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Combinatorial Probability and Independent Events Notes for GATE CS

Combinatorial Probability and Independent Events notes for GATE CS: 30 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice

combinatorial probability and independent events notes

Chapter Roadmap: Probability and Statistics

Chapter Journey: Probability and Statistics 1 Binomial Models & Repeated Bernoulli Trials Foundation of discrete distributions. Focus on fixed trials and binary outcomes. 2 Classical Counting Probability Permutations, combinations, and sample spaces. Focus on equally likely outcomes. 3 Event Algebra and Independence Union, intersection, conditional probability, and Bayes theorem. Goal: Master the transition from simple counting to complex event dependencies.

The Bernoulli Trial: Binary Randomness

What is a Bernoulli Trial? A Bernoulli trial is a random experiment with exactly two possible outcomes: Success (S) Probability p Failure (F) Probability q = 1 - p Key Characteristics Binary Outcome: Only two results are possible. Fixed Probability: p remains constant for every trial. Independence: The outcome of one trial does not affect the next. Intuition: Think of a light switch. It is either ON or OFF. That is a Bernoulli state.

Repeated Bernoulli Trials

Repeated Bernoulli Trials When we perform n independent Bernoulli trials, we establish a specific structure: Fixed Number of Trials: n is predetermined. Independence: Outcome of trial i does not influence trial j . Constant Probability: p is the same for all trials. Sequence Example (n = 3) Possible outcomes for 3 coin tosses (H=Success, T=Failure): HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Each specific sequence has a probability calculated by multiplying individual probabilities due to independence. P (H H T) = p \cdot p \cdot q = p^{2} q

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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 Notes for GATE CS

Combinatorial Probability and Independent Events notes for GATE CS: 30 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Probability and Statistics

Chapter Journey: Probability and Statistics

Binomial Models & Repeated Bernoulli Trials

Foundation of discrete distributions. Focus on fixed trials and binary outcomes.

Classical Counting Probability

Permutations, combinations, and sample spaces. Focus on equally likely outcomes.

Event Algebra and Independence

Union, intersection, conditional probability, and Bayes theorem.

Goal: Master the transition from simple counting to complex event dependencies.

The Bernoulli Trial: Binary Randomness

What is a Bernoulli Trial?

A Bernoulli trial is a random experiment with exactly two possible outcomes:

Success (

S

)

Probability

p

Failure (

F

)

Probability

q = 1 - p

Key Characteristics

Binary Outcome: Only two results are possible.
Fixed Probability: $p$ remains constant for every trial.
Independence: The outcome of one trial does not affect the next.

Intuition: Think of a light switch. It is either ON or OFF. That is a Bernoulli state.

Repeated Bernoulli Trials

When we perform $n$ independent Bernoulli trials, we establish a specific structure:

Fixed Number of Trials: $n$ is predetermined.
Independence: Outcome of trial $i$ does not influence trial $j$ .
Constant Probability: $p$ is the same for all trials.

Sequence Example ( $n = 3$ )

Possible outcomes for 3 coin tosses (H=Success, T=Failure):

    HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
  

Each specific sequence has a probability calculated by multiplying individual probabilities due to independence.

P (H H T) = p \cdot p \cdot q = p^{2} q

Example: Rolling a Die Thrice

Problem: Rolling a Die Thrice

A fair six-faced die is rolled thrice. What is the probability of rolling a 6 exactly once?

Identify Parameters

$n = 3$ (3 rolls)
$k = 1$ (exactly one 6)
$p = \frac{1}{6}$ (probability of rolling a 6)
$q = 1 - p = \frac{5}{6}$ (probability of not rolling a 6)

Apply Binomial Formula

P (X = 1) = (1 3) (\frac{1}{6})^{1} (\frac{5}{6})^{3 - 1}

Calculate Terms

$(1 3) = 3$
$(\frac{1}{6})^{1} = \frac{1}{6}$
$(\frac{5}{6})^{2} = \frac{25}{36}$

P (X = 1) = 3 \times \frac{1}{6} \times \frac{25}{36} = \frac{75}{216} = \frac{25}{72}

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 notes

Chapter Roadmap: Probability and Statistics

Chapter Journey: Probability and Statistics

The Bernoulli Trial: Binary Randomness

What is a Bernoulli Trial?

Key Characteristics

Repeated Bernoulli Trials

Repeated Bernoulli Trials

Sequence Example (n=3)

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

Combinatorial Probability and Independent Events Notes for GATE CS

Chapter Roadmap: Probability and Statistics

Chapter Journey: Probability and Statistics

The Bernoulli Trial: Binary Randomness

What is a Bernoulli Trial?

Key Characteristics

Repeated Bernoulli Trials

Repeated Bernoulli Trials

Sequence Example (n=3)

Example: Rolling a Die Thrice

Problem: Rolling a Die Thrice

Identify Parameters

Apply Binomial Formula

Calculate Terms

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

More notes in this unit

Sequence Example ( $n = 3$ )

Sequence Example ( $n = 3$ )