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Conditional Probability and Bayes' Theorem Short Notes for GATE DA

Conditional Probability and Bayes' Theorem short notes for GATE DA: 1 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice

conditional probability and bayes theorem short notes

Last-Minute Sheet: Conditional Probability and Bayes

Last-Minute Sheet Core formulas P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (A \cap B) = P (A ∣ B) P (B) = P (B ∣ A) P (A) Total probability & Bayes P (A) = i \sum P (A ∣ B_{i}) P (B_{i}) P (B_{j} ∣ A) = \frac{P ( A ∣ B _{j} ) P ( B _{j} )}{\sum _{i} P ( A ∣ B _{i} ) P ( B _{i} )} Exam checklist Forward (cause \to effect): total probability. Backward (effect \to cause): Bayes. Never swap P (A ∣ B) and P (B ∣ A) . In medical/source problems, always write prior, likelihood, evidence, posterior in that order.

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Question 1

Level 1: Warm-up

Which of the following statements correctly describes the primary use of the law of total probability?

Question 2

Level 1: Warm-up

Three boxes contain balls as follows: Box-1 has 2 black and 1 white ball, Box-2 has 1 black and 2 white balls, Box-3 has 3 black and 3 white balls. A box is chosen and one ball is drawn from it. Which option correctly lists P (white ∣ Box-1), P (white ∣ Box-2), P (white ∣ Box-3) in that order?

Question 3

Level 1: Warm-up

A factory has three machines M_{1}, M_{2}, M_{3} that produce items. The probabilities of selecting each machine are P (M_{1}) = 0.5, P (M_{2}) = 0.3, and P (M_{3}) = 0.2 . The probabilities of producing a defective item D given the machine are P (D ∣ M_{1}) = 0.02, P (D ∣ M_{2}) = 0.03, and P (D ∣ M_{3}) = 0.04 . Match each quantity in List I to its correct numerical value in List II. List I: (P) P (M_{2}) (Q) P (D ∣ M_{3}) (R) P (M_{1} \cap D) (S) P (D) List II: (1) 0.027 (2) 0.04 (3) 0.01 (4) 0.3

Question 4

Level 1: Warm-up

Let A and B be two events with P (A \cap B) = 0.2 and P (A) = 0.5 . The minimum possible value of P (A ∣ B) is

Question 5

Level 1: Warm-up

A geologist is searching for a specific mineral M . The land is divided into three soil types: Sandy, Clay, and Loam. The probabilities of the soil types are P (Sandy) = 0.5, P (Clay) = 0.3, and P (Loam) = 0.2 . The probabilities of finding the mineral M given the soil type are P (M ∣ Sandy) = 0.1, P (M ∣ Clay) = 0.4, and P (M ∣ Loam) = 0.6 . Match each quantity in List I to its correct numerical value in List II. List I: (P) P (Clay) (Q) P (M ∣ Loam) (R) P (Sandy \cap M) (S) P (M) List II: (1) 0.29 (2) 0.30 (3) 0.05 (4) 0.60

Question 6

Level 1: Warm-up

Let A and B be two events with P (A) = 0.4 and P (B) = 0.8 . The minimum possible value of P (A ∣ B) is

Question 7

Level 1: Warm-up

A cloud system routes incoming requests to three servers: S_{1}, S_{2}, S_{3} . The probabilities of routing to each server are P (S_{1}) = 0.4, P (S_{2}) = 0.4, and P (S_{3}) = 0.2 . The probabilities of a request timing out T given the server are P (T ∣ S_{1}) = 0.05, P (T ∣ S_{2}) = 0.10, and P (T ∣ S_{3}) = 0.20 . Match each quantity in List I to its correct numerical value in List II. List I: (P) P (S_{2}) (Q) P (T ∣ S_{3}) (R) P (S_{1} \cap T) (S) P (T) List II: (1) 0.02 (2) 0.20 (3) 0.10 (4) 0.40

Question 8

Level 1: Warm-up

Let A and B be two events with P (A) = 0.9 and P (B) = 0.5 . The minimum possible value of P (A ∣ B) is

Question 9

Level 1: Warm-up

If P (A ∣ B) = 40% and P (B) = 0.5, then P (A \cap B) =______ (round off to two decimal places).

Question 10

Level 1: Warm-up

A Naive Bayes classifier has three classes y_{1}, y_{2}, y_{3} . For a specific feature vector x, which of the following CANNOT be a valid assignment of likelihoods P (x ∣ y_{1}), P (x ∣ y_{2}), P (x ∣ y_{3}) ?

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Conditional Probability and Bayes' Theorem Short Notes for GATE DA

Conditional Probability and Bayes' Theorem short notes for GATE DA: 1 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Last-Minute Sheet: Conditional Probability and Bayes

Last-Minute Sheet

Core formulas

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (A \cap B) = P (A ∣ B) P (B) = P (B ∣ A) P (A)

Total probability & Bayes

P (A) = i \sum P (A ∣ B_{i}) P (B_{i})

P (B_{j} ∣ A) = \frac{P ( A ∣ B _{j} ) P ( B _{j} )}{\sum _{i} P ( A ∣ B _{i} ) P ( B _{i} )}

Exam checklist

Forward (cause → effect): total probability.
Backward (effect → cause): Bayes.
Never swap $P (A ∣ B)$ and $P (B ∣ A)$ .
In medical/source problems, always write prior, likelihood, evidence, posterior in that order.

Conditional Probability and Bayes' Theorem: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Probability and Statistics MCQ

Which of the following statements correctly describes the primary use of the law of total probability?

A.
It computes the probability of a cause given an observed effect.
B.
It computes the overall probability of an event by summing over a partition of the sample space.
C.
It reverses the direction of conditioning using prior and likelihood.
D.
It computes the joint probability of two independent events as the product of their marginals.

Correct Answer:

Step-by-Step Solution

Key idea: The law of total probability is the forward-direction tool — it combines conditional pieces across a partition to get an unconditional probability.

Step 1: Recall the formula: $P (A) = \sum_{i} P (A ∣ B_{i}) P (B_{i})$ , where ${B_{i}}$ is a partition.

Step 2: Evaluate each option:

A: "cause given effect" describes Bayes' theorem (backward inference), not total probability.
B: "overall probability by summing over a partition" matches the formula exactly.
C: "reverses conditioning" is Bayes' theorem again.
D: "joint of independent events" is $P (A \cap B) = P (A) P (B)$ , a different rule entirely.

Answer: B

Trap path: Confusing total probability with Bayes' theorem is the most common error. Remember: total probability goes forward (cause → effect), Bayes goes backward (effect → cause).

Verification: The formula $P (A) = \sum_{i} P (A ∣ B_{i}) P (B_{i})$ literally sums over a partition to get the overall $P (A)$ , which is exactly what option B states.

Question 2 · Probability and Statistics MCQ

Three boxes contain balls as follows: Box-1 has 2 black and 1 white ball, Box-2 has 1 black and 2 white balls, Box-3 has 3 black and 3 white balls. A box is chosen and one ball is drawn from it. Which option correctly lists $P (white ∣ Box-1)$ , $P (white ∣ Box-2)$ , $P (white ∣ Box-3)$ in that order?

A.
$\frac{1}{3}, \frac{2}{3}, \frac{1}{2}$
B.
$\frac{2}{3}, \frac{1}{3}, \frac{1}{2}$
C.
$\frac{1}{3}, \frac{1}{2}, \frac{2}{3}$
D.
$\frac{1}{2}, \frac{2}{3}, \frac{1}{3}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct conditional-probability reading question — count favourable over total inside each box.

Step 1: For Box-1 (2 black, 1 white), total = 3, white = 1, so $P (white ∣ Box-1) = \frac{1}{3}$ .

Step 2: For Box-2 (1 black, 2 white), total = 3, white = 2, so $P (white ∣ Box-2) = \frac{2}{3}$ .

Step 3: For Box-3 (3 black, 3 white), total = 6, white = 3, so $P (white ∣ Box-3) = \frac{3}{6} = \frac{1}{2}$ .

Step 4: The ordered triple is $\frac{1}{3}, \frac{2}{3}, \frac{1}{2}$ .

Answer: A

Trap path: Swapping Box-1 and Box-2 compositions gives $\frac{2}{3}, \frac{1}{3}, \frac{1}{2}$ (option B). This happens when you misread which box has more white balls.

Verification: $\frac{1}{3} + \frac{2}{3}$ are complementary only within their own boxes — each fraction is independently computed from its box's composition.

Question 3 · Probability and Statistics MCQ

A factory has three machines $M_{1}, M_{2}, M_{3}$ that produce items. The probabilities of selecting each machine are $P (M_{1}) = 0.5$ , $P (M_{2}) = 0.3$ , and $P (M_{3}) = 0.2$ . The probabilities of producing a defective item $D$ given the machine are $P (D ∣ M_{1}) = 0.02$ , $P (D ∣ M_{2}) = 0.03$ , and $P (D ∣ M_{3}) = 0.04$ .

Match each quantity in List I to its correct numerical value in List II.

List I:

(P) $P (M_{2})$

(Q) $P (D ∣ M_{3})$

(R) $P (M_{1} \cap D)$

(S) $P (D)$

List II:

(1) $0.027$

(2) $0.04$

(3) $0.01$

(4) $0.3$

A.
P $o$ 4, Q $o$ 2, R $o$ 3, S $o$ 1
B.
P $o$ 2, Q $o$ 4, R $o$ 3, S $o$ 1
C.
P $o$ 4, Q $o$ 3, R $o$ 2, S $o$ 1
D.
P $o$ 4, Q $o$ 2, R $o$ 1, S $o$ 3

Correct Answer:

Step-by-Step Solution

Key idea: In a partition setup, each symbol has a fixed role. Match the definition to the calculation.

Step 1: $P (M_{2})$ is the prior probability of selecting machine 2. Given directly as $0.3$ . So P $\to$ 4.

Step 2: $P (D ∣ M_{3})$ is the likelihood of a defect given machine 3. Given directly as $0.04$ . So Q $\to$ 2.

Step 3: $P (M_{1} \cap D)$ is the joint probability. By the multiplication rule, $P (M_{1} \cap D) = P (M_{1}) \times P (D ∣ M_{1}) = 0.5 \times 0.02 = 0.01$ . So R $\to$ 3.

Step 4: $P (D)$ is the total probability of a defect. By the law of total probability:

$P (D) = P (M_{1}) P (D ∣ M_{1}) + P (M_{2}) P (D ∣ M_{2}) + P (M_{3}) P (D ∣ M_{3})$

$P (D) = (0.5 \times 0.02) + (0.3 \times 0.03) + (0.2 \times 0.04) = 0.01 + 0.009 + 0.008 = 0.027$ . So S $\to$ 1.

Answer: A

Question 4 · Probability and Statistics MCQ

Let $A$ and $B$ be two events with $P (A \cap B) = 0.2$ and $P (A) = 0.5$ . The minimum possible value of $P (A ∣ B)$ is

A.
0.20
B.
0.40
C.
0.50
D.
0.80

Correct Answer:

Step-by-Step Solution

Key idea: The conditional probability $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ . To minimize this fraction, we must maximize the denominator $P (B)$ .

Step 1: By definition, $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{0.2}{P ( B )}$ .

Step 2: We need to find the maximum possible value of $P (B)$ . Since $P (A \cap B) \leq P (B) \leq 1$ , the maximum value of $P (B)$ is $1$ .

Step 3: Substitute $P (B) = 1$ into the formula:

Minimum $P (A ∣ B) = \frac{0.2}{1} = 0.20$ .

Answer: A

Question 5 · Probability and Statistics MCQ

A geologist is searching for a specific mineral $M$ . The land is divided into three soil types: Sandy, Clay, and Loam. The probabilities of the soil types are $P (Sandy) = 0.5$ , $P (Clay) = 0.3$ , and $P (Loam) = 0.2$ . The probabilities of finding the mineral $M$ given the soil type are $P (M ∣ Sandy) = 0.1$ , $P (M ∣ Clay) = 0.4$ , and $P (M ∣ Loam) = 0.6$ .

Match each quantity in List I to its correct numerical value in List II.

List I:

(P) $P (Clay)$

(Q) $P (M ∣ Loam)$

(R) $P (Sandy \cap M)$

(S) $P (M)$

List II:

(1) $0.29$

(2) $0.30$

(3) $0.05$

(4) $0.60$

A.
P $o$ 2, Q $o$ 4, R $o$ 3, S $o$ 1
B.
P $o$ 4, Q $o$ 2, R $o$ 3, S $o$ 1
C.
P $o$ 2, Q $o$ 3, R $o$ 4, S $o$ 1
D.
P $o$ 4, Q $o$ 3, R $o$ 2, S $o$ 1

Correct Answer:

Step-by-Step Solution

Key idea: In a partition setup, each symbol has a fixed role. Match the definition to the calculation.

Step 1: $P (Clay)$ is the prior probability of the Clay soil type. Given directly as $0.3$ . So P $\to$ 2.

Step 2: $P (M ∣ Loam)$ is the likelihood of finding the mineral given Loam soil. Given directly as $0.6$ . So Q $\to$ 4.

Step 3: $P (Sandy \cap M)$ is the joint probability. By the multiplication rule, $P (Sandy \cap M) = P (Sandy) \times P (M ∣ Sandy) = 0.5 \times 0.1 = 0.05$ . So R $\to$ 3.

Step 4: $P (M)$ is the total probability of finding the mineral. By the law of total probability:

$P (M) = P (Sandy) P (M ∣ Sandy) + P (Clay) P (M ∣ Clay) + P (Loam) P (M ∣ Loam)$

$P (M) = (0.5 \times 0.1) + (0.3 \times 0.4) + (0.2 \times 0.6) = 0.05 + 0.12 + 0.12 = 0.29$ . So S $\to$ 1.

Answer: A

Question 6 · Probability and Statistics MCQ

Let $A$ and $B$ be two events with $P (A) = 0.4$ and $P (B) = 0.8$ . The minimum possible value of $P (A ∣ B)$ is

A.
0.20
B.
0.40
C.
0.25
D.
0.50

Correct Answer:

Step-by-Step Solution

Key idea: The conditional probability $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ . To minimize this fraction, we must minimize the numerator $P (A \cap B)$ .

Step 1: By definition, $P (A ∣ B) = \frac{P ( A \cap B )}{0.8}$ .

Step 2: We need to find the minimum possible value of $P (A \cap B)$ . By the inclusion-exclusion principle, $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ . Since $P (A \cup B) \leq 1$ , we have:

$P (A) + P (B) - P (A \cap B) \leq 1 ⟹ 0.4 + 0.8 - P (A \cap B) \leq 1 ⟹ 1.2 - P (A \cap B) \leq 1 ⟹ P (A \cap B) \geq 0.2$ .

Step 3: Substitute the minimum $P (A \cap B) = 0.2$ into the formula:

Minimum $P (A ∣ B) = \frac{0.2}{0.8} = 0.25$ .

Answer: C

Question 7 · Probability and Statistics MCQ

A cloud system routes incoming requests to three servers: $S_{1}, S_{2}, S_{3}$ . The probabilities of routing to each server are $P (S_{1}) = 0.4$ , $P (S_{2}) = 0.4$ , and $P (S_{3}) = 0.2$ . The probabilities of a request timing out $T$ given the server are $P (T ∣ S_{1}) = 0.05$ , $P (T ∣ S_{2}) = 0.10$ , and $P (T ∣ S_{3}) = 0.20$ .

Match each quantity in List I to its correct numerical value in List II.

List I:

(P) $P (S_{2})$

(Q) $P (T ∣ S_{3})$

(R) $P (S_{1} \cap T)$

(S) $P (T)$

List II:

(1) $0.02$

(2) $0.20$

(3) $0.10$

(4) $0.40$

A.
P $o$ 4, Q $o$ 2, R $o$ 1, S $o$ 3
B.
P $o$ 2, Q $o$ 4, R $o$ 1, S $o$ 3
C.
P $o$ 4, Q $o$ 1, R $o$ 2, S $o$ 3
D.
P $o$ 4, Q $o$ 2, R $o$ 3, S $o$ 1

Correct Answer:

Step-by-Step Solution

Key idea: In a partition setup, each symbol has a fixed role. Match the definition to the calculation.

Step 1: $P (S_{2})$ is the prior probability of routing to server 2. Given directly as $0.4$ . So P $\to$ 4.

Step 2: $P (T ∣ S_{3})$ is the likelihood of a timeout given server 3. Given directly as $0.2$ . So Q $\to$ 2.

Step 3: $P (S_{1} \cap T)$ is the joint probability. By the multiplication rule, $P (S_{1} \cap T) = P (S_{1}) \times P (T ∣ S_{1}) = 0.4 \times 0.05 = 0.02$ . So R $\to$ 1.

Step 4: $P (T)$ is the total probability of a timeout. By the law of total probability:

$P (T) = P (S_{1}) P (T ∣ S_{1}) + P (S_{2}) P (T ∣ S_{2}) + P (S_{3}) P (T ∣ S_{3})$

$P (T) = (0.4 \times 0.05) + (0.4 \times 0.10) + (0.2 \times 0.20) = 0.02 + 0.04 + 0.04 = 0.10$ . So S $\to$ 3.

Answer: A

Question 8 · Probability and Statistics MCQ

Let $A$ and $B$ be two events with $P (A) = 0.9$ and $P (B) = 0.5$ . The minimum possible value of $P (A ∣ B)$ is

A.
0.50
B.
0.40
C.
0.80
D.
0.90

Correct Answer:

Step-by-Step Solution

Key idea: The conditional probability $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ . To minimize this fraction, we must minimize the numerator $P (A \cap B)$ .

Step 1: By definition, $P (A ∣ B) = \frac{P ( A \cap B )}{0.5}$ .

Step 2: We need to find the minimum possible value of $P (A \cap B)$ . By the inclusion-exclusion principle, $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ . Since $P (A \cup B) \leq 1$ , we have:

$P (A) + P (B) - P (A \cap B) \leq 1 ⟹ 0.9 + 0.5 - P (A \cap B) \leq 1 ⟹ 1.4 - P (A \cap B) \leq 1 ⟹ P (A \cap B) \geq 0.4$ .

Step 3: Substitute the minimum $P (A \cap B) = 0.4$ into the formula:

Minimum $P (A ∣ B) = \frac{0.4}{0.5} = 0.80$ .

Answer: C

Question 9 · Probability and Statistics NAT

If $P (A ∣ B) = 40%$ and $P (B) = 0.5$ , then $P (A \cap B) =$ ______ (round off to two decimal places).

Correct Answer:

0.20

Step-by-Step Solution

Key idea: This is a direct application of the multiplication rule $P (A \cap B) = P (A ∣ B) P (B)$ .

Step 1: Convert the percentage to a decimal. $P (A ∣ B) = 40% = 0.40$ .

Step 2: Apply the multiplication rule:

$P (A \cap B) = P (A ∣ B) \times P (B) = 0.40 \times 0.5 = 0.20$

Answer: 0.20

Trap path: Using $40$ instead of $0.40$ gives $40 \times 0.5 = 20.00$ , which is not a valid probability. Always convert percentages to decimals before multiplying.

Verification: $P (A \cap B) = 0.20 \leq P (B) = 0.5$ ✓, and $P (A \cap B) = 0.20 \leq 1$ ✓.

Question 10 · Probability and Statistics MCQ

A Naive Bayes classifier has three classes $y_{1}, y_{2}, y_{3}$ . For a specific feature vector $x$ , which of the following CANNOT be a valid assignment of likelihoods $P (x ∣ y_{1}), P (x ∣ y_{2}), P (x ∣ y_{3})$ ?

A.
$0.2, 0.3, 0.5$
B.
$0.6, 0.6, 0.6$
C.
$0.9, 0.9, 0.9$
D.
$- 0.1, 0.5, 0.6$

Correct Answer:

Step-by-Step Solution

Key idea: Likelihoods $P (x ∣ y_{i})$ are probabilities of observing $x$ given class $y_{i}$ . Each must be a valid probability, meaning it must lie in the interval $[0, 1]$ .

Step 1: Check option A: $0.2, 0.3, 0.5$ . All values are in $[0, 1]$ . Valid.

Step 2: Check option B: $0.6, 0.6, 0.6$ . All values are in $[0, 1]$ . Valid.

Step 3: Check option C: $0.9, 0.9, 0.9$ . All values are in $[0, 1]$ . Valid.

Step 4: Check option D: $- 0.1, 0.5, 0.6$ . The value $- 0.1$ is negative. Probabilities cannot be negative. Invalid.

Note: Unlike prior probabilities, likelihoods for a fixed $x$ across different classes do NOT need to sum to 1.

Answer: D

conditional probability and bayes theorem short notes

Last-Minute Sheet: Conditional Probability and Bayes

Last-Minute Sheet

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Conditional Probability and Bayes' Theorem Short Notes for GATE DA

Last-Minute Sheet: Conditional Probability and Bayes

Last-Minute Sheet

Conditional Probability and Bayes' Theorem: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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