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Probability and Statistics Short Notes for GATE DA

GATE DA Probability and Statistics: 3 units and 11 chapters, weightage from 35 previous year questions across 3 papers, a study order by exam weight and 165 p

A question from this chapter

Question 1

Level 1: Warm-up

A bag contains 5 red and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that at least one ball is red?

Question 2

Level 1: Warm-up

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

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 and Statistics Short Notes for GATE DA

GATE DA Probability and Statistics: 3 units and 11 chapters, weightage from 35 previous year questions across 3 papers, a study order by exam weight and 165 practice questions.

About Probability and Statistics Short Notes

Quick revision sheets for Probability and Statistics in GATE DA. Every chapter is condensed into key formulas, shortcuts and common traps so you can revise 11 chapters fast before the exam.

GATE DA Probability and Statistics Unit-wise Weightage from Past Papers

We counted every GATE DA Probability and Statistics previous year question in our bank (35 questions from 3 papers) and grouped them by unit.

Unit	Chapters	PYQs	Share of section	Avg per paper
Probability	4	11	31%	3.7
Random Variables	5	20	57%	6.7
Statistical Inference	2	4	11%	1.3

Suggested Probability and Statistics Study Order for GATE DA

Random Variables: 57% of past Probability and Statistics questions, about 6.7 per paper.
Probability: 31% of past Probability and Statistics questions, about 3.7 per paper.
Statistical Inference: 11% of past Probability and Statistics questions, about 1.3 per paper.

Start where the marks are. Units at the top of this list have appeared most often in past GATE DA papers.

Units in GATE DA Probability and Statistics

All Probability and Statistics chapters

One Solved Question from Each Probability and Statistics Chapter

Question 1 · Basic Probability, Events and Counting MCQ

A bag contains 5 red and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that at least one ball is red?

A.
$r a c 2528$
B.
$r a c 1528$
C.
$r a c 58$
D.
$r a c 27$

Correct Answer:

Step-by-Step Solution

Key idea: The phrase "at least one" is a strong signal to use the complement rule.

Step 1: Identify the total number of ways to draw 2 balls from 8. $n (S) = (2 8) = \frac{8 \times 7}{2} = 28$ .

Step 2: Identify the complement event. "At least one red" is the complement of "No red balls" (i.e., both balls are blue).

Step 3: Calculate the number of ways to draw 2 blue balls from the 3 available. $n (E^{c}) = (2 3) = 3$ .

Step 4: Calculate the probability of the complement. $P (No red) = \frac{3}{28}$ .

Step 5: Apply the complement rule. $P (At least one red) = 1 - P (No red) = 1 - \frac{3}{28} = \frac{25}{28}$ .

Answer: $\frac{25}{28}$

Question 2 · Conditional Probability and Bayes' Theorem 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 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