(a) The probability distribution for the number of students being absent in a class on a Saturday is as follow

Question 160 of 165

X

(i) Calculate

p

(ii) Calculate the mean of the number of absent students on Saturday.

(OR)

(b) For the vacancy advertised in the newspaper, 3000 candidates submitted their applications. From the data it was revealed that two third of the total applicants were females and other were males. The selection for the job was done through a written test. The performance of the applicants indicates that the probability of a male getting a distinction in written test is 0.4 and that a female getting a distinction is 0.35. Find the probability that the candidate chosen at random will have a distinction in the written test.

Tripura TbseCBSE Class XII Board 2025Subjective· 3mImportance★★★★★

97% · 160/165 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)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. …

Part (b)Concept understanding — Discrete Probability Distribution

Discrete Probability Distribution: From Intuition to Precision

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. …

Start your 14-day free trial to unlock the full solution →

Part (a): $\sum P=1$ gives $p=\tfrac17$ , and the mean is $3$ absent students. Part (b): by the law of total probability, $P(\text{distinction})=\tfrac13(0.4)+\tfrac23(0.35)=\tfrac{11}{30}$ .

Part (a)

The values are $X=0,2,4,5$ with $P(X)=p,2p,3p,p$ .

(i) A valid probability distribution must have total probability $1$ :

p+2p+3p+p=7p=1\Rightarrow p=\frac{1}{7}.

(ii) The mean (expected number absent): …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ4 — The probability of finding a green signal on a busy crossing X is 30%.…→Next questionQ6 — (a) A die with numbers 1 to 6 is biased such that P(2) = (3)/(10) and …

Q.(a) The probability distribution for the number of students being absent in a class on a Saturday is as follows: $X$ : 0, 2, 4, 5 $P(X)$ : $p$ , $2p$ , $3p$ , $p$ where $X$ is the number of students absent.

Part (a)Concept understanding — Discrete Probability Distribution

Discrete Probability Distribution: From Intuition to Precision

The Intuition

The Precise Statement

A Concrete Example

Part (b)Concept understanding — Discrete Probability Distribution

Discrete Probability Distribution: From Intuition to Precision

The Intuition

The Precise Statement

A Concrete Example

Part (a)

Part (a)

Method: Distribution normalisation with a mean, and the law of total probability

Steps

Common Mistakes

Step 1 — Total-probability axiom

Step 2 — Substitute and sum

Step 3 — Solve

Q.(a) The probability distribution for the number of students being absent in a class on a Saturday is as follows: XXX: 0, 2, 4, 5 P(X)P(X)P(X): ppp, 2p2p2p, 3p3p3p, ppp where XXX is the number of students absent.

Part (a)

🎓Unlock everything free for 14 days

Q.(a) The probability distribution for the number of students being absent in a class on a Saturday is as follows: $X$ : 0, 2, 4, 5 $P(X)$ : $p$ , $2p$ , $3p$ , $p$ where $X$ is the number of students absent.

Unlock everything free for 14 days