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Basic Probability, Events and Counting Notes for GATE DA

Basic Probability, Events and Counting notes for GATE DA: 26 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions

basic probability events and counting notes

Chapter Roadmap: Probability, Events & Counting

The Classical Definition: Probability as a Ratio

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Order Matters: Permutation vs Combination

.smart-card { font-family: system-ui, -apple-system, sans-serif; background-color: var(--color-bg); color: var(--color-text); padding: 24px; border-radius: 12px; border: 1px solid var(--color-border); max-width: 600px; margin: 0 auto; } .comparison-table { width: 100%; border-collapse: collapse; margin-bottom: 24px; font-size: 0.9rem; } .comparison-table th { text-align: left; padding: 12px; border-bottom: 2px solid var(--color-border); color: var(--color-muted); font-weight: 600; } .comparison-table td { padding: 12px; border-bottom: 1px solid var(--color-border); vertical-align: top; } .formula-highlight { background-color: color-mix(in srgb, var(--color-primary), transparent 90%); padding: 16px; border-radius: 8px; text-align: center; margin-bottom: 20px; } .rule-thumb { font-size: 0.9rem; color: var(--color-muted); border-left: 3px solid var(--color-primary); padding-left: 12px; } When to Use What? Feature Permutation Combination Question Arrangement / Ordering Selection / Grouping Order Matters? Yes No Formula \frac{n !}{( n - r )!} \frac{n !}{r ! ( n - r )!}^{n} C_{r} = (r n) = \frac{n !}{r ! ( n - r )!} Rule of Thumb: If the problem says "choose," "select," or "form a group," use Combinations. If it says "arrange," "order," or "rank," use Permutations.

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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

A fair coin is tossed n times. What is the minimum value of n such that the probability of getting at least one head is strictly greater than \frac{15}{16} ?

Question 3

Level 1: Warm-up

A box contains 5 bulbs, of which 2 are defective and 3 are good. If 3 bulbs are chosen at random, what is the probability that at least one is defective?

Question 4

Level 1: Warm-up

A standard deck of 52 playing cards is shuffled thoroughly. One card is drawn at random. What is the probability that the drawn card is a face card (Jack, Queen, or King)?

Question 5

Level 1: Warm-up

Let S = {1, 2, 3, 4, 5} . Consider the following statements: (I) The number of non-empty subsets of S whose product of elements is even is 2^{5} - 2^{3} . (II) The number of non-empty subsets of S whose product of elements is odd is 2^{3} . Which of the statements is/are true?

Question 6

Level 1: Warm-up

Let S = {1, 2, 3} . Consider the following statements: (I) The number of non-empty subsets of S whose product of elements is odd is 2^{2} = 4 . (II) The number of non-empty subsets of S whose product of elements is even is 4. Which of the statements is/are true?

Question 7

Level 1: Warm-up

Let S = {1, 2, 3} . Consider the following statements regarding the subsets of S : (I) The number of subsets of S whose sum of elements is even is 4. (II) The number of non-empty subsets of S whose sum of elements is even is 4. Which of the statements is/are true?

Question 8

Level 1: Warm-up

Let S = {1, 2, 3, 4} . Consider the following statements regarding the subsets of S : (I) The number of subsets of S whose sum of elements is even is 8. (II) The number of non-empty subsets of S whose sum of elements is even is 8. Which of the statements is/are true?

Question 9

Level 1: Warm-up

A fair six-sided die is rolled. Let A be the event of getting a prime number and B be the event of getting an even number. What is the number of elements in A \cup B ?

Question 10

Level 1: Warm-up

A fair six-sided die is rolled twice. What is the number of outcomes in which the value of the first roll is strictly greater than the value of the second roll?

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Basic Probability, Events and Counting Notes for GATE DA

Basic Probability, Events and Counting notes for GATE DA: 26 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Probability, Events & Counting

Chapter Journey

1. Counting-Based Probability

The Foundation. Learn to count favorable vs total outcomes using permutations and combinations.

2. Coin Toss & Elementary Operations

Defining the Space. Sample spaces, unions, intersections, and complements.

3. Conditional Probability & Independence

Updating Beliefs. How new information changes probability. Bayes' Theorem.

4. Random Variables & Distributions

Quantifying Outcomes. Discrete/continuous variables, expectation, variance.

Why start here? In competitive exams, many problems look like probability questions but are actually counting puzzles in disguise. If you cannot count the sample space correctly, the probability formula yields incorrect results.

The Classical Definition: Probability as a Ratio

The Classical Approach

P (E) = \frac{n ( E )}{n ( S )}

Favorable Outcomes / Total Outcomes

Finite Sample Space: The total number of outcomes must be countable.
Equally Likely: No outcome is favored over another. (Fair coin vs Loaded coin).

The Real Challenge

Calculating $P (E)$ is rarely about plugging numbers. It is almost always a Counting Problem. We use Permutations and Combinations to find $n (S)$ and $n (E)$ .

Order Matters: Permutation vs Combination

When to Use What?

Feature	Permutation	Combination
Question	Arrangement / Ordering	Selection / Grouping
Order Matters?	Yes	No
Formula	$\frac{n !}{( n - r )!}$	$\frac{n !}{r ! ( n - r )!}$

^{n} C_{r} = (r n) = \frac{n !}{r ! ( n - r )!}

Rule of Thumb: If the problem says "choose," "select," or "form a group," use Combinations. If it says "arrange," "order," or "rank," use Permutations.

Strategy: The Complement Rule

The Power of "At Least One"

Directly counting events with phrases like "at least one" often requires summing many separate cases. Instead, use the complement.

P (E) = 1 - P (E^{c})

Common Pattern

Target: Probability that at least one element satisfies condition

C

Complement: Probability that no element satisfies condition

C

P (At least one) = 1 - P (None)

Basic Probability, Events and Counting: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Probability and Statistics 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 · Probability and Statistics MCQ

A fair coin is tossed $n$ times. What is the minimum value of $n$ such that the probability of getting at least one head is strictly greater than $\frac{15}{16}$ ?

A.
3
B.
4
C.
5
D.
6

Correct Answer:

Step-by-Step Solution

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

Step 1: Express the probability using the complement. $P (At least one head) = 1 - P (No heads)$ .

Step 2: Calculate $P (No heads)$ . The probability of getting a tail on a single toss is $\frac{1}{2}$ . For $n$ tosses, $P (No heads) = (\frac{1}{2})^{n}$ .

Step 3: Set up the inequality. We want $1 - (\frac{1}{2})^{n} > \frac{15}{16}$ .

Step 4: Solve for $n$ .

$(\frac{1}{2})^{n} < 1 - \frac{15}{16} = \frac{1}{16}$ .

Since $\frac{1}{16} = (\frac{1}{2})^{4}$ , we need $(\frac{1}{2})^{n} < (\frac{1}{2})^{4}$ .

This implies $n > 4$ .

Step 5: Find the minimum integer. The smallest integer strictly greater than 4 is 5.

Answer: 5

Question 3 · Probability and Statistics MCQ

A box contains 5 bulbs, of which 2 are defective and 3 are good. If 3 bulbs are chosen at random, what is the probability that at least one is defective?

A.
$r a c 35$
B.
$r a c 610$
C.
$r a c 110$
D.
$r a c 910$

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 choose 3 bulbs from 5. $n (S) = (3 5) = 10$ .

Step 2: Identify the complement event. "At least one defective" is the complement of "No defective bulbs" (i.e., all 3 are good).

Step 3: Calculate the number of ways to choose 3 good bulbs from the 3 available. $n (E^{c}) = (3 3) = 1$ .

Step 4: Calculate the probability of the complement. $P (None defective) = \frac{1}{10}$ .

Step 5: Apply the complement rule. $P (At least one defective) = 1 - \frac{1}{10} = \frac{9}{10}$ .

Answer: $\frac{9}{10}$

Question 4 · Probability and Statistics MCQ

A standard deck of 52 playing cards is shuffled thoroughly. One card is drawn at random. What is the probability that the drawn card is a face card (Jack, Queen, or King)?

A.
$r a c 313$
B.
$r a c 413$
C.
$r a c 113$
D.
$r a c 14$

Correct Answer:

Step-by-Step Solution

Key idea: This is a classical probability problem where we count favorable outcomes over total outcomes.

Step 1: Identify total outcomes. A standard deck has 52 cards, so $n (S) = 52$ .

Step 2: Identify favorable outcomes. Face cards are Jack, Queen, and King. There are 3 face cards in each of the 4 suits, so $n (E) = 3 \times 4 = 12$ .

Step 3: Calculate probability. $P (E) = \frac{n ( E )}{n ( S )} = \frac{12}{52} = \frac{3}{13}$ .

Answer: $\frac{3}{13}$

Question 5 · Probability and Statistics MCQ

Let $S = {1, 2, 3, 4, 5}$ . Consider the following statements:

(I) The number of non-empty subsets of $S$ whose product of elements is even is $2^{5} - 2^{3}$ .

(II) The number of non-empty subsets of $S$ whose product of elements is odd is $2^{3}$ .

Which of the statements is/are true?

A.
Only (I)
B.
Only (II)
C.
Both (I) and (II)
D.
Neither (I) nor (II)

Correct Answer:

Step-by-Step Solution

Key idea: A product is odd if and only if all elements are odd. Use the complement rule.

Step 1: Identify odd numbers in $S$ . The odd numbers are ${1, 3, 5}$ , so there are 3 odd numbers.

Step 2: Count subsets with odd product. A subset has an odd product if and only if all its elements are odd. The number of non-empty subsets of ${1, 3, 5}$ is $2^{3} - 1 = 7$ . So statement (II) claims $2^{3} = 8$ , which is FALSE (it includes the empty set).

Step 3: Count subsets with even product. Total non-empty subsets of $S$ is $2^{5} - 1 = 31$ . Subsets with even product = Total - Subsets with odd product = $31 - 7 = 24$ . Statement (I) claims $2^{5} - 2^{3} = 32 - 8 = 24$ , which is TRUE.

Answer: Only (I)

Question 6 · Probability and Statistics MCQ

Let $S = {1, 2, 3}$ . Consider the following statements:

(I) The number of non-empty subsets of $S$ whose product of elements is odd is $2^{2} = 4$ .

(II) The number of non-empty subsets of $S$ whose product of elements is even is 4.

Which of the statements is/are true?

A.
Only (I)
B.
Only (II)
C.
Both (I) and (II)
D.
Neither (I) nor (II)

Correct Answer:

Step-by-Step Solution

Key idea: A subset product is odd if and only if all elements in the subset are odd. We must carefully count non-empty subsets.

Step 1: Identify the elements of $S$ . $S = {1, 2, 3}$ . The odd numbers are ${1, 3}$ (count = 2). The even number is ${2}$ (count = 1).

Step 2: Evaluate Statement (I). A subset has an odd product if it contains only odd numbers. The subsets of ${1, 3}$ are $\emptyset, {1}, {3}, {1, 3}$ . There are $2^{2} = 4$ such subsets. However, the problem specifies "non-empty" subsets. The empty set $\emptyset$ must be excluded. So, the number of non-empty subsets with an odd product is $4 - 1 = 3$ . Statement (I) claims it is 4, so it is FALSE (it ignores the non-empty constraint).

Step 3: Evaluate Statement (II). The total number of non-empty subsets of $S$ is $2^{3} - 1 = 7$ . The number of non-empty subsets with an even product is Total - Odd = $7 - 3 = 4$ . Statement (II) claims it is 4, so it is TRUE.

Answer: Only (II)

Question 7 · Probability and Statistics MCQ

Let $S = {1, 2, 3}$ . Consider the following statements regarding the subsets of $S$ :

(I) The number of subsets of $S$ whose sum of elements is even is 4.

(II) The number of non-empty subsets of $S$ whose sum of elements is even is 4.

Which of the statements is/are true?

A.
Only (I)
B.
Only (II)
C.
Both (I) and (II)
D.
Neither (I) nor (II)

Correct Answer:

Step-by-Step Solution

Key idea: We must carefully count subsets based on the parity of their sum, paying close attention to whether the empty set is included.

Step 1: List all subsets of $S = {1, 2, 3}$ and calculate their sums.

$\emptyset$ : sum = 0 (even)
${1}$ : sum = 1 (odd)
${2}$ : sum = 2 (even)
${3}$ : sum = 3 (odd)
${1, 2}$ : sum = 3 (odd)
${1, 3}$ : sum = 4 (even)
${2, 3}$ : sum = 5 (odd)
${1, 2, 3}$ : sum = 6 (even)

Step 2: Evaluate Statement (I). The subsets with an even sum are $\emptyset, {2}, {1, 3}, {1, 2, 3}$ . There are exactly 4 such subsets. Statement (I) is TRUE.

Step 3: Evaluate Statement (II). The non-empty subsets with an even sum are ${2}, {1, 3}, {1, 2, 3}$ . There are exactly 3 such subsets. Statement (II) claims there are 4, so it is FALSE (it incorrectly includes the empty set).

Answer: Only (I)

Question 8 · Probability and Statistics MCQ

Let $S = {1, 2, 3, 4}$ . Consider the following statements regarding the subsets of $S$ :

(I) The number of subsets of $S$ whose sum of elements is even is 8.

(II) The number of non-empty subsets of $S$ whose sum of elements is even is 8.

Which of the statements is/are true?

A.
Only (II)
B.
Only (I)
C.
Both (I) and (II)
D.
Neither (I) nor (II)

Correct Answer:

Step-by-Step Solution

Key idea: We must carefully count subsets based on the parity of their sum, paying close attention to whether the empty set is included.

Step 1: List all subsets of $S = {1, 2, 3, 4}$ and calculate their sums.

$\emptyset$ : sum = 0 (even)
${1}$ : 1 (odd), ${2}$ : 2 (even), ${3}$ : 3 (odd), ${4}$ : 4 (even)
${1, 2}$ : 3 (odd), ${1, 3}$ : 4 (even), ${1, 4}$ : 5 (odd), ${2, 3}$ : 5 (odd), ${2, 4}$ : 6 (even), ${3, 4}$ : 7 (odd)
${1, 2, 3}$ : 6 (even), ${1, 2, 4}$ : 7 (odd), ${1, 3, 4}$ : 8 (even), ${2, 3, 4}$ : 9 (odd)
${1, 2, 3, 4}$ : 10 (even)

Step 2: Evaluate Statement (I). The subsets with an even sum are $\emptyset, {2}, {4}, {1, 3}, {2, 4}, {1, 2, 3}, {1, 3, 4}, {1, 2, 3, 4}$ . There are exactly 8 such subsets. Statement (I) is TRUE.

Step 3: Evaluate Statement (II). The non-empty subsets with an even sum are the 8 listed above, excluding the empty set. There are exactly 7 such subsets. Statement (II) claims there are 8, so it is FALSE.

Answer: Only (I)

Question 9 · Probability and Statistics MCQ

A fair six-sided die is rolled. Let $A$ be the event of getting a prime number and $B$ be the event of getting an even number. What is the number of elements in $A \cup B$ ?

A.
4
B.
5
C.
6
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: This requires finding the union of two events using the principle of inclusion-exclusion or direct casework.

Step 1: Identify the sample space. $S = {1, 2, 3, 4, 5, 6}$ .

Step 2: Identify event $A$ (prime numbers). The primes on a die are 2, 3, 5. So $A = {2, 3, 5}$ and $n (A) = 3$ .

Step 3: Identify event $B$ (even numbers). The evens on a die are 2, 4, 6. So $B = {2, 4, 6}$ and $n (B) = 3$ .

Step 4: Find the intersection $A \cap B$ . The only number that is both prime and even is 2. So $A \cap B = {2}$ and $n (A \cap B) = 1$ .

Step 5: Apply the inclusion-exclusion formula. $n (A \cup B) = n (A) + n (B) - n (A \cap B) = 3 + 3 - 1 = 5$ .

Answer: 5

Question 10 · Probability and Statistics MCQ

A fair six-sided die is rolled twice. What is the number of outcomes in which the value of the first roll is strictly greater than the value of the second roll?

A.
10
B.
15
C.
21
D.
30

Correct Answer:

Step-by-Step Solution

Key idea: This requires systematic casework to count pairs $(a, b)$ where $a > b$ .

Step 1: Identify the sample space. Each roll has 6 outcomes, so there are $6 \times 6 = 36$ total ordered pairs.

Step 2: Casework for $a > b$ .

If $b = 1$ , $a$ can be 2, 3, 4, 5, 6 (5 outcomes).
If $b = 2$ , $a$ can be 3, 4, 5, 6 (4 outcomes).
If $b = 3$ , $a$ can be 4, 5, 6 (3 outcomes).
If $b = 4$ , $a$ can be 5, 6 (2 outcomes).
If $b = 5$ , $a$ can be 6 (1 outcome).
If $b = 6$ , $a$ cannot be strictly greater (0 outcomes).

Step 3: Sum the favorable outcomes. $5 + 4 + 3 + 2 + 1 = 15$ .

Answer: 15

basic probability events and counting notes

Chapter Roadmap: Probability, Events & Counting

1. Counting-Based Probability

2. Coin Toss & Elementary Operations

3. Conditional Probability & Independence

4. Random Variables & Distributions

The Classical Definition: Probability as a Ratio

The Real Challenge

Order Matters: Permutation vs Combination

When to Use What?

Try a question

Login to view the complete notes

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

Basic Probability, Events and Counting Notes for GATE DA

Chapter Roadmap: Probability, Events & Counting

1. Counting-Based Probability

2. Coin Toss & Elementary Operations

3. Conditional Probability & Independence

4. Random Variables & Distributions

The Classical Definition: Probability as a Ratio

The Real Challenge

Order Matters: Permutation vs Combination

When to Use What?

Strategy: The Complement Rule

Common Pattern

Basic Probability, Events and Counting: 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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