(a) Let X denote the number of hours a Class 12 student studies during a randomly selected school day. The pro

Question

x_i

(i) Find the value of k.

(ii) Determine the probability that the student studied for at least 2 hours.

(iii) Determine the probability that the student studied for at most 2 hours.

(OR)

(b) A river near a small town floods and overflows twice in every 10 years on an average. Assuming that the Poisson distribution is appropriate, what is the mean expectation ? Also, calculate the probability of 3 or less overflows and floods in a 10-year interval. [Given

e^{-2} = 0.13534

]

CBSECBSE Class XII Board 2024Subjective· 5mImportance★★★★★

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Concept understanding — Discrete Probability Distribution

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:

What are all the possible outcomes? (The sample space)
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.

P(X = x) = p(x), \quad \text{where } 0 \leq p(x) \leq 1 \text{ and } \sum_{x} p(x) = 1

A Concrete Example

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

Outcome (x)	How it happens	Probability P(x)
0	TT	$\frac{1}{4}$
1	HT, TH	$\frac{2}{4} = \frac{1}{2}$
2	HH	$\frac{1}{4}$

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

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$\sum P=1\Rightarrow0.1+6k=1\Rightarrow k=\tfrac{3}{20}$ ; then $P(X\ge2)=5k=\tfrac34$ and $P(X\le2)=0.1+3k=\tfrac{11}{20}$ .
$\lambda=2$ , $P(X\le3)=e^{-2}(1+2+2+\tfrac43)=0.8571$ .

\sum_i P(X=x_i)=1

Part (a): $P(0)=0.1,\ P(1)=k,\ P(2)=2k,\ P(3)=k(5-3)=2k,\ P(4)=k(5-4)=k$ .

$x_i$	0	1	2	3	4
$P(X=x_i)$	$0.1$	$k$	$2k$	$2k$	$k$

(i) Total probability $=1$ : $0.1+k+2k+2k+k=1\Rightarrow0.1+6k=1\Rightarrow6k=0.9\Rightarrow k=\dfrac{3}{20}=0.15$ .
(ii) $P(X\ge2)=P(2)+P(3)+P(4)=2k+2k+k=5k=5\times\dfrac{3}{20}=\dfrac{15}{20}=\dfrac{3}{4}$ .
(iii) $P(X\le2)=P(0)+P(1)+P(2)=0.1+k+2k=0.1+3k=0.1+3(0.15)=0.1+0.45=0.55=\dfrac{11}{20}$ .

Part (b): flood overflows twice per $10$ years on average, Poisson, interval $=10$ years.

Mean expectation: $\lambda=2$ . …

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