This MCQ module is based on: Foundations of Probability
TOPIC 20 OF 25
Foundations of Probability
🎓 Class 9
Mathematics
CBSE
Theory
Ch 7 — The Mathematics of Maybe: Introduction to Probability
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Foundations of Probability
Targeting Class 9 level in Probability, with Intermediate difficulty.
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7.1 The Idea of Chance — Why Probability?
Will it rain tomorrow? Will India win the next cricket match? Will the bus arrive on time? Every day we make decisions based on uncertainty. Mathematics gives us a tool to measure uncertainty — that tool is probability?.
Long before formal mathematics, people relied on words like likely, unlikely, possible, probably, certainly, never. These words are useful, but they are vague. Two people may disagree on what "likely" means. Probability replaces vague words with a single number between 0 and 1.
Definition — Probability
Probability is a numerical measure of how likely an event is to occur. It is always a number between 0 and 1 (inclusive).
- 0 means the event is impossible (will never happen).
- 1 means the event is certain (will always happen).
- Values in between mean the event is more or less likely.
Fig. 7.1 — The probability scale: from 0 (impossible) to 1 (certain).
Words and Numbers — Linking Everyday Language to Numbers
Consider the following statements. We can place each one on the probability scale:
| Event | Word | Approx. Probability |
|---|---|---|
| The Sun will rise tomorrow | Certain | 1 |
| Tossing a coin and getting Heads | Equally likely | 1/2 |
| Rolling a 6 on a fair die | Unlikely | 1/6 |
| Picking a red ball from a bag of all blue balls | Impossible | 0 |
| It rains in Cherrapunji in July | Very likely | ~0.95 |
7.2 Random Experiments, Outcomes and Events
To talk about probability precisely, we need three building blocks: a random experiment?, its outcomes?, and the events? we are interested in.
Three Key Definitions
1. Random experiment: An action whose result cannot be predicted with certainty (e.g., tossing a coin, rolling a die, drawing a card).2. Outcome: A single possible result of one trial of the experiment.
3. Event: A collection of one or more outcomes that satisfy a chosen description.
Example 1 — Tossing a coin
The experiment is "toss a fair coin once." The two possible outcomes are Head (H) and Tail (T). The event "getting a Head" consists of just the outcome H.
Fig. 7.2 — A coin toss has exactly two outcomes: Head or Tail.
Example 2 — Rolling a die
The experiment is "roll a fair six-sided die once." The possible outcomes are 1, 2, 3, 4, 5, 6. The event "rolling an even number" is the collection {2, 4, 6}.
Fig. 7.3 — A die has six outcomes. Even outcomes (2, 4, 6) are highlighted in blue; odd outcomes (1, 3, 5) in yellow.
Sample Space
Definition — Sample Space
The sample space, written as S, is the set of all possible outcomes of a random experiment.
- Coin toss: S = {H, T}, so |S| = 2
- Die roll: S = {1, 2, 3, 4, 5, 6}, so |S| = 6
- Drawing one card from a standard deck: |S| = 52
Example 3 — Tossing two coins
If we toss two coins (or one coin twice), the outcomes are pairs. The sample space is S = {HH, HT, TH, TT}, so |S| = 4.
Fig. 7.4 — Probability tree for tossing two coins. Each path from left to right gives one outcome of the sample space.
7.3 Three Special Kinds of Events
Certain, Impossible and Equally Likely
Certain Event
An event that always happens. Probability = 1.
Example: rolling a number less than 7 on a die.
Example: rolling a number less than 7 on a die.
Impossible Event
An event that never happens. Probability = 0.
Example: rolling a 7 on an ordinary die.
Example: rolling a 7 on an ordinary die.
Equally Likely
Two or more outcomes that have the same chance of happening.
Example: H and T on a fair coin.
Example: H and T on a fair coin.
Important — fairness assumption
When we say a coin or die is fair (or unbiased), we are saying that every outcome of the sample space is equally likely. Without fairness, the simple counting formula in Section 7.4 cannot be used.
Likely and Unlikely Events
An event with probability greater than 1/2 is likely. An event with probability less than 1/2 is unlikely. An event with probability exactly 1/2 is on the boundary — neither likely nor unlikely.
Try this: Classify each event as certain, impossible, equally likely, likely or unlikely: (a) drawing a king from a deck of 52 cards, (b) tossing a coin and getting H or T, (c) the next baby born will be either a boy or a girl, (d) it will snow in Chennai in May, (e) drawing a vowel from the word "PROBABILITY".
Suggested classification
(a) Unlikely (4/52 ≈ 0.077). (b) Certain (one of H or T must happen). (c) Certain (assuming only two recorded sexes). (d) Impossible (or near impossible). (e) The word "PROBABILITY" has letters P, R, O, B, A, B, I, L, I, T, Y. Vowels = O, A, I, I → 4 out of 11 letters → unlikely (≈ 0.36).
7.4 Classical (Theoretical) Probability
When all the outcomes in the sample space are equally likely, we can compute probability simply by counting.
Classical Probability Formula
For an event E in a sample space where every outcome is equally likely:
\[ P(E) \;=\; \dfrac{\text{Number of outcomes favourable to } E}{\text{Total number of outcomes in the sample space}} \;=\; \dfrac{n(E)}{n(S)} \]
Example 4 — Rolling a die
A fair die is rolled once. Find the probability that the number rolled is (i) 4, (ii) an even number, (iii) less than 5, (iv) a 7.
Solution
Sample space S = {1, 2, 3, 4, 5, 6}, so n(S) = 6.
(i) E = {4}, n(E) = 1. \(P(E) = \dfrac{1}{6}\).
(ii) E = {2, 4, 6}, n(E) = 3. \(P(E) = \dfrac{3}{6} = \dfrac{1}{2}\).
(iii) E = {1, 2, 3, 4}, n(E) = 4. \(P(E) = \dfrac{4}{6} = \dfrac{2}{3}\).
(iv) E = { } (empty). \(P(E) = \dfrac{0}{6} = 0\) — impossible event.
(i) E = {4}, n(E) = 1. \(P(E) = \dfrac{1}{6}\).
(ii) E = {2, 4, 6}, n(E) = 3. \(P(E) = \dfrac{3}{6} = \dfrac{1}{2}\).
(iii) E = {1, 2, 3, 4}, n(E) = 4. \(P(E) = \dfrac{4}{6} = \dfrac{2}{3}\).
(iv) E = { } (empty). \(P(E) = \dfrac{0}{6} = 0\) — impossible event.
Example 5 — Drawing a coloured ball
A bag contains 5 red, 3 green and 2 blue balls. One ball is drawn at random. Find the probability that it is (i) red, (ii) not red, (iii) green or blue.
Solution
Total balls = 5 + 3 + 2 = 10, so n(S) = 10.
(i) Favourable (red) = 5. \(P(\text{red}) = \dfrac{5}{10} = \dfrac{1}{2}\).
(ii) Not red = 3 + 2 = 5. \(P(\text{not red}) = \dfrac{5}{10} = \dfrac{1}{2}\).
(iii) Green or blue = 3 + 2 = 5. \(P = \dfrac{1}{2}\).
Note: \(P(\text{red}) + P(\text{not red}) = 1\) — this is a useful property of complementary events.
(i) Favourable (red) = 5. \(P(\text{red}) = \dfrac{5}{10} = \dfrac{1}{2}\).
(ii) Not red = 3 + 2 = 5. \(P(\text{not red}) = \dfrac{5}{10} = \dfrac{1}{2}\).
(iii) Green or blue = 3 + 2 = 5. \(P = \dfrac{1}{2}\).
Note: \(P(\text{red}) + P(\text{not red}) = 1\) — this is a useful property of complementary events.
Example 6 — Drawing a card
One card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability that it is (i) an ace, (ii) a heart, (iii) the queen of spades, (iv) a face card (J, Q, K).
Solution
n(S) = 52.
(i) Aces = 4. \(P = \dfrac{4}{52} = \dfrac{1}{13}\).
(ii) Hearts = 13. \(P = \dfrac{13}{52} = \dfrac{1}{4}\).
(iii) Queen of spades = 1. \(P = \dfrac{1}{52}\).
(iv) Face cards = 3 (J, Q, K) × 4 suits = 12. \(P = \dfrac{12}{52} = \dfrac{3}{13}\).
(i) Aces = 4. \(P = \dfrac{4}{52} = \dfrac{1}{13}\).
(ii) Hearts = 13. \(P = \dfrac{13}{52} = \dfrac{1}{4}\).
(iii) Queen of spades = 1. \(P = \dfrac{1}{52}\).
(iv) Face cards = 3 (J, Q, K) × 4 suits = 12. \(P = \dfrac{12}{52} = \dfrac{3}{13}\).
7.5 Two Important Properties
Property 1 — Probability is bounded
For any event E: \[ 0 \le P(E) \le 1 \]
Probability cannot be negative and cannot exceed 1.
Property 2 — Complementary events
If E is an event, the complement "not E" (written \(\bar E\)) is the event that E does not happen.
\[ P(E) + P(\bar E) = 1 \quad\Longrightarrow\quad P(\bar E) = 1 - P(E) \]
Example 7 — Using the complement
The probability that it will rain tomorrow is 0.7. What is the probability that it will not rain?
\(P(\text{no rain}) = 1 - 0.7 = 0.3\).
Activity 7.1 — Building a Probability Table
L3 Apply
Materials: One coin, one six-sided die, an opaque bag with 4 red and 6 white marbles (or paper chits), notebook.
Predict before you do: Without computing, write your guess for each probability below. Will your guesses match the formulas?
- For the coin, list the sample space and write \(P(H)\) and \(P(T)\).
- For the die, list the sample space and write \(P(\text{prime})\), where primes from 1–6 are 2, 3, 5.
- For the marble bag, list the colours, count favourable cases and write \(P(\text{red})\) and \(P(\text{white})\).
- For each, also compute \(P(\bar E)\) and verify that \(P(E) + P(\bar E) = 1\).
- Compare with your predictions. Where were you closest? Where were you furthest off?
Coin: S = {H, T}, P(H) = P(T) = 1/2. P(H) + P(T) = 1. ✓
Die — primes: Primes ≤ 6 are {2, 3, 5}, so n(E) = 3, n(S) = 6, P = 3/6 = 1/2. P(not prime) = 1/2.
Marbles: n(S) = 10. P(red) = 4/10 = 2/5. P(white) = 6/10 = 3/5. Sum = 1. ✓
Competency-Based Questions
Scenario: The student council of Vidyalaya is forming a 1-member quiz team by drawing a name slip at random from a box. The box contains 9 girls' name slips and 6 boys' name slips. Two of the girls — Anika and Diya — and one of the boys — Karan — are house captains.
Q1. What is the probability that the chosen student is a girl?
L3 ApplyAnswer: (a) 3/5. Total slips = 15, girls = 9. \(P = 9/15 = 3/5\).
Q2. The school principal claims, "There is more than a 50% chance that a house captain will be picked." Analyse whether this claim is correct.
L4 AnalyseAnswer: House captains = Anika, Diya, Karan = 3 slips. \(P(\text{captain}) = 3/15 = 1/5 = 20\%\). Since 20% < 50%, the principal's claim is not correct.
Q3. After the first draw, the chosen slip is set aside without replacement. A second slip is drawn for an alternate. Evaluate: does the probability of drawing a girl on the second draw still equal 3/5? Justify.
L5 EvaluateAnswer: No. After one draw, the box has 14 slips left, and the composition depends on what was drawn. If a girl was drawn first, P(girl on draw 2) = 8/14 = 4/7. If a boy was drawn first, P(girl on draw 2) = 9/14. The probability is conditional on the first draw, so the simple 3/5 no longer applies.
Q4. Design a fairer selection method (still using random drawing) so that boys and girls have equal individual chance of being picked, even though the box currently has more girl slips. Describe one method clearly.
L6 CreateOne possible design: First toss a fair coin: if Head → draw from a box containing only girls' slips, if Tail → draw from a box containing only boys' slips. Then \(P(\text{a particular girl}) = \tfrac{1}{2} \times \tfrac{1}{9} = \tfrac{1}{18}\) and \(P(\text{a particular boy}) = \tfrac{1}{2} \times \tfrac{1}{6} = \tfrac{1}{12}\) — still unequal! A better design: stratified sampling — pick one girl and one boy each by independent draws and then toss a coin to choose which one gets the slot. This gives every individual the same probability of \(1/(2\times 9)+\ldots\). Many valid designs exist; the key idea is to separate the two groups.
Assertion–Reason Questions
Assertion (A): The probability of an impossible event is 0.
Reason (R): The number of favourable outcomes for an impossible event is zero, while n(S) is a positive number.
Reason (R): The number of favourable outcomes for an impossible event is zero, while n(S) is a positive number.
Answer: (a) — Using the formula \(P = n(E)/n(S) = 0/n(S) = 0\). R correctly explains A.
Assertion (A): The probability of an event can be 1.5.
Reason (R): Probability is always a real number.
Reason (R): Probability is always a real number.
Answer: (d) — A is false: probability lies in [0, 1], so 1.5 is impossible. R is true (probability is a real number, but it is restricted to [0, 1]).
Assertion (A): If \(P(E) = 0.6\), then \(P(\bar E) = 0.4\).
Reason (R): \(E\) and \(\bar E\) together cover the entire sample space, and they do not overlap.
Reason (R): \(E\) and \(\bar E\) together cover the entire sample space, and they do not overlap.
Answer: (a) — Both statements are true and R is the precise reason \(P(E) + P(\bar E) = 1\), so \(P(\bar E) = 1 - 0.6 = 0.4\).
Frequently Asked Questions
What is an experiment, outcome and event in Class 9 probability?
An experiment is any procedure with a well-defined collection of possible results, such as tossing a coin or rolling a die. An outcome is one possible result of the experiment. An event is any collection (subset) of outcomes — for instance 'getting an even number' on a die.
What is a sample space in Class 9 Maths Chapter 7?
The sample space, denoted S, is the set of all possible outcomes of an experiment. For tossing a coin, S = {Head, Tail}. For rolling a die, S = {1, 2, 3, 4, 5, 6}. Every event is a subset of the sample space.
What is the formula for theoretical probability in Class 9?
P(E) = (number of outcomes favourable to E) / (total number of equally likely outcomes). For example, the probability of getting an even number when rolling a fair die is 3/6 = 1/2 because three outcomes (2, 4, 6) are favourable out of six equally likely outcomes.
Why must P(E) always lie between 0 and 1 in Class 9?
P(E) is a fraction whose numerator (favourable outcomes) cannot exceed its denominator (total outcomes), and neither can be negative. So 0 <= P(E) <= 1. P(E) = 0 means E is impossible; P(E) = 1 means E is certain.
What is the probability of getting a head when tossing a fair coin?
There is one favourable outcome (Head) out of two equally likely outcomes ({Head, Tail}). So P(Head) = 1/2 = 0.5.
Who developed the early theory of probability?
The mathematical theory of probability was launched in the 1650s by the French mathematicians Blaise Pascal and Pierre de Fermat in their letters about gambling problems, and later formalised by Pierre-Simon Laplace in 1812 in his Theorie Analytique des Probabilites.
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