According to an educational board survey, it was observed that class XII students apply at least one to four w

Question

P(X = x)

(i) Determine the value of k.

(ii) What is the probability that Mahesh will get admission in the college, given that he applied at least 3 weeks ahead of application deadline ?

(iii)

(a) Calculate the mathematical expectation of number of weeks taken by a student to apply ahead of a college’s application deadline.

(OR)

(iii)

(b) To promote early admissions, the college is offering scholarships to the students for applying ahead of deadline as follows : ₹ 50,000 for applying 4 weeks ahead ₹ 20,000 for applying 3 weeks ahead ₹ 12,000 for applying 2 weeks ahead and ₹ 9,600 for applying 1 week ahead Determine the expected scholarship offered by the college.

CBSECBSE Class XII Board 2025Subjective· 4mImportance★★★★★

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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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$\sum P=1\Rightarrow k=\tfrac14$ ; then $P(X{=}3)+P(X{=}4)=\tfrac78$ ; $E(X)=\tfrac{10}{3}$ weeks; expected scholarship $=$ ₹ 33,900.

\sum_x P(X=x)=1

First list the probabilities using the given rule:

$P(0)=0,\quad P(1)=\dfrac{k}{6},\quad P(2)=\dfrac{2k}{6}=\dfrac{k}{3},\quad P(3)=\dfrac{3(1-k)}{6}=\dfrac{1-k}{2},\quad P(4)=\dfrac{4k}{2}=2k.$

(i) Value of $k$

Sum all probabilities to $1$ : $0+\dfrac{k}{6}+\dfrac{2k}{6}+\dfrac{3(1-k)}{6}+\dfrac{4k}{2}=1$ .
Multiply through by $6$ : $k+2k+3(1-k)+12k=6$ .
Simplify: $k+2k-3k+12k+3=6\Rightarrow 12k+3=6\Rightarrow 12k=3$ .
So $k=\dfrac{3}{12}=\dfrac14$ .

With $k=\tfrac14$ : $P(0)=0,\ P(1)=\dfrac1{24},\ P(2)=\dfrac1{12},\ P(3)=\dfrac38,\ P(4)=\dfrac12$ (these sum to $1$ ).

(ii) Probability of admission given he applied at least 3 weeks ahead

"At least 3 weeks ahead" covers $X=3$ and $X=4$ .
The admission probability is $P(X=3)+P(X=4)=\dfrac38+\dfrac12=\dfrac38+\dfrac48=\dfrac78$ .

(iii)(a) Mathematical expectation $E(X)$

$E(X)=\sum x\,P(X=x)=0\cdot0+1\cdot\dfrac1{24}+2\cdot\dfrac1{12}+3\cdot\dfrac38+4\cdot\dfrac12$ .
Convert to twenty-fourths: $\dfrac1{24}+\dfrac{4}{24}+\dfrac{27}{24}+\dfrac{48}{24}=\dfrac{80}{24}=\dfrac{10}{3}$ .
So $E(X)=\dfrac{10}{3}\approx3\dfrac13$ weeks. …

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