GANITA MANJARI · GRADE 9 · PART I · CHAPTER 7

Chapter Summary

Think first. Use AI purposefully. Verify always.

Concept Map

Key Ideas

A random experiment has identifiable possible outcomes, but its next result is uncertain under the model. A probability describes likelihood, not a promise about a single trial.
Probability lies from 0 to 1. In these finite models, 0 is impossible and 1 is certain. At 1/2 an event and its complement have equal chances. The complement contains all outcomes outside the event.
Experimental probability is the observed event count divided by the positive number of trials. It reports this record; different records can give different proportions.
For a finite model with equally likely elementary outcomes, P(E) = n(E)/n(S). The elementary outcomes must retain distinctions that make selection equally likely, such as separately labelled cards.
The theoretical probability of heads in independent fair tosses is 1/2. Observed proportions fluctuate. In long runs they tend to be closer to the model probability; no fixed trial count guarantees exact equality or steadily improving closeness.
A sample provides evidence about a population, not certainty. How people are selected matters. A larger sample drawn through a biased method can still be unrepresentative.
S contains all possible recorded outcomes, without duplicates. n(S) counts its elements. An event E is a subset of S. Choose enough detail for the question; grouped outcomes need not be equally likely.
A tree lists complete paths through sequential trials. Multiply conditional branch probabilities along a path. Add probabilities of disjoint paths for an event. Multi-step experiments can be independent or dependent.
Past independent outcomes do not change the next trial’s model probability. Removing an object can change the next draw’s probabilities. Fairness, replacement and uniformity are assumptions that must be stated—not guessed.

Formula Bank

0 \leq P (E) \leq 1

P_{exp} (E) = \frac{event count}{trial count}

Use a positive trial count. This is an observed ratio.

P (E) = \frac{n (E)}{n (S)}

The counting formula requires equally likely elementary outcomes. n(S) counts possible outcomes, not experimental trials or survey respondents.

P (E) + P (not E) = 1

On a tree multiply conditional branch probabilities along a path; add disjoint paths. For uniform landing, probability is the favourable-area fraction.

Visual Recall

Build a complete sample space

Recorded experiment

HHHTTHTT

n(S) = 4. Every possible recorded outcome appears once.

Uniform probabilities require fair independent coin tosses and, for cards, uniform independent selection.

All eleven visual lessons →

Important Connections

Fractions express both observed proportions and model probabilities, but they answer different questions. Grouping elementary outcomes changes their counts and may create unequal probabilities.

Replacement restores the original collection; removal changes it. A tree represents either situation when its branches match the actual conditions.

A 1/2 chance balances an event with its complement. It does not identify the chance of one particular alternative if several alternatives exist.

Common Mistakes

Two possible outcomes alone do not make them equally likely. A car starting and not starting need not have equal chances.
If a match can be drawn, P(win) = 1/2 means P(not win) = 1/2; it does not by itself mean P(lose) = 1/2.
An observed frequency is not a guarantee about the next trial or evidence that the exact model probability equals that frequency.
The colour outcomes {red, blue} are not equally likely when there are 3 red and 7 blue equally selectable marbles.
A longer run need not be closer at every step. A run of heads does not require an immediate balancing run of tails.
n(S), the number of possible outcomes, is not the number of people in a statistical survey. Neither a large sample nor an estimate guarantees the true population count.
Recording only the number of heads gives {0,1,2}, but these values are not equally likely for two independent fair tosses.
Without replacement, the second denominator decreases and the count of the colour removed decreases. Using the original fractions again is incorrect.
A geometric area ratio requires uniform landing over the whole region. Merely saying a point lands somewhere does not justify that assumption.
Gambler’s Fallacy: an opposite outcome is not due after a run of independent results.
Do not omit HT or TH, and do not count duplicate entries as extra outcomes.
An event can contain several outcomes; a certain event need not be a single elementary outcome.
Observed relative frequency need not equal theoretical probability after a few trials.

Mini Examples

CHANCE · One random toss of a fair coin; record H or T.

Given

One random toss of a fair coin; record H or T.

To Find

Possible outcomes and the probability of heads

Reasoning

Model: S = {H, T}; the two elementary outcomes are equally likely.

Solution

Favourable outcomes for heads: {H}.
Probability calculation: 1 favourable outcome out of 2 gives $\frac{1}{2}$ .

Result

P(H) = $\frac{1}{2}$ ; the next outcome is still uncertain.

SCALE · Six equally selectable cards; three purple and three green.

Given

Six equally selectable cards; three purple and three green.

To Find

Probability of drawing purple and probability of not drawing purple

Reasoning

The event and its complement partition all six cards.

Solution

P(purple) = $\frac{3}{6}$ = $\frac{1}{2}$ .
P(not purple) = 1 − $\frac{1}{2}$ = $\frac{1}{2}$ .

Result

An even chance of purple versus not purple.

OBSERVE · A 4 is recorded in 8 of 50 die rolls (Example 2).

Given

A 4 is recorded in 8 of 50 die rolls (Example 2).

To Find

Experimental probability of rolling a 4

Reasoning

Observed relative frequency = event count ÷ trial count.

Solution

Event count is 8; total trials are 50.
$\frac{8}{50}$ = $\frac{4}{25}$ = 0.16 = 16%.

Result

Experimental probability is 0.16 for this record.

EXPECT · Eleven equally selectable cards carry the letters of PROBABILITY (Example 4).

Given

Eleven equally selectable cards carry the letters of PROBABILITY (Example 4).

To Find

Probability of selecting a B card

Reasoning

Sample-space model: the eleven card positions, including both separate B cards.

Solution

The B cards are positions 4 and 6; n(E) = 2.
There are 11 cards; n(S) = 11.
P(B) = $\frac{2}{11}$ ≈ 0.182.

Result

P(B) = $\frac{2}{11}$ , approximately 18.2%.

COMPARE · In 12 fair-die rolls a 3 appears 3 times (Exercise 7.2 Q6).

Given

In 12 fair-die rolls a 3 appears 3 times (Exercise 7.2 Q6).

To Find

Experimental and theoretical probabilities of a 3

Reasoning

Experimental uses observations; theoretical uses the fair-die model.

Solution

Observed: $\frac{3}{12}$ = $\frac{1}{4}$ .
Theoretical: $\frac{1}{6}$ .
The difference is consistent with short-run variability; it does not force the next outcome.

Result

$\frac{1}{4}$ observed versus $\frac{1}{6}$ theoretical; neither implies a guaranteed next roll.

SAMPLE · 20 of 50 surveyed students prefer mangoes; school population 1500 (Example 5).

Given

20 of 50 surveyed students prefer mangoes; school population 1500 (Example 5).

To Find

Estimated number of students in the school who prefer mangoes

Reasoning

Use the sample proportion only as an estimate, with representativeness as an assumption.

Solution

Sample share: $\frac{20}{50}$ = 0.4.
Estimated count: 0.4 × 1500 = 600.
Sampling across grades would provide better coverage than using one class alone.

Result

About 600 students, as an estimate—not an exact known count.

SPACE · Two independent fair coin tosses; record the ordered results.

Given

Two independent fair coin tosses; record the ordered results.

To Find

Probability of at least one head

Reasoning

S = {HH, HT, TH, TT}; each ordered outcome has probability $\frac{1}{4}$ .

Solution

Favourable event E = {HH, HT, TH}.
n(E) = 3; n(S) = 4.
P(E) = $\frac{3}{4}$ .

Result

At least one head has probability $\frac{3}{4}$ .

BRANCH · 4 red and 5 blue balls; two random draws without replacement (starred Q10).

Given

4 red and 5 blue balls; two random draws without replacement (starred Q10).

To Find

Probability of red followed by blue

Reasoning

First draw: R has probability $\frac{4}{9}$ . After R: 3 red and 5 blue remain.

Solution

Favourable colour path: RB.
Probability calculation: ( $\frac{4}{9}$ ) × ( $\frac{5}{8}$ ) = $\frac{20}{72}$ = $\frac{5}{18}$ .

Result

P(RB) = $\frac{5}{18}$ .

REASON · A fair die has shown 6 on three consecutive independent rolls (Example 6).

Given

A fair die has shown 6 on three consecutive independent rolls (Example 6).

To Find

Probability that the next roll is 6

Reasoning

Independent rolls retain the same six equally likely outcomes.

Solution

The next sample space remains {1,2,3,4,5,6}.
One of six outcomes is a 6.
P(next 6) = $\frac{1}{6}$ .

Result

The probability remains $\frac{1}{6}$ ; no opposite result is due.

5-Question Quick Check

Progress: $\frac{0}{5}$ checked correctly

Q1. If P(E) = 0.3, find P(not E).Your answer (number or fraction such as 1/6)

Attempts: 0

Q2. In 50 rolls, a die shows 4 eight times. Enter the observed relative frequency of 4.Your answer (number or fraction such as 1/6)

Attempts: 0

Q3. A fair six-sided die is rolled. Enter P(result greater than 4) as a fraction.Your answer (number or fraction such as 1/6)

Attempts: 0

Q4. From the same bag, draw twice without replacement. Enter P(red then blue).Your answer (number or fraction such as 1/6)

Attempts: 0

Q5. After six independent fair tosses all show H, enter P(T on the next toss).Your answer (number or fraction such as 1/6)

Attempts: 0

Concept Map

Key Ideas

Formula Bank

Visual Recall

Build a complete sample space

Important Connections

Common Mistakes

Mini Examples

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5-Question Quick Check

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