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Probability Short Notes for GATE DA

GATE DA Probability: 4 chapters, 11 previous year questions (31% of Probability and Statistics), 165 practice questions and one solved question from each chap

A question from this chapter

Question 1

Level 3: Exam Standard

Match the random experiment in List I with the probability of the stated event in List II. In each case, a computer program generates a uniformly random non-negative integer solution to the given equation. List I (P) x_{1} + x_{2} + x_{3} = 8; event: all x_{i} \geq 1 (Q) x_{1} + x_{2} + x_{3} + x_{4} = 7; event: all x_{i} \geq 1 (R) x_{1} + x_{2} + x_{3} + x_{4} = 13; event: all x_{i} \geq 2 (S) x_{1} + x_{2} + x_{3} = 10; event: all x_{i} \geq 1 List II (1) \frac{1}{6} (2) \frac{7}{15} (3) \frac{6}{11} (4) \frac{1}{10}

Question 2

Level 3: Exam Standard

Let B_{1}, B_{2}, B_{3} form a partition of the sample space with P (B_{1}) = 0.2, P (B_{2}) = 0.3, and P (B_{3}) = 0.5 . For an event A, it is given that P (A ∣ B_{1}) = 0.6 and P (A ∣ B_{2}) = 0.4 . If P (A ∣ B_{3}) = q where 0 \leq q \leq 1, which of the following statements about P (B_{1} ∣ A) is correct for all valid q ?

Question 3

Level 1: Warm-up

A student incorrectly calculates the probability of the first success in the first two trials as p + p = 2 p = 0.4 by overcounting. Using the correct geometric distribution with p = 0.2, what is the minimum number of trials k required for the cumulative probability P (T \leq k) to be at least 0.36 ?

Question 4

Level 1: Warm-up

Two independent random variables X \sim U (0, 2) and Y \sim U (0, 3) are given. What is the area of the joint sample space rectangle in the X Y -plane?

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Probability Short Notes for GATE DA

GATE DA Probability: 4 chapters, 11 previous year questions (31% of Probability and Statistics), 165 practice questions and one solved question from each chapter.

About Probability Short Notes

Quick revision sheets for Probability in GATE DA. Every chapter is condensed into key formulas, shortcuts and common traps so you can revise 4 chapters fast before the exam.

Probability Weightage in GATE DA

Probability accounts for 11 of 35 Probability and Statistics previous year questions in our bank (31%), about 3.7 per paper across 3 papers.

Probability Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Basic Probability, Events and Counting	Counting-Based Probability and Combinatorial Events, Coin Toss and Elementary Event Operations, Poisson Limit and Distribution Approximation	5	45%	60
Conditional Probability and Bayes' Theorem	Conditional Probability, Bayes' Theorem and Posterior Classification	4	36%	63
Independence and Expected Waiting Time	Independent Trials and Expected Waiting Time	1	9%	21
Continuous Probability and Geometric Probability	Continuous Uniform Distributions and Geometric Probability	1	9%	21

One Solved Question from Each Probability Chapter

Question 1 · Basic Probability, Events and Counting MCQ

Match the random experiment in List I with the probability of the stated event in List II. In each case, a computer program generates a uniformly random non-negative integer solution to the given equation.

List I

(P) $x_{1} + x_{2} + x_{3} = 8$ ; event: all $x_{i} \geq 1$

(Q) $x_{1} + x_{2} + x_{3} + x_{4} = 7$ ; event: all $x_{i} \geq 1$

(R) $x_{1} + x_{2} + x_{3} + x_{4} = 13$ ; event: all $x_{i} \geq 2$

(S) $x_{1} + x_{2} + x_{3} = 10$ ; event: all $x_{i} \geq 1$

List II

(1) $\frac{1}{6}$

(2) $\frac{7}{15}$

(3) $\frac{6}{11}$

(4) $\frac{1}{10}$

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

Correct Answer:

Step-by-Step Solution

Key idea: This is a stars-and-bars probability question. For $x_{1} + \dots + x_{k} = n$ with all $x_{i} \geq 0$ , the total number of solutions is $(k - 1 n + k - 1)$ . If we require all $x_{i} \geq m$ , substitute $y_{i} = x_{i} - m$ to get $y_{1} + \dots + y_{k} = n - k m$ with $y_{i} \geq 0$ , giving $(k - 1 n - k m + k - 1)$ favorable solutions.

Step 1: Compute (P). Here $n = 8, k = 3, m = 1$ . Total solutions: $(2 8 + 2) = (2 10) = 45$ . Favorable ( $y_{i} \geq 0, \sum y_{i} = 5$ ): $(2 5 + 2) = (2 7) = 21$ . Probability: $\frac{21}{45} = \frac{7}{15}$ . So P matches (2).

Step 2: Compute (Q). Here $n = 7, k = 4, m = 1$ . Total: $(3 7 + 3) = (3 10) = 120$ . Favorable ( $\sum y_{i} = 3$ ): $(3 3 + 3) = (3 6) = 20$ . Probability: $\frac{20}{120} = \frac{1}{6}$ . So Q matches (1).

Step 3: Compute (R). Here $n = 13, k = 4, m = 2$ . Total: $(3 13 + 3) = (3 16) = 560$ . Favorable ( $\sum y_{i} = 13 - 8 = 5$ ): $(3 5 + 3) = (3 8) = 56$ . Probability: $\frac{56}{560} = \frac{1}{10}$ . So R matches (4).

Step 4: Compute (S). Here $n = 10, k = 3, m = 1$ . Total: $(2 10 + 2) = (2 12) = 66$ . Favorable ( $\sum y_{i} = 7$ ): $(2 7 + 2) = (2 9) = 36$ . Probability: $\frac{36}{66} = \frac{6}{11}$ . So S matches (3).

Answer: P-2, Q-1, R-4, S-3, which is option A.

Question 2 · Conditional Probability and Bayes' Theorem MCQ

Let $B_{1}, B_{2}, B_{3}$ form a partition of the sample space with $P (B_{1}) = 0.2$ , $P (B_{2}) = 0.3$ , and $P (B_{3}) = 0.5$ . For an event $A$ , it is given that $P (A ∣ B_{1}) = 0.6$ and $P (A ∣ B_{2}) = 0.4$ . If $P (A ∣ B_{3}) = q$ where $0 \leq q \leq 1$ , which of the following statements about $P (B_{1} ∣ A)$ is correct for all valid $q$ ?

A.
$P (B_{1} ∣ A) > \frac{1}{2}$ for all $q \in [0, 1]$
B.
$P (B_{1} ∣ A) \leq \frac{1}{2}$ for all $q \in [0, 1]$
C.
$P (B_{1} ∣ A) = \frac{1}{4}$ for all $q \in [0, 1]$
D.
$P (B_{1} ∣ A) \geq \frac{1}{4}$ for all $q \in [0, 1]$

Correct Answer:

Step-by-Step Solution

Key idea: This is a bounding question on a Bayes posterior with a free parameter. The posterior $P (B_{1} ∣ A)$ is a decreasing function of $q$ , so its maximum occurs at $q = 0$ .

Step 1: Write the total probability of $A$ .

$P (A) = P (A ∣ B_{1}) P (B_{1}) + P (A ∣ B_{2}) P (B_{2}) + P (A ∣ B_{3}) P (B_{3}) = 0.6 (0.2) + 0.4 (0.3) + q (0.5) = 0.24 + 0.5 q$

Step 2: Write the posterior.

$P (B_{1} ∣ A) = \frac{P ( A ∣ B _{1} ) P ( B _{1} )}{P ( A )} = \frac{0.12}{0.24 + 0.5 q}$

Step 3: Find the range. The denominator $0.24 + 0.5 q$ is increasing in $q$ , so the fraction is decreasing in $q$ .

At $q = 0$ : $P (B_{1} ∣ A) = \frac{0.12}{0.24} = \frac{1}{2}$ .

At $q = 1$ : $P (B_{1} ∣ A) = \frac{0.12}{0.74} = \frac{6}{37} \approx 0.162$ .

So $P (B_{1} ∣ A) \in [\frac{6}{37}, \frac{1}{2}]$ for $q \in [0, 1]$ .

Step 4: Check each option.

A) $> \frac{1}{2}$ : False, the maximum is exactly $\frac{1}{2}$ .

B) $\leq \frac{1}{2}$ : True, since the maximum is $\frac{1}{2}$ .

C) $= \frac{1}{4}$ : False, the value varies with $q$ .

D) $\geq \frac{1}{4}$ : False, at $q = 1$ the value is $\frac{6}{37} \approx 0.162 < 0.25$ .

Answer: B.

Question 3 · Independence and Expected Waiting Time MCQ

A student incorrectly calculates the probability of the first success in the first two trials as $p + p = 2 p = 0.4$ by overcounting. Using the correct geometric distribution with $p = 0.2$ , what is the minimum number of trials $k$ required for the cumulative probability $P (T \leq k)$ to be at least $0.36$ ?

A.
2
B.
1
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: The cumulative distribution function (CDF) for a geometric distribution is $P (T \leq k) = 1 - (1 - p)^{k}$ .

Step 1: Identify the target cumulative probability: $P (T \leq k) \geq 0.36$ .

Step 2: Substitute the CDF formula: $1 - (1 - p)^{k} \geq 0.36$ .

Step 3: Plug in $p = 0.2$ : $1 - (0.8)^{k} \geq 0.36$ .

Step 4: Rearrange: $(0.8)^{k} \leq 1 - 0.36 = 0.64$ .

Step 5: Test integer values for $k$ :

For $k = 1$ : $0. 8^{1} = 0.8$ (Not $\leq 0.64$ )
For $k = 2$ : $0. 8^{2} = 0.64$ (Satisfies $\leq 0.64$ )

Answer: The minimum number of trials $k$ is 2.

Question 4 · Continuous Probability and Geometric Probability MCQ

Two independent random variables $X \sim U (0, 2)$ and $Y \sim U (0, 3)$ are given. What is the area of the joint sample space rectangle in the $X Y$ -plane?

A.
5
B.
6
C.
2.5
D.
1.5

Correct Answer:

Step-by-Step Solution

Key idea: For independent uniform random variables, the joint sample space is a rectangle with dimensions equal to the interval lengths.

Step 1: Identify the intervals:

$X \in [0, 2]$ , so the width is $2 - 0 = 2$
$Y \in [0, 3]$ , so the height is $3 - 0 = 3$

Step 2: The joint sample space is the rectangle $[0, 2] \times [0, 3]$ .

Step 3: Calculate the area:

$Area = width \times height = 2 \times 3 = 6$

Answer: B

A question from this chapter

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Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Probability Short Notes for GATE DA

About Probability Short Notes

Probability Weightage in GATE DA

Probability Chapter Matrix

More from Probability and Statistics

One Solved Question from Each Probability Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution