

Let a random variable $X$ take values $0, 1, 2, 3$ with $P (X = 0) = P (X = 1) = p$ , $P (X = 2) = P (X = 3)$ and $E (X^{2}) = 2 E (X)$ . Then the value of $8 p - 1$ is :

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 :