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Combinatorial Probability and Independent Events Short Notes for GATE CS

Combinatorial Probability and Independent Events short notes for GATE CS: 3 study cards covering concepts, formulas, shortcuts and exam traps, plus solved pra

combinatorial probability and independent events short notes

Quick Revision: Binomial Models

Summary Checklist Conditions: Independent trials, Binary outcomes, Constant p . Formula: P (X = k) = (k n) p^{k} (1 - p)^{n - k} Parameters: n : Total trials k : Desired successes p : Probability of success Common Pitfall: Do not use for "without replacement" cases. Property: \sum_{k = 0}^{n} P (X = k) = 1 Exam Tip: If p = 0.5, the distribution is symmetric. (k n) (0.5)^{n} .

Classical Counting Checklist

Summary Checklist Equally Likely: Verify your sample space outcomes have equal probability. Distinct Objects: Always treat multiple dice/coins as distinct (Sample space = 6^{n}, not 21). Complementary Counting: For "at least one/two", calculate 1 - P (all distinct) . Replacement Check: With replacement \to N^{k} . Without replacement \to^{N} P_{k} . Algebraic Sets: Solve the constraint equation first to define the exact elements of S . Final Formula: P (A) = \frac{Favorable Outcomes}{Total Equally Likely Outcomes}

Exam Day Checklist: Event Algebra

Exam Day Checklist: Event Algebra Mutually Exclusive vs Independent: Mutually exclusive means P (A \cap B) = 0 . Independent means P (A \cap B) = P (A) P (B) . They are mutually exclusive ONLY if they cannot be independent (assuming non-zero probabilities). "At Least One" Problems: Always calculate 1 - P (none) . Do not sum the individual cases. Given P (A), P (B), P (A \cup B) : Step 1: Find P (A \cap B) = P (A) + P (B) - P (A \cup B) . Step 2: Use this to find any conditional probability or check independence. Permutation Symmetry: If Event 1 depends on subset X and Event 2 depends on disjoint subset Y, they are independent. Multiply their individual probabilities. Three Events Independence: Pairwise independence \neq = Mutual independence. You must verify all 4 conditions (3 pairs + 1 triple intersection).

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Question 1

Level 1: Warm-up

A fair coin is tossed 4 times independently. What is the probability of obtaining exactly 2 heads?

Question 2

Level 1: Warm-up

Assertion (A): If P (A \cap B) = P (A) \times P (B), then events A and B are independent. Reason (R): Independent events cannot occur at the same time.

Question 3

Level 1: Warm-up

Assertion (A): If events A and B are independent, then P (A ∣ B) = P (A) . Reason (R): For independent events, P (A \cap B) = P (A) + P (B) .

Question 4

Level 1: Warm-up

A test has 3 multiple-choice questions. Each question has 4 options, and a student guesses randomly on all questions. What is the probability of getting at least 2 correct?

Question 5

Level 1: Warm-up

A bag contains 3 red and 2 blue balls. Balls are drawn one by one without replacement. Let X be the number of red balls drawn in 3 draws. Which of the following statements about X is true?

Question 6

Level 1: Warm-up

For a binomial distribution with n = 4 and p = 0.5, rank the probabilities P (X = 0), P (X = 2), and P (X = 4) in ascending order.

Question 7

Level 1: Warm-up

Assertion (A): The probability of getting a sum of 7 when two unbiased dice are rolled is 1/12 . Reason (R): There are 6 favorable outcomes for a sum of 7 out of 36 total outcomes.

Question 8

Level 1: Warm-up

When six unbiased dice are rolled simultaneously, how many favorable outcomes are there for the event that all six dice show distinct numbers?

Question 9

Level 1: Warm-up

Assertion (A): The probability of getting a sum of 6 when two unbiased dice are rolled is 5%. Reason (R): There are 5 favorable outcomes out of 36 total outcomes.

Question 10

Level 1: Warm-up

When n unbiased dice are rolled simultaneously, the number of favorable outcomes for the event that all n dice show distinct numbers is exactly 720. What is the value of n ?

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Combinatorial Probability and Independent Events Short Notes for GATE CS

Combinatorial Probability and Independent Events short notes for GATE CS: 3 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Quick Revision: Binomial Models

Summary Checklist

Conditions: Independent trials, Binary outcomes, Constant $p$ .
Formula:
$P (X = k) = (k n) p^{k} (1 - p)^{n - k}$
Parameters:
- $n$ : Total trials
- $k$ : Desired successes
- $p$ : Probability of success
Common Pitfall: Do not use for "without replacement" cases.
Property: $\sum_{k = 0}^{n} P (X = k) = 1$

Exam Tip: If

p = 0.5

, the distribution is symmetric.

(k n) (0.5)^{n}

Classical Counting Checklist

Summary Checklist

Equally Likely: Verify your sample space outcomes have equal probability.
Distinct Objects: Always treat multiple dice/coins as distinct (Sample space = $6^{n}$ , not 21).
Complementary Counting: For "at least one/two", calculate $1 - P (all distinct)$ .
Replacement Check: With replacement $\to N^{k}$ . Without replacement $\to^{N} P_{k}$ .
Algebraic Sets: Solve the constraint equation first to define the exact elements of $S$ .

Final Formula:

P (A) = \frac{Favorable Outcomes}{Total Equally Likely Outcomes}

Exam Day Checklist: Event Algebra

Mutually Exclusive vs Independent: Mutually exclusive means $P (A \cap B) = 0$ . Independent means $P (A \cap B) = P (A) P (B)$ . They are mutually exclusive ONLY if they cannot be independent (assuming non-zero probabilities).
"At Least One" Problems: Always calculate $1 - P (none)$ . Do not sum the individual cases.
Given $P (A), P (B), P (A \cup B)$ : Step 1: Find $P (A \cap B) = P (A) + P (B) - P (A \cup B)$ . Step 2: Use this to find any conditional probability or check independence.
Permutation Symmetry: If Event 1 depends on subset $X$ and Event 2 depends on disjoint subset $Y$ , they are independent. Multiply their individual probabilities.
Three Events Independence: Pairwise independence $\neq =$ Mutual independence. You must verify all 4 conditions (3 pairs + 1 triple intersection).

Combinatorial Probability and Independent Events: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Engineering Mathematics MCQ

A fair coin is tossed 4 times independently. What is the probability of obtaining exactly 2 heads?

A.
$\frac{1}{16}$
B.
$\frac{3}{8}$
C.
$\frac{1}{4}$
D.
$\frac{1}{2}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct binomial probability calculation with $n = 4$ , $k = 2$ , $p = 0.5$ .

Step 1: Identify parameters: $n = 4$ (tosses), $k = 2$ (heads), $p = 0.5$ (probability of heads).

Step 2: Apply binomial formula: $P (X = 2) = (2 4) (0.5)^{2} (0.5)^{4 - 2}$

Step 3: Calculate binomial coefficient: $(2 4) = \frac{4 !}{2 ! 2 !} = 6$

Step 4: Calculate probability: $P (X = 2) = 6 \times (0.5)^{2} \times (0.5)^{2} = 6 \times 0.25 \times 0.25 = 6 \times 0.0625 = 0.375 = \frac{3}{8}$

Answer: $\frac{3}{8}$

Question 2 · Engineering Mathematics MCQ

Assertion (A): If

P (A \cap B) = P (A) \times P (B)

, then events

A

and

B

are independent.
Reason (R): Independent events cannot occur at the same time.

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

Correct Answer:

Step-by-Step Solution

Key idea: This is an assertion-reason problem testing the definition of independence versus mutual exclusivity.

Step 1: Evaluate Assertion (A): $P (A \cap B) = P (A) \times P (B)$ is the mathematical definition of independence. So A is TRUE.

Step 2: Evaluate Reason (R): "Independent events cannot occur at the same time" describes mutually exclusive events, not independent events.

Mutually exclusive: $P (A \cap B) = 0$ (cannot occur together)
Independent: $P (A \cap B) = P (A) P (B)$ (occurrence of one doesn't affect the other)
So R is FALSE.

Step 3: In fact, if $A$ and $B$ are independent with $P (A) > 0$ and $P (B) > 0$ , then $P (A \cap B) = P (A) P (B) > 0$ , meaning they CAN occur together.

Answer: A is true but R is false.

Common trap: Students confuse independence with mutual exclusivity, thinking independent events cannot happen together.

Question 3 · Engineering Mathematics MCQ

Assertion (A): If events

A

and

B

are independent, then

P (A ∣ B) = P (A)

.
Reason (R): For independent events,

P (A \cap B) = P (A) + P (B)

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

Correct Answer:

Step-by-Step Solution

Key idea: This is an assertion-reason problem testing the correct formula for independent events.

Step 1: Evaluate Assertion (A): Independence means $P (A ∣ B) = P (A)$ by definition. So A is TRUE.

Step 2: Evaluate Reason (R): For independent events, $P (A \cap B) = P (A) P (B)$ , not $P (A) + P (B)$ . The formula $P (A) + P (B)$ applies to mutually exclusive events' union, not independent events' intersection. So R is FALSE.

Step 3: Since A is true and R is false, the answer is "A is true but R is false."

Answer: A is true but R is false.

Common trap: Students confuse the independence formula $P (A \cap B) = P (A) P (B)$ with the mutually exclusive union formula $P (A \cup B) = P (A) + P (B)$ .

Question 4 · Engineering Mathematics MCQ

A test has 3 multiple-choice questions. Each question has 4 options, and a student guesses randomly on all questions. What is the probability of getting at least 2 correct?

A.
$\frac{5}{32}$
B.
$\frac{9}{64}$
C.
$\frac{1}{64}$
D.
$\frac{27}{64}$

Correct Answer:

Step-by-Step Solution

Key idea: "At least 2" means we need casework: $P (X \geq 2) = P (X = 2) + P (X = 3)$ .

Step 1: Identify parameters: $n = 3$ (questions), $p = \frac{1}{4} = 0.25$ (probability of guessing correctly).

Step 2: Calculate $P (X = 2)$ :

$P (X = 2) = (2 3) (0.25)^{2} (0.75)^{1} = 3 \times 0.0625 \times 0.75 = 0.140625 = \frac{9}{64}$

Step 3: Calculate $P (X = 3)$ :

$P (X = 3) = (3 3) (0.25)^{3} (0.75)^{0} = 1 \times 0.015625 \times 1 = 0.015625 = \frac{1}{64}$

Step 4: Add the cases:

$P (X \geq 2) = \frac{9}{64} + \frac{1}{64} = \frac{10}{64} = \frac{5}{32}$

Answer: $\frac{5}{32}$

Question 5 · Engineering Mathematics MCQ

A bag contains 3 red and 2 blue balls. Balls are drawn one by one without replacement. Let $X$ be the number of red balls drawn in 3 draws. Which of the following statements about $X$ is true?

A.
$X$ follows a binomial distribution
B.
$P (X = 2) = (2 3) (\frac{3}{5})^{2} (\frac{2}{5})$
C.
$X$ can take values 0, 1, 2, 3
D.
The draws are independent trials

Correct Answer:

Step-by-Step Solution

Key idea: Drawing without replacement violates the independence assumption required for binomial distribution. Step 1: Analyze the setup: 3 red, 2 blue balls, drawing 3 without replacement. Step 2: Check binomial conditions: - Fixed number of trials: Yes,

n = 3

draws - Binary outcome: Yes, red or blue - Independence: No - drawing without replacement changes probabilities - Constant

p

: No - probability of red changes after each draw Step 3: Evaluate each option: - Option A: False - not binomial because trials are not independent - Option B: False - this formula assumes binomial with constant

p = 3/5

, but

p

changes - Option C: True -

X

can be 0, 1, 2, or 3 red balls (we draw 3 balls total) - Option D: False - draws are dependent (without replacement) Answer:

X

can take values 0, 1, 2, 3

Question 6 · Engineering Mathematics MCQ

For a binomial distribution with $n = 4$ and $p = 0.5$ , rank the probabilities $P (X = 0)$ , $P (X = 2)$ , and $P (X = 4)$ in ascending order.

A.
$P (X = 0) < P (X = 4) < P (X = 2)$
B.
$P (X = 0) = P (X = 4) < P (X = 2)$
C.
$P (X = 2) < P (X = 0) < P (X = 4)$
D.
$P (X = 4) < P (X = 2) < P (X = 0)$

Correct Answer:

Step-by-Step Solution

Key idea: When $p = 0.5$ , the binomial distribution is symmetric about $n /2$ .

Step 1: Calculate $P (X = 0) = (0 4) (0.5)^{0} (0.5)^{4} = 1/16$ .

Step 2: Calculate $P (X = 4) = (4 4) (0.5)^{4} (0.5)^{0} = 1/16$ .

Step 3: Calculate $P (X = 2) = (2 4) (0.5)^{2} (0.5)^{2} = 6/16$ .

Step 4: Compare the values: $1/16 = 1/16 < 6/16$ .

Answer: $P (X = 0) = P (X = 4) < P (X = 2)$ .

Question 7 · Engineering Mathematics MCQ

Assertion (A): The probability of getting a sum of 7 when two unbiased dice are rolled is $1/12$ .

Reason (R): There are 6 favorable outcomes for a sum of 7 out of 36 total outcomes.

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.

Correct Answer:

Step-by-Step Solution

Key idea: Verify the assertion by calculating the probability from the favorable and total outcomes given in the reason.

Step 1: Check Reason (R): For two dice, total outcomes = 36. Favorable for sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 outcomes. R is true.

Step 2: Check Assertion (A): Probability = Favorable / Total = 6 / 36 = 1/6.

Step 3: Compare A with the calculated value: A claims 1/12, but the true value is 1/6. So A is false.

Answer: A is false but R is true.

Question 8 · Engineering Mathematics MCQ

When six unbiased dice are rolled simultaneously, how many favorable outcomes are there for the event that all six dice show distinct numbers?

A.
6
B.
36
C.
720
D.
46656

Correct Answer:

Step-by-Step Solution

Key idea: "All distinct" for 6 dice means we are arranging the 6 unique faces. This is a permutation of 6 items.

Step 1: Identify the condition: 6 dice, all showing different numbers (1, 2, 3, 4, 5, 6).

Step 2: Calculate favorable outcomes: This is the number of ways to arrange 6 distinct items, which is $6!$ .

Step 3: Compute $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ .

Answer: 720.

Question 9 · Engineering Mathematics MCQ

Assertion (A): The probability of getting a sum of 6 when two unbiased dice are rolled is 5%.

Reason (R): There are 5 favorable outcomes out of 36 total outcomes.

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.

Correct Answer:

Step-by-Step Solution

Key idea: Verify the assertion by calculating the probability from the favorable and total outcomes.

Step 1: Check Reason (R): For two dice, total outcomes = 36. Favorable for sum 6: (1,5), (2,4), (3,3), (4,2), (5,1). There are 5 outcomes. R is true.

Step 2: Check Assertion (A): Probability = Favorable / Total = 5 / 36.

Step 3: Convert 5/36 to a percentage: $5/36 \approx 0.1388 = 13.88%$ .

Step 4: Compare A with the calculated value: A claims 5%, but the true value is 13.88%. So A is false.

Answer: A is false but R is true.

Question 10 · Engineering Mathematics MCQ

When $n$ unbiased dice are rolled simultaneously, the number of favorable outcomes for the event that all $n$ dice show distinct numbers is exactly 720. What is the value of $n$ ?

A.
4
B.
5
C.
6
D.
7

Correct Answer:

Step-by-Step Solution

Key idea: The number of favorable outcomes for all distinct is $6! / (6 - n)!$ .

Step 1: We are given the favorable outcomes = 720.

Step 2: Recognize that $720 = 6!$ .

Step 3: This means we are using all 6 faces, so $n = 6$ .

Answer: 6.

combinatorial probability and independent events short notes

Quick Revision: Binomial Models

Summary Checklist

Classical Counting Checklist

Summary Checklist

Exam Day Checklist: Event Algebra

Exam Day Checklist: Event Algebra

Try a question

Login to view the complete short 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

Combinatorial Probability and Independent Events Short Notes for GATE CS

Quick Revision: Binomial Models

Summary Checklist

Classical Counting Checklist

Summary Checklist

Exam Day Checklist: Event Algebra

Exam Day Checklist: Event Algebra

Combinatorial Probability and Independent Events: 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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