GANITA MANJARI · GRADE 9 · PART I · CHAPTER 7

REASON — Fairness, dependence and careful conclusions

Think first. Use AI purposefully. Verify always.

DISCOVER

Begin with your own prediction.

After six independent fair tosses show heads, are tails now due? How is that different from removing balls from a bag?

Write or sketch a first attempt. Use the model to test your assumptions before accepting a rule.

Understand

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.

Key Ideas

P (T) = \frac{1}{2}

A geometric area ratio requires uniform landing over the whole region. Merely saying a point lands somewhere does not justify that assumption.

Visualise

Is the opposite result due?

Suppose the past six results were H H H H H H. For independent fair tosses, predict the next tail probability.

Demonstration seed

Fresh tosses: 0. Latest simulated result: not started. These fresh trials are not forced to offset the hypothetical history.

Change the replacement rule

Replace the first ball before the second draw

Four equally selectable numbered balls · without replacement. Read each route from Start to its final outcome.

Start

First: 1 · $\frac{1}{4}$

↳ Second: 2 · $\frac{1}{3}$
Outcome: 1, 2
↳ Second: 3 · $\frac{1}{3}$
Outcome: 1, 3
↳ Second: 4 · $\frac{1}{3}$
Outcome: 1, 4

First: 2 · $\frac{1}{4}$

↳ Second: 1 · $\frac{1}{3}$
Outcome: 2, 1
↳ Second: 3 · $\frac{1}{3}$
Outcome: 2, 3
↳ Second: 4 · $\frac{1}{3}$
Outcome: 2, 4

First: 3 · $\frac{1}{4}$

↳ Second: 1 · $\frac{1}{3}$
Outcome: 3, 1
↳ Second: 2 · $\frac{1}{3}$
Outcome: 3, 2
↳ Second: 4 · $\frac{1}{3}$
Outcome: 3, 4

First: 4 · $\frac{1}{4}$

↳ Second: 1 · $\frac{1}{3}$
Outcome: 4, 1
↳ Second: 2 · $\frac{1}{3}$
Outcome: 4, 2
↳ Second: 3 · $\frac{1}{3}$
Outcome: 4, 3

(1, 2)

(1, 3)

(1, 4)

(2, 1)

(2, 3)

(2, 4)

(3, 1)

(3, 2)

(3, 4)

(4, 1)

(4, 2)

(4, 3)

n(S) = 12. 4 × 3 = 12; the four repeated-label pairs are excluded.

Each pair has probability $\frac{1}{12}$ under uniform selection at each draw.

Uniform landing and area

Rectangle: 3 m × 2 m. Dashed diameter: 1 m; radius: 0.5 m. The circle is wholly inside the rectangle.

Assume uniform landing over the rectangle. Each local simulated point uses uniform horizontal and vertical coordinates. Boundary conventions do not change the continuous-area probability.

Demonstration seed

P = \frac{π}{24} \approx 0.131

Area of circle ÷ area of rectangle: ( $\frac{π}{4}$ ) ÷ 6. Without uniform landing, this ratio is not justified.

Method

Identify the model and its assumptions before calculating. Check whether the state changes. For a uniformly distributed landing point, compare areas rather than counting a finite list of points.

Worked Example

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.

AIMS Method

A — Attempt independently
Predict and describe what is being recorded.
I — Investigate the idea
Identify the model, evidence and assumptions.
M — Make connections
Connect outcomes, events, fractions and tree paths.
S — Scrutinise your reasoning
Check completeness, likelihoods and changing conditions.

Think first. Use AI purposefully. Verify always.

Focused Practice

Explain why drawing two numbered balls without replacement cannot have outcomes $(1, 1)$ , $(2, 2)$ , $(3, 3)$ or $(4, 4)$ .

Your reasoning

Continue with MIXED MASTERY →

Check & Feedback

Explain the model, state an assumption and identify one conclusion the information does not justify.

Explain your conclusion

Textbook connection: Ganita Manjari · Grade 9 · Part I · Chapter 7 · printed page 164