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Theoretical Approach to Probability

🎓 Class 10 Mathematics CBSE Theory Ch 14 — Probability ⏱ ~25 min
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This MCQ module is based on: Theoretical Approach to Probability

This mathematics assessment will be based on: Theoretical Approach to Probability
Targeting Class 10 level in Probability, with Intermediate difficulty.

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14.1 Probability — A Theoretical Approach

In everyday language we often use words such as "probably", "likely", "chances are", or "most certainly". These capture an informal sense of uncertainty. Probability? is the branch of mathematics that makes such intuition precise by measuring uncertainty with a number between 0 and 1.

Consider tossing a fair coin. It is symmetrical, so there is no reason for either face to turn up more often than the other. We call the two outcomes equally likely. Similarly when a balanced die is rolled, each of the six faces is equally likely to land on top; when a card is drawn from a well-shuffled deck of 52 cards, each card is equally likely to be picked.

Equally Likely Outcomes
Two or more outcomes of an experiment are said to be equally likely if, on the basis of symmetry or other physical reasoning, there is no reason to expect one of them more often than another in a large number of trials.

Random Experiments and Events

A random experiment? is one whose outcome cannot be predicted with certainty but whose set of possible outcomes is well-defined. Tossing a coin, rolling a die, and drawing a card are all random experiments.

  • The set of all possible outcomes is called the sample space.
  • Any subset of the sample space is an event?.
  • An outcome that belongs to an event is called a favourable outcome for that event.

For example, when a die is thrown, the sample space is {1, 2, 3, 4, 5, 6}. The event "an even number shows up" is the subset {2, 4, 6} — it has 3 favourable outcomes out of 6.

Classical (Theoretical) Definition of Probability

Definition
If a random experiment has \(n\) possible outcomes that are equally likely, and an event E occurs in \(m\) of these outcomes, then the probability of E is \[ P(E) = \dfrac{\text{Number of outcomes favourable to E}}{\text{Total number of possible outcomes}} = \dfrac{m}{n}. \]

This is known as the classical definition and was proposed by Pierre-Simon Laplace in his famous 1812 treatise Théorie Analytique des Probabilités.

Historical Note
Probability theory was born in the 16th century as an Italian physician and mathematician, Gerolamo Cardano (1501–1576), wrote the first book on the subject — Liber de Ludo Aleae ("The Book on Games of Chance"). Later, Blaise Pascal and Pierre de Fermat (1654) exchanged letters on gambling problems posed by Chevalier de Méré; their correspondence is considered the formal beginning of the subject. Jakob Bernoulli (1654–1705) and Pierre-Simon Laplace (1749–1827) gave the modern shape to the theory.

Example 1 — A Coin is Tossed

Find the probability of getting a head when a coin is tossed once.

Solution. Possible outcomes: head (H), tail (T). So \(n=2\). Let E be the event "getting a head". Then \(m=1\).

\(P(E) = P(\text{head}) = \dfrac{1}{2}\)

Similarly, \(P(\text{tail})=\tfrac{1}{2}\).

Fair Coin — Sample Space H P(H) = 1/2 T P(T) = 1/2
Two equally likely outcomes: each has probability 1/2.

Example 2 — A Bag of Marbles

A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the (i) yellow ball? (ii) red ball? (iii) blue ball?

Solution. \(n=3\) equally likely outcomes. In each case \(m=1\). So \(P(\text{yellow})=P(\text{red})=P(\text{blue})=\tfrac{1}{3}\).

In-text: Notice that \(P(\text{yellow})+P(\text{red})+P(\text{blue})=\tfrac{1}{3}+\tfrac{1}{3}+\tfrac{1}{3}=1\). The probabilities of all possible outcomes in any experiment always add up to 1.

Range of Probability — 0 to 1

An event that cannot happen in the given experiment is an impossible event; its probability is 0. An event that is bound to happen is a sure (or certain) event; its probability is 1.

Property
For any event E, \(0 \le P(E) \le 1\). The complement of E, written \(\bar{\text{E}}\) or E', consists of all outcomes not in E, and satisfies \[ P(\text{E})+P(\bar{\text{E}})=1\quad\Rightarrow\quad P(\bar{\text{E}})=1-P(\text{E}). \]
0 — Impossible
0.25 — Unlikely
0.5 — Even
0.75 — Likely
1 — Certain

Example 3 — Rolling a Die

A die is thrown once. Find the probability of getting (i) a prime number (ii) a number lying between 2 and 6 (iii) an odd number.

Solution. Sample space = {1, 2, 3, 4, 5, 6}, \(n=6\).

  • Prime numbers in {1..6} are 2, 3, 5 → \(m=3\); \(P=\tfrac{3}{6}=\tfrac{1}{2}\).
  • Numbers strictly between 2 and 6: 3, 4, 5 → \(m=3\); \(P=\tfrac{1}{2}\).
  • Odd numbers: 1, 3, 5 → \(m=3\); \(P=\tfrac{1}{2}\).

Example 4 — Drawing a Card

One card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card will (i) be an ace (ii) not be an ace.

Solution. \(n=52\). There are 4 aces.

(i) \(P(\text{ace})=\dfrac{4}{52}=\dfrac{1}{13}\)

(ii) \(P(\text{not ace})=1-P(\text{ace})=1-\dfrac{1}{13}=\dfrac{12}{13}\)
A Standard Deck of 52 Cards ♥ Hearts (13) Red ♦ Diamonds (13) Red ♠ Spades (13) Black ♣ Clubs (13) Black Each suit: Ace, 2, 3, …, 10, J, Q, K → 13 cards. Total = 4 × 13 = 52.
4 suits × 13 ranks = 52 cards. Face cards (J, Q, K) = 12; Aces = 4.

Example 5 — More Card Questions

From a well-shuffled deck of 52 cards, one card is drawn at random. Find the probability that the card drawn is (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds.

\(n=52\) in every part.
(i) Red kings: king of hearts + king of diamonds = 2. \(P=2/52=1/26\).
(ii) Face cards (J,Q,K in each of 4 suits) = 12. \(P=12/52=3/13\).
(iii) Red face cards = 6. \(P=6/52=3/26\).
(iv) Jack of hearts = 1. \(P=1/52\).
(v) Spades = 13. \(P=13/52=1/4\).
(vi) Queen of diamonds = 1. \(P=1/52\).

Example 6 — Two Coins

Two coins are tossed simultaneously. Find the probability of getting (i) at least one head (ii) exactly one tail (iii) no heads.

Sample space = {HH, HT, TH, TT}, \(n=4\).

(i) At least one head = {HH, HT, TH}, \(m=3\). \(P=3/4\).

(ii) Exactly one tail = {HT, TH}, \(m=2\). \(P=2/4=1/2\).

(iii) No heads = {TT}, \(m=1\). \(P=1/4\).
Two Coins — Tree of Outcomes Start H (1/2) T (1/2) H T HH — 1/4 HT — 1/4 TH — 1/4 TT — 1/4
Probability tree for tossing two coins. Each path has probability 1/2 × 1/2 = 1/4.
Activity — Experiment vs Theory
L3 Apply
Materials: 1 coin, pen, tally sheet.
Predict: If you toss a coin 100 times, how many heads do you expect? How close will the empirical ratio be to 1/2?
  1. Toss the coin 10 times and record the number of heads.
  2. Continue in batches of 10, up to 100 total tosses. Keep a running tally of heads.
  3. After each batch, compute the empirical probability = (heads so far)/(tosses so far).
  4. Plot the values on a graph of empirical probability vs number of tosses.
After 10 or 20 tosses the ratio may swing between 0.3 and 0.7. By 100 tosses it usually settles near 0.5. This illustrates the Law of Large Numbers: the experimental probability approaches the theoretical probability as the number of trials grows.

Competency-Based Questions

Scenario: A survey of 40 students of a class X is taken. 25 like football and 30 like cricket; some like both. One student is chosen at random from the class of 40. (Total students surveyed = 40; all 40 like at least one of the two sports.)
Q1. How many students like both football and cricket? Hence find P(the chosen student likes only football).
L3 Apply
Using |F ∪ C| = |F| + |C| − |F ∩ C|: 40 = 25 + 30 − x ⇒ x = 15. Students liking only football = 25 − 15 = 10. P(only football) = 10/40 = 1/4.
Q2. Analyse: Two students argue — one says P(likes cricket) = 30/40 and the other says 30/55 (since 25 + 30 = 55 sport-likings). Who is correct and why?
L4 Analyse
The first student. In the classical definition, the sample space is the set of students (size 40), not sport-likings. 30/55 wrongly counts students who like both sports twice. Correct probability = 30/40 = 3/4.
Q3. Evaluate: For any event E, P(E) + P(not E) = 1. Use this to quickly find P(the chosen student does NOT like football).
L5 Evaluate
P(likes football) = 25/40 = 5/8. So P(does not like football) = 1 − 5/8 = 3/8. This matches 15/40 (students liking only cricket) = 3/8. ✓
Q4. Create a small experiment (≤ 10 outcomes) in which exactly one event has probability 1/3 and a different event has probability 3/4, and they share some favourable outcomes. Show the sample space clearly.
L6 Create
Sample space: outcomes numbered 1–12 (roll a 12-sided die). Event A = {1, 2, 3, 4}; P(A) = 4/12 = 1/3. Event B = {1, 2, 3, 4, 5, 6, 7, 8, 9} (any outcome 1–9); P(B) = 9/12 = 3/4. They share outcomes 1, 2, 3, 4 (A ⊂ B). Many valid constructions exist.

Assertion–Reason Questions

Assertion (A): The probability of any event lies in the closed interval [0, 1].
Reason (R): The number of favourable outcomes cannot exceed the total number of outcomes and cannot be negative.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(a) — 0 ≤ m ≤ n forces 0 ≤ m/n ≤ 1. R directly explains A.
Assertion (A): When a die is rolled, P(number > 6) = 0.
Reason (R): The event "a number greater than 6" has no favourable outcome in the sample space {1, 2, 3, 4, 5, 6}.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(a) — Impossible event has probability 0; R is the direct reason.
Assertion (A): If the probability of rain today is 0.7, the probability of no rain today is 0.3.
Reason (R): For any event E and its complement, P(E) + P(not E) = 1.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(a) — Complementary events sum to 1; the statement follows directly.

Frequently Asked Questions

What is theoretical probability?
Theoretical (or classical) probability of an event E is P(E) = (number of favourable outcomes) / (total equally likely outcomes). It is computed from the structure of the experiment without performing trials.
What are equally likely outcomes?
Outcomes are equally likely when each has the same chance of occurring - for example, any face of a fair die, or any card drawn from a well-shuffled deck.
What is the sample space?
The sample space is the set of all possible outcomes of a random experiment. Tossing a coin once gives sample space {Head, Tail}; rolling a die gives {1,2,3,4,5,6}.
What is an elementary event?
An elementary event is an event containing exactly one outcome of the sample space. The sum of probabilities of all elementary events of an experiment is 1.
What are sure and impossible events?
A sure (or certain) event is one that always happens - its probability is 1. An impossible event can never happen - its probability is 0. Thus 0 <= P(E) <= 1 for any event.
What is the probability of the complementary event?
For any event E, its complement 'not E' satisfies P(E) + P(not E) = 1. So P(not E) = 1 - P(E), which is often the quickest way to solve 'at least' or 'not' problems.
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