Four candidates A, B, C, D have applied for the assignment to coach a school cricket team. If A is twice as li

NCERT Exemplar · Q9

Q. Four candidates A, B, C, D have applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that

(a) C will be selected?

(b) A will not be selected?

Uttarakhand UbseShort· 3mImportance★★★★★est

63% · 59/93 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 →

Concept understanding — Classical Probability

Classical Probability: The "Fair Game" Definition

Imagine you roll a fair six-sided die. Before it lands, you know there are exactly six possible outcomes — 1, 2, 3, 4, 5, or 6 — and you have no reason to believe any one face is more likely than another. That gut feeling of "all outcomes are equally likely" is the entire foundation of classical probability.

The Intuition

Classical probability was born from games of chance — dice, coins, cards. In these settings, the physical symmetry of the objects (a balanced die, a fair coin) guarantees that no outcome is favoured. So the probability of an event is simply:

Number of ways the event can happen, divided by the total number of possible outcomes.

If you want the chance of rolling an even number on a die, count the evens: 2, 4, 6 — that's 3 ways. Total outcomes: 6. So probability = $3/6 = 1/2$ .

This is the "counting" approach. It works beautifully when the underlying experiment is symmetric and finite.

The Precise Statement

P(E) = \frac{\text{Number of outcomes favourable to event } E}{\text{Total number of equally likely outcomes}}

This is called the classical definition (or a priori definition) of probability. It was formalised by Pierre-Simon Laplace in the 18th century.

Three conditions must hold for this definition to apply:

Finite sample space — there are only a fixed, countable number of possible outcomes.
Equally likely outcomes — each outcome has the same chance of occurring (the "fairness" condition).
Mutually exclusive outcomes — no two outcomes can happen at the same time.

Watch out

The biggest mistake students make is applying classical probability to situations where outcomes are not equally likely. For example: "I can either pass or fail the exam — two outcomes, so probability of passing is $1/2$ ." That's nonsense, because passing and failing are not equally likely. The die works only because the die is fair.

A Simple Example

Problem: A bag contains 3 red marbles and 2 blue marbles. You pick one marble at random. What is the probability it is red?

Step 1 — Identify the sample space: There are 5 marbles total. If the marbles are physically identical except for colour, and you pick without looking, each marble is equally likely to be chosen. So total outcomes = 5.

Step 2 — Count favourable outcomes: 3 marbles are red. So favourable outcomes = 3.

Step 3 — Apply the formula:

P(\text{red}) = \frac{3}{5}

That's it. No deeper theory needed for this case.

When Classical Probability Fails

Classical probability cannot handle: …

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

Assign relative weights to each candidate based on the given likelihood ratios, convert to probabilities by normalizing, then compute the required probabilities. C has probability $\frac{2}{9}$ and A will not be selected with probability $\frac{5}{9}$ .

The problem describes how likely each candidate is relative to the others. Classical probability tells us that when outcomes have different likelihoods, we assign weights proportional to those likelihoods, then normalize so the total probability equals 1. Think of it as dividing a pie: if A deserves twice as much as B, we give A two slices for every one slice B gets, then make sure all slices add up to the whole pie.

Let's translate the given relationships into mathematical weights.

Setting up the relative weights

We need a common reference. Let's say D has weight $w$ .

C is twice as likely as D:

If $P(D) = w$ , then $P(C) = 2w$ .
B and C have about the same chance:

This means $P(B) = P(C) = 2w$ .
A is twice as likely as B:

Since $P(B) = 2w$ , we have $P(A) = 2 \times 2w = 4w$ .

Now we have weights:

$P(A) = 4w$
$P(B) = 2w$
$P(C) = 2w$
$P(D) = w$

Normalizing to find actual probabilities

The sum of all probabilities must equal 1 (someone will definitely be selected):

P(A) + P(B) + P(C) + P(D) = 1

4w + 2w + 2w + w = 1

9w = 1

w = \frac{1}{9}

Now substitute back to find each probability:

$P(A) = 4w = \frac{4}{9}$
$P(B) = 2w = \frac{2}{9}$
$P(C) = 2w = \frac{2}{9}$
$P(D) = w = \frac{1}{9}$

Tip

When dealing with relative likelihoods, always pick the simplest candidate (usually the one with the smallest weight) as your unit, then express everyone else in terms of that unit.

Part (a): Probability that C will be selected

From our calculation above: …

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

←Previous questionQ8 — A team of medical students doing their internship have to assist durin…→Next questionQ10 — One of the four persons John, Rita, Aslam or Gurpreet will be promoted…

Q.Four candidates A, B, C, D have applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that

🎓Unlock everything free for 14 days

Unlock everything free for 14 days