Q.If a random variable X has the probability distribution $P(X = x) = \begin{cases} k, & \text{if } x = 0 \\ 2k, & \text{if } x = 1 \text{ or } 2 \\ 0, & \text{otherwise,} \end{cases}$ then the value of k is (A) $\dfrac{1}{3}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{1}{6}$ (D) $\dfrac{1}{4}$

Discrete Probability Distribution: From Intuition to Precision

Imagine you're about to roll a fair six-sided die. Before it lands, you know something important: the outcome will be one of six numbers — 1, 2, 3, 4, 5, or 6. You also know that each number is equally likely. That's your first taste of a discrete probability distribution: a complete description of what can happen and how likely each possibility is.

The word "discrete" means separate, countable. The outcomes are like individual points — you can list them. This is different from something like "the height of a randomly chosen student," which can be any value in a continuous range. Here, we're dealing with things you can count: number of heads in three coin tosses, the sum of two dice, the number of customers arriving at a shop in an hour.

The Intuition

A discrete probability distribution answers two questions:

1. What are all the possible outcomes? (The sample space)
2. What probability does each outcome carry? (The chance it occurs)

The key rule: the probabilities of all possible outcomes must add up to exactly 1. That makes sense — something has to happen, and the total chance of all possibilities is certainty.

Think of a spinner divided into slices. Each slice is an outcome, and the size of the slice is its probability. The whole circle is 1 (or 100%). That's your distribution.

The Precise Statement

Formally, a discrete probability distribution is a function $P$ that assigns a probability to each possible outcome $x$ in a countable set $X$ (the sample space), such that:

• For every outcome $x$, $0 \leq P(x) \leq 1$
• The sum over all outcomes is exactly 1: $\sum_{x \in X} P(x) = 1$

The function $P$ is called the probability mass function (PMF). It gives the "mass" or weight of probability at each discrete point.

A Concrete Example

Consider tossing a fair coin twice. Let $X$ be the number of heads.

Check: $\frac{1}{4} + \frac{1}{2} + \frac{1}{4} = 1$. That's a valid discrete probability distribution. …