Class 12 Mathematics PROBABILITY Notes

Probability Notes | Nepal eNotes

Unit 14

Statistics and Probability

Class 12 Mathematics

Probability

Class 12 Mathematics – Probability Notes PDF

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NEB / CDC Focus

The Grade 12 focus of this unit is solving probability problems using combinations and solving problems involving conditional probability. The basic event notation, multiplication rule, and independence tests below are included because they are required to solve those problems correctly.

1. Probability Basics

Random Experiment

An experiment whose exact outcome cannot be predicted with certainty in advance is called a random experiment.

Sample Space

The set of all possible outcomes is the sample space S. An event A is a subset of S.

For equally likely finite outcomes,

P(A)=n(A)/n(S).
0≤P(A)≤1,   P(S)=1,   P(A′)=1−P(A).

1.1 Addition Rule

P(A∪B)=P(A)+P(B)−P(A∩B).

If A and B are mutually exclusive, then P(A∩B)=0, so

P(A∪B)=P(A)+P(B).
Events Inside a Sample Space S ABA∩B
Figure 1: Union includes both events; intersection is their overlap.

2. Probability Using Combinations

When selection order does not matter, use combinations to count favorable and total selections.

ⁿCᵣ = n!/[r!(n−r)!].

If all groups of size r are equally likely,

P(A)=number of favorable combinations / total combinations.
Worked Example

A committee of 3 is selected from 5 boys and 4 girls. Find the probability that exactly 2 girls are selected.

Total committees: ⁹C₃=84.

Favorable: ⁴C₂·⁵C₁=6·5=30.

P=30/84=5/14.

Combination-Based Probability Count all selectionsn(S) Count favorablen(A) Take ratioP(A)=n(A)/n(S)
Figure 2: Use nCr when the order of selected objects is irrelevant.
Combination or Permutation?

A committee, hand of cards, chosen group, or sample usually ignores order, so combinations are appropriate. If positions or order matter, a permutation model may be needed.

3. Conditional Probability

Conditional Probability

The probability of event A given that B has already occurred is denoted P(A|B).

P(A|B)=P(A∩B)/P(B),   P(B)>0.

Likewise,

P(B|A)=P(A∩B)/P(A),   P(A)>0.
Conditional Probability Restricts the Sample Space Original space SGiven BA∩B Once B is known, B becomes the effective sample space.
Figure 3: Conditioning on B means probabilities are recalculated within B.
Worked Example

A card is drawn from a standard deck. Given that it is a face card, find the probability that it is a king.

There are 12 face cards and 4 kings, all kings being face cards.

P(King | Face)=4/12=1/3.

4. Multiplication Rule of Probability

Rearranging the conditional-probability formula gives

P(A∩B)=P(B)P(A|B)=P(A)P(B|A).

This formula is particularly important in sequential experiments.

Probability Tree for Sequential Events A: P(A)A′ B: P(B|A)B′ Multiply probabilities along a path: P(A∩B)=P(A)P(B|A).
Figure 4: Sequential probabilities are naturally represented by tree branches.
Worked Example

A bag has 3 red and 2 blue balls. Two balls are drawn without replacement. Find the probability both are red.

P(R₁)=3/5,   P(R₂|R₁)=2/4.

P(R₁∩R₂)=(3/5)(2/4)=3/10.

5. Independent and Dependent Events

Independent Events

Events A and B are independent when occurrence of one does not change the probability of the other.

P(A|B)=P(A)P(B|A)=P(B)P(A∩B)=P(A)P(B).
Dependent Events

Events are dependent when occurrence of one changes the probability of the other. Drawing without replacement is a common example.

ConceptCan both occur?Main probability condition
IndependentYes, usuallyP(A∩B)=P(A)P(B)
Mutually exclusiveNoP(A∩B)=0
Do Not Confuse These

Mutually exclusive events are not generally independent. If two non-zero-probability events cannot occur together, knowing one occurred makes the other impossible.

6. Without-Replacement Problems

Without replacement, the total number of objects and category counts change after each draw. Therefore later probabilities are conditional on earlier outcomes.

Example

From 5 red and 3 white balls, two are drawn without replacement. Find the probability of one red followed by one white.

P(R then W)=(5/8)(3/7)=15/56.

If the question asks for one red and one white in any order, include both orders:

P(RW or WR)=P(RW)+P(WR).
With vs Without Replacement With replacementcomposition restoredsuccessive draws may be independent Without replacementcomposition changessuccessive draws are dependent
Figure 5: Replacement determines whether the sample composition changes.

7. Worked Examples

Example 1: Combination Probability

Five cards are chosen from 8 red and 4 black cards. Find the probability of exactly 3 red cards.

P=[⁸C₃·⁴C₂]/¹²C₅=(56·6)/792=14/33.

Example 2: Conditional Probability from Given Data

If P(A∩B)=0.18 and P(B)=0.30, find P(A|B).

P(A|B)=0.18/0.30=0.6.

Example 3: Find Intersection

If P(A)=0.5 and P(B|A)=0.4, then

P(A∩B)=0.5×0.4=0.2.

Example 4: Independent Events

If P(A)=0.6, P(B)=0.5, and A,B are independent:

P(A∩B)=0.3.

Example 5: At Least One

If two fair coins are tossed, probability of at least one head is

1−P(no head)=1−1/4=3/4.

Probability Method Checklist Selection?Use nCr Given event?Conditional Sequential?Multiply path At least one?Complement
Figure 6: Recognize the structure before choosing a probability formula.

8. Problem-Solving Strategy

  1. Define the event clearly.
  2. Check whether outcomes are equally likely.
  3. For unordered selections, count using combinations.
  4. If the problem says “given that,” identify the conditioning event.
  5. For sequential events, use the multiplication rule and update counts after each draw.
  6. Check whether replacement makes events independent or dependent.
  7. For “at least one,” consider using the complement.
  8. Ensure the final probability lies between 0 and 1.

9. Common Mistakes

Wrong denominator

In P(A|B), B becomes the effective sample space.

Ignoring order condition

Use combinations only when order truly does not matter.

Replacement overlooked

Without replacement, probabilities change after each draw.

Independent vs exclusive

These are different ideas and use different conditions.

Missing second order

For “one of each” in sequence, include all allowed orders unless the order is specified.

No complement check

“At least one” is often easiest as 1−P(none).

10. Formula Sheet

P(A)=n(A)/n(S)P(A′)=1−P(A) P(A∪B)=P(A)+P(B)−P(A∩B) ⁿCᵣ=n!/[r!(n−r)!] P(A|B)=P(A∩B)/P(B) P(A∩B)=P(A)P(B|A)=P(B)P(A|B) Independent: P(A∩B)=P(A)P(B)

11. Important Exam Questions

Short

  1. Define conditional probability.
  2. Distinguish independent and mutually exclusive events.
  3. State the multiplication rule.
  4. State the combination formula.

Numerical

  1. Solve probability of committee/card selection using combinations.
  2. Find conditional probability from event probabilities.
  3. Solve without-replacement ball/card problems.
  4. Test whether two events are independent.
  5. Solve “at least one” problems using complements.

12. One-Minute Revision

  • Probability always lies between 0 and 1.
  • Use nCr for unordered selections.
  • P(A|B)=P(A∩B)/P(B).
  • P(A∩B)=P(A)P(B|A).
  • Independent events satisfy P(A∩B)=P(A)P(B).
  • Without replacement usually creates dependence.
  • Mutually exclusive does not mean independent.
  • For “at least one,” try the complement.

13. Diagram Practice

  • Venn diagram of union/intersection.
  • Combination-probability flow.
  • Conditional sample-space diagram.
  • Sequential probability tree.
  • With/without-replacement comparison.
  • Probability method checklist.

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