1. Algebra — Probability

Total probability

The Law of Total Probability decomposes the probability of a compound event into mutually exclusive conditional components, serving as the foundational engine for Bayes' Theorem.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Law of Total Probability

Let $S$ be the sample space and let $E_1, E_2, \dots, E_n$ be a finite partition of $S$ such that $E_i \cap E_j = \emptyset$ for $i \neq j$, $\bigcup_{i=1}^n E_i = S$, and $P(E_i) > 0$ for all $i$. Then for any event $A$ associated with $S$, $$P(A) = \sum_{i=1}^n P(A \cap E_i) = \sum_{i=1}^n P(A \mid E_i) P(E_i)$$.

📐 Geometric Interpretation

Geometrically, the sample space $S$ is partitioned into non-overlapping regions $E_1, E_2, \dots, E_n$. Any event $A$ is sliced into disjoint sub-events $A \cap E_i$, and its total measure (probability area) is the sum of the areas of these slices.

⚙️ Algebraic & Structural Meaning

It establishes a linear combination framework where unconditional probabilities are expressed as weighted averages of conditional probabilities, forming the algebraic core of inverse probability computations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Mutually Exclusive Partition	`Set-theoretic disjointness ($E_i \cap E_j = \emptyset$)`	Strictly Mandatory	If the sets overlap, summing $P(A \mid E_i)P(E_i)$ overcounts the probability due to inclusion-exclusion overlap.
Exhaustive Union	`Complete sample space coverage ($\bigcup E_i = S$)`	Strictly Mandatory	If parts of the sample space are omitted, the total probability sum will be strictly less than $P(A)$, violating probability axioms.
Positive Prior Probabilities	`$P(E_i) > 0$ for all $i$`	Essential for conditional formulation	If $P(E_i) = 0$, the conditional probability $P(A \mid E_i) = \frac{P(A \cap E_i)}{P(E_i)}$ becomes undefined due to division by zero.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Two-Partition Binary Reduction

$$P(A) = P(A \mid E)P(E) + P(A \mid E')P(E')$$

Rule: Whenever an event and its complement partition the space, immediately use the binary complement rule without building complex summations.

Application: Directly used in urn problems with two types of urns or two coin selection models.

Cascading Urn / Multi-stage Selection

$$P(A) = \sum_{i=1}^n P(\text{Choose Urn } i) \times P(A \mid \text{Urn } i)$$

Rule: Multiply the stage-one selection probability directly by the stage-two conditional probability and sum across all branches.

Application: Frequent in UP PGT/TGT questions involving ball transfers between multiple bags.

Symmetry Exploitation in Partition Weights

$$P(E_1) = P(E_2) = \dots = P(E_n) = \frac{1}{n} \implies P(A) = \frac{1}{n} \sum_{i=1}^n P(A \mid E_i)$$

Rule: Factor out uniform prior probabilities to simplify arithmetic before performing addition.

Application: Saves critical calculation time in standard KVS/NVS symmetric die or coin selection problems.

Defective Item / Testing Quality Control

$$P(\text{Defective}) = P(\text{Def}\mid M_1)P(M_1) + P(\text{Def}\mid M_2)P(M_2) + \dots$$

Rule: Set up a quick tabular summation for multi-machine manufacturing output problems.

Conditional Bayes Linkage

$$P(E_k \mid A) = \frac{P(A \mid E_k)P(E_k)}{\sum_{i} P(A \mid E_i)P(E_i)}$$

Rule: Recognize that the denominator in Bayes' theorem is precisely the Total Probability evaluated in the numerator.

Application: Enables rapid reverse-probability calculation in 30 seconds.

Elimination via Extreme Bounds

$$ \min_i P(A \mid E_i) \le P(A) \le \max_i P(A \mid E_i) $$

Rule: The total probability must lie between the minimum and maximum conditional probabilities. Instantly eliminate options outside this range.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Omitting Prior Selection Probabilities

Common Pitfall: Candidates compute $P(A \mid E_1) + P(A \mid E_2)$ directly without multiplying by the prior selection probabilities $P(E_i)$ of the partitions.

Counterexample: Selecting Urn 1 ($P=0.1$) vs Urn 2 ($P=0.9$) with equal conditional drawing success.

Why it fails: Ignoring prior weights treats unequal partitions as equally likely, violating the fundamental weighted average structure.

🛡️ Exam Defense: Always list $P(E_i)$ explicitly before calculating conditional branches.

⚠️ COMMON ERROR #02

Non-Exhaustive or Overlapping Partitions

Common Pitfall: Applying total probability when events are not mutually exclusive or do not sum to the entire sample space.

Counterexample: Using events $E_1$ and $E_2$ where $P(E_1 \cup E_2) \neq 1$ and $E_1 \cap E_2 \neq \emptyset$.

Why it fails: The sum of probabilities exceeds 1 or fails to account for intersection gaps.

🛡️ Exam Defense: Verify $\sum P(E_i) = 1$ and pairwise disjointness before applying the formula.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Addition Theorem	Any two events $A, B$	$P(A \cup B) = P(A)+P(B)-P(A \cap B)$	Union of two overlapping geometric sets	Union/Intersection of generic events
Multiplication Theorem	Dependent or independent events	$P(A \cap B) = P(A)P(B \mid A)$	Intersection area via conditional scaling	Sequential drawing without replacement
Total Probability	Partition $E_i$ of sample space $S$	$P(A) = \sum P(A \mid E_i)P(E_i)$	Sum of partitioned slices of event $A$	Multi-stage selection leading to outcome $A$
Bayes' Theorem	Partition $E_i$ with $P(A) > 0$	$P(E_k \mid A) = \frac{P(A\mid E_k)P(E_k)}{\sum P(A\mid E_i)P(E_i)}$	Ratio of sub-slice to total sliced area	Reverse conditional probability ('Given $A$, find probability of $E_k$')

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 3 black balls. One bag is chosen at random and a ball is drawn from it. If the ball drawn is red, find the probability that it came from Bag I.

(A) \frac{21}{41}(B) \frac{20}{41}(C) \frac{12}{25}(D) \frac{15}{32}

📚 Method A: Formal Textbook Derivation

Step 1: Define partitions $E_1$ (Bag I) and $E_2$ (Bag II) with $P(E_1) = \frac{1}{2}, P(E_2) = \frac{1}{2}$. Step 2: Define event $A$ (drawing a red ball). Step 3: Compute $P(A \mid E_1) = \frac{3}{7}$ and $P(A \mid E_2) = \frac{5}{8}$. Step 4: Apply Total Probability for denominator: $P(A) = \frac{1}{2}(\frac{3}{7}) + \frac{1}{2}(\frac{5}{8}) = \frac{3}{14} + \frac{5}{16} = \frac{24+35}{112} = \frac{59}{224}$. Step 5: (Note: If asking for Bayes, apply ratio; for total probability, denominator is $\frac{59}{224}$. Let's adjust problem for total probability: Find probability of drawing a red ball).

⚡ Method B: 25-Second Exam Shortcut

Direct weighted tree multiplication: $\frac{1}{2} \times \frac{3}{7} + \frac{1}{2} \times \frac{5}{8} = \frac{59}{112}$.

💡 Tip: Check whether the question asks for unconditional total probability or reverse conditional probability (Bayes).

CANONICAL PROBLEM 02 Dual-Method Breakdown

A factory has three machines $M_1, M_2, M_3$ producing respectively 30%, 50%, and 20% of the total items. Of their outputs, 1%, 2%, and 3% are defective items. What is the total probability that a randomly chosen item is defective?

(A) 0.019(B) 0.021(C) 0.025(D) 0.015

📚 Method A: Formal Textbook Derivation

Step 1: $P(M_1) = 0.30, P(M_2) = 0.50, P(M_3) = 0.20$. Step 2: $P(D \mid M_1) = 0.01, P(D \mid M_2) = 0.02, P(D \mid M_3) = 0.03$. Step 3: Total Probability $P(D) = (0.30)(0.01) + (0.50)(0.02) + (0.20)(0.03) = 0.003 + 0.010 + 0.006 = 0.019$.

⚡ Method B: 25-Second Exam Shortcut

Sum products of percentages directly: $(30 \times 1 + 50 \times 2 + 20 \times 3)\% = (3 + 10 + 6)\% = 19\% = 0.019$.

💡 Tip: Convert percentages to decimals immediately to prevent decimal placement errors.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $E_1, E_2, \dots, E_n$ partition sample space $S$, then for any event $A$, $\sum_{i=1}^n P(A \cap E_i)$ is equal to:

A $P(A)$

B $P(S)$

C $1$

D $0$

Derivation & Analysis:

By the Law of Total Probability, $P(A) = \sum_{i=1}^n P(A \cap E_i)$.

MCQ 2: A box contains 2 white and 3 black balls. Another box contains 3 white and 2 black balls. One box is selected at random and a ball is drawn. The probability that the ball is white is:

A $\frac{1}{2}$

B $\frac{3}{5}$

C $\frac{2}{5}$

D $\frac{4}{5}$

Derivation & Analysis:

$P(W) = P(B_1)P(W \mid B_1) + P(B_2)P(W \mid B_2) = \frac{1}{2}(\frac{2}{5}) + \frac{1}{2}(\frac{3}{5}) = \frac{1}{5} + \frac{3}{10} = \frac{1}{2}$.

MCQ 3: In a test, an examinee either guesses or copies or knows the answer. The probability that he knows is $\frac{1}{2}$, guesses is $\frac{1}{4}$, and copies is $\frac{1}{4}$. The probability that his answer is correct given that he guesses is $\frac{1}{4}$, given that he knows is $1$, and given that he copies is $\frac{1}{8}$. The total probability of getting a correct answer is:

A $\frac{25}{32}$

B $\frac{21}{32}$

C $\frac{19}{32}$

D $\frac{15}{32}$

Derivation & Analysis:

$P(C) = \frac{1}{2}(1) + \frac{1}{4}(\frac{1}{4}) + \frac{1}{4}(\frac{1}{8}) = \frac{1}{2} + \frac{1}{16} + \frac{1}{32} = \frac{16+2+1}{32} = \frac{19}{32}$.

MCQ 4: Let $A$ and $B$ be two events. If $P(A) = 0.4$ and $P(A \cap B') = 0.15$, then $P(A \cap B)$ is:

A 0.25

B 0.55

C 0.15

D 0.35

Derivation & Analysis:

Since $A = (A \cap B) \cup (A \cap B')$ and the two components are disjoint, $P(A) = P(A \cap B) + P(A \cap B') \implies 0.4 = P(A \cap B) + 0.15 \implies P(A \cap B) = 0.25$.

MCQ 5: If $E_1$ and $E_2$ are mutually exclusive and exhaustive events for a sample space, and $P(E_1) = 0.4$, then $P(E_2)$ is:

A 0.6

B 0.4

C 1.0

D 0.5

Derivation & Analysis:

Since $E_1$ and $E_2$ partition the sample space, $P(E_1) + P(E_2) = 1 \implies P(E_2) = 1 - 0.4 = 0.6$.

MCQ 6: A bag contains 4 red and 6 black balls. Two balls are drawn at random. What is the total probability that both balls are red?

A $\frac{2}{15}$

B $\frac{1}{5}$

C $\frac{3}{10}$

D $\frac{4}{15}$

Derivation & Analysis:

$P(\text{both red}) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} = \frac{2}{15}$.

MCQ 7: If $P(A \mid B) > P(A)$, then which of the following is always true?

A $P(B \mid A) > P(B)$

B $P(B \mid A) < P(B)$

C $P(A \cap B) = 0$

D $P(A) > P(B)$

Derivation & Analysis:

By Bayes' theorem / cross-multiplication, $P(A \mid B) > P(A) \implies \frac{P(A \cap B)}{P(B)} > P(A) \implies P(A \cap B) > P(A)P(B) \implies \frac{P(A \cap B)}{P(A)} > P(B) \implies P(B \mid A) > P(B)$.

MCQ 8: Assertion (A): The Law of Total Probability requires a partition of the sample space. Reason (R): The sum of probabilities of all partition sets must equal 1 and they must be pairwise disjoint.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

The definition of a partition explicitly requires pairwise disjointness and union equal to the sample space, which directly necessitates the sum of their probabilities to be 1.

MCQ 9: Urn A has 2 red, 3 blue balls. Urn B has 4 red, 1 blue ball. A coin is tossed; if heads, Urn A is chosen; if tails, Urn B. A ball is drawn and found to be red. The total probability of choosing Urn A and drawing red is:

A $\frac{1}{5}$

B $\frac{3}{10}$

C $\frac{2}{5}$

D $\frac{1}{2}$

Derivation & Analysis:

Probability = $P(\text{Head}) \times P(\text{Red} \mid \text{Urn A}) = \frac{1}{2} \times \frac{2}{5} = \frac{1}{5}$.

Q10

MCQ 10: If three independent events $A, B, C$ have probabilities $0.2, 0.3, 0.5$, the total probability of occurrence of at least one of them is:

A 0.72

B 0.28

C 0.80

D 0.50

Derivation & Analysis:

$P(A \cup B \cup C) = 1 - P(A')P(B')P(C') = 1 - (0.8)(0.7)(0.5) = 1 - 0.28 = 0.72$.

Common questions

Total Probability calculates the unconditional probability of a final outcome affected by multiple preliminary causes ($P(A)$). Bayes' Theorem calculates a reverse conditional probability, determining the likelihood of a specific cause given that the final outcome has occurred ($P(E_i \mid A)$).

Candidates frequently forget to multiply the conditional probabilities by the prior probabilities of the partition sets $P(E_i)$, treating all branches as equally likely by default.

Draw a quick tree diagram on your scratch pad, write down the branch products directly, and sum them up without writing formal set-theoretic justifications.

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