Bernoulli Trials

Master Bernoulli trials and repeated experiments for JEE - The foundation of binomial distribution and success-failure experiments

11 min read

Bernoulli Trials

The Penalty Shootout Challenge

In a penalty shootout, a soccer player takes 5 shots. She scores on each with probability 0.8 (independent of previous shots). What’s the probability she scores on the first 3 and misses the last 2?

This is a classic Bernoulli trial problem—each shot is an independent trial with two outcomes (success/failure). Understanding Bernoulli trials is the foundation for binomial distribution and crucial for JEE probability!

What is a Bernoulli Trial?

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

  • Success (with probability p)
  • Failure (with probability q = 1 - p)

Named after Swiss mathematician Jacob Bernoulli (1654-1705).

Examples of Bernoulli Trials

ExperimentSuccessFailurep
Coin tossHeadsTails0.5
Free throwScoreMiss0.7 (varies)
True/False questionCorrectWrong0.5 (guessing)
Quality testPassFail0.95 (typical)
Medical testPositiveNegativeVaries
Customer purchaseBuyDon’t buyVaries

Single Bernoulli Trial

Bernoulli Distribution

For a single Bernoulli trial, if X = 1 (success) or 0 (failure):

$$P(X = 1) = p$$ $$P(X = 0) = q = 1 - p$$

PMF:

$$P(X = x) = p^x (1-p)^{1-x}, \quad x \in \{0, 1\}$$

Mean: $E(X) = p$

Variance: $\text{Var}(X) = pq = p(1-p)$

Repeated Bernoulli Trials

Definition

Repeated Bernoulli trials (or Bernoulli process) consists of:

Properties of Repeated Bernoulli Trials
  1. Two outcomes: Each trial has only success or failure
  2. Fixed probability: P(success) = p is same for all trials
  3. Independence: Trials don’t affect each other
  4. Identical: Each trial is conducted under identical conditions

Mnemonic: TFII - Two outcomes, Fixed p, Independent, Identical

Notation

  • n = number of trials
  • p = probability of success on each trial
  • q = 1 - p = probability of failure on each trial
  • X = number of successes in n trials

Probability of Specific Sequences

For a specific sequence of successes and failures:

Probability of Specific Sequence

If a sequence has r successes and (n - r) failures in a specific order:

$$P(\text{specific sequence}) = p^r \times q^{n-r} = p^r (1-p)^{n-r}$$

Key: Order matters for specific sequences!

Example: 3 Coin Tosses

Specific sequence HHT (2 heads, 1 tail):

$$P(HHT) = (0.5)(0.5)(0.5) = (0.5)^3 = \frac{1}{8}$$

Different sequence HTH (also 2 heads, 1 tail):

$$P(HTH) = (0.5)(0.5)(0.5) = (0.5)^3 = \frac{1}{8}$$

Same probability because same number of successes and failures!

Probability of r Successes (Any Order)

When we don’t care about order, just the total number of successes:

Probability of r Successes in n Trials
$$P(X = r) = \binom{n}{r} p^r q^{n-r}$$

where:

  • $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ = number of ways to choose positions for r successes
  • $p^r$ = probability of r successes
  • $q^{n-r}$ = probability of (n-r) failures

This is the Binomial Distribution!

Why $\binom{n}{r}$?

  • Each specific sequence with r successes has probability $p^r q^{n-r}$
  • There are $\binom{n}{r}$ different sequences with r successes
  • Total probability = (number of sequences) × (probability of each sequence)

Important Probabilities

At Least r Successes

At Least r Successes
$$P(X \geq r) = \sum_{k=r}^{n} \binom{n}{k} p^k q^{n-k}$$

Alternative (often easier):

$$P(X \geq r) = 1 - P(X < r) = 1 - P(X \leq r-1)$$

At Most r Successes

At Most r Successes
$$P(X \leq r) = \sum_{k=0}^{r} \binom{n}{k} p^k q^{n-k}$$

Exactly r Successes in First m Trials

For consecutive successes at the beginning:

First m Trials

P(exactly r successes in first m trials) = $\binom{m}{r} p^r q^{m-r}$

This is independent of what happens in remaining (n - m) trials!

Question: What’s the probability that the first success occurs on the k-th trial?

Geometric Distribution
$$P(\text{first success on trial } k) = q^{k-1} \times p = (1-p)^{k-1} p$$

Interpretation: (k-1) failures, then 1 success

Expected number of trials until first success:

$$E(k) = \frac{1}{p}$$

Example: Rolling Until Getting a 6

$$P(\text{first 6 on roll 4}) = \left(\frac{5}{6}\right)^3 \times \frac{1}{6} = \frac{125}{1296}$$

Expected rolls until first 6: $E(k) = \frac{1}{1/6} = 6$ rolls

Negative Binomial Distribution (Advanced)

Question: What’s the probability that the r-th success occurs on the k-th trial?

Negative Binomial
$$P(\text{r-th success on trial } k) = \binom{k-1}{r-1} p^r q^{k-r}$$

Interpretation:

  • First (k-1) trials have exactly (r-1) successes
  • The k-th trial is a success

Interactive Visualization

Memory Tricks

Memory Tricks

1. “Bernoulli = Binary”

  • Bernoulli has Binary outcomes (2 only)

2. “TFII” for Conditions

  • Two outcomes
  • Fixed probability
  • Independent trials
  • Identical conditions

3. Specific vs General

  • Specific sequence: Just multiply probabilities → $p^r q^{n-r}$
  • Any order: Multiply by number of arrangements → $\binom{n}{r} p^r q^{n-r}$

4. “Choose positions, then probabilities”

  • First: Choose where successes occur ($\binom{n}{r}$)
  • Then: Multiply by probability of that pattern ($p^r q^{n-r}$)

5. Geometric Distribution

  • “Wait for success” → Geometric
  • Expected wait = $\frac{1}{p}$ (reciprocal of probability)

6. At Least vs At Most

  • At least r: $P(X \geq r) = 1 - P(X \leq r-1)$ (use complement!)
  • At most r: $P(X \leq r)$ (sum from 0 to r)

Common Mistakes to Avoid

Common Mistakes

  1. Forgetting independence: Can’t use Bernoulli if trials affect each other (e.g., drawing without replacement)

  2. Wrong probability for sequence: P(SSFF) = $p^2 q^2$, NOT $p + p + q + q$

  3. Confusing specific vs any order:

    • P(exactly HHT) = $(1/2)^3$ (specific sequence)
    • P(exactly 2 heads in 3 tosses) = $\binom{3}{2}(1/2)^3$ (any order)

  4. “At least” calculation: P(X ≥ 3) means 3, 4, 5, …, n (not just 3!)

  5. Changing p: If probability changes between trials, not Bernoulli!

  6. Not checking independence: Phone calls, customer arrivals might not be independent

  7. Geometric confusion: “First success on trial k” needs (k-1) failures first

Solved Examples

Example 1: Specific Sequence (Penalty Kicks)

Problem: A player takes 5 penalty kicks with P(score) = 0.8. Find probability of sequence: Score, Score, Score, Miss, Miss.

Solution:

Specific sequence SSSFF:

$$P(SSSFF) = (0.8)(0.8)(0.8)(0.2)(0.2) = (0.8)^3 (0.2)^2$$ $$= 0.512 \times 0.04 = 0.02048$$

Answer: 0.02048 or about 2.05%

Example 2: Any Order (Penalty Kicks Continued)

Problem: Same setup. Find probability of exactly 3 scores in any order.

Solution:

Number of ways to arrange 3 scores in 5 shots: $\binom{5}{3} = 10$

Each arrangement has probability $(0.8)^3 (0.2)^2$

$$P(X = 3) = \binom{5}{3} (0.8)^3 (0.2)^2 = 10 \times 0.512 \times 0.04 = 0.2048$$

Answer: 0.2048 or about 20.5%

Example 3: At Least (True/False Test)

Problem: A student guesses on 6 true/false questions. Find probability of passing (at least 4 correct).

Solution:

$p = 0.5$, $n = 6$, need $P(X \geq 4)$

Method 1 (Direct):

$$P(X \geq 4) = P(X=4) + P(X=5) + P(X=6)$$ $$P(X=4) = \binom{6}{4} (0.5)^6 = 15 \times \frac{1}{64} = \frac{15}{64}$$ $$P(X=5) = \binom{6}{5} (0.5)^6 = 6 \times \frac{1}{64} = \frac{6}{64}$$ $$P(X=6) = \binom{6}{6} (0.5)^6 = 1 \times \frac{1}{64} = \frac{1}{64}$$ $$P(X \geq 4) = \frac{15 + 6 + 1}{64} = \frac{22}{64} = \frac{11}{32}$$

Method 2 (Complement):

$$P(X \geq 4) = 1 - P(X \leq 3)$$ $$P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$$ $$= \frac{1 + 6 + 15 + 20}{64} = \frac{42}{64}$$ $$P(X \geq 4) = 1 - \frac{42}{64} = \frac{22}{64} = \frac{11}{32}$$

Answer: 11/32 ≈ 0.344 or about 34.4%

Example 4: Geometric Distribution (Rolling a Die)

Problem: A die is rolled repeatedly. Find: (a) P(first 6 on roll 3) (b) P(first 6 within 3 rolls) (c) Expected rolls until first 6

Solution:

(a) First 6 on roll 3: Need non-6, non-6, then 6

$$P(\text{first 6 on roll 3}) = \left(\frac{5}{6}\right)^2 \times \frac{1}{6} = \frac{25}{36} \times \frac{1}{6} = \frac{25}{216}$$

(b) First 6 within 3 rolls: P(on roll 1, 2, or 3)

$$P = \frac{1}{6} + \frac{5}{6} \times \frac{1}{6} + \left(\frac{5}{6}\right)^2 \times \frac{1}{6}$$ $$= \frac{1}{6}\left[1 + \frac{5}{6} + \frac{25}{36}\right] = \frac{1}{6} \times \frac{36 + 30 + 25}{36} = \frac{91}{216}$$

Alternative (complement): 1 - P(no 6 in first 3)

$$= 1 - \left(\frac{5}{6}\right)^3 = 1 - \frac{125}{216} = \frac{91}{216}$$

(c) Expected rolls:

$$E = \frac{1}{p} = \frac{1}{1/6} = 6 \text{ rolls}$$

Example 5: First m Trials (JEE Pattern)

Problem: A coin is tossed 10 times. Find probability of: (a) Exactly 3 heads in first 5 tosses (b) Exactly 3 heads in first 5 tosses AND exactly 2 heads in last 5 tosses

Solution:

(a) First 5 tosses only:

Treat as binomial with n = 5, r = 3, p = 0.5

$$P = \binom{5}{3} (0.5)^5 = 10 \times \frac{1}{32} = \frac{10}{32} = \frac{5}{16}$$

(What happens in tosses 6-10 doesn’t matter!)

(b) Both conditions:

Events are independent (different tosses), so multiply:

$$P(\text{3 in first 5}) = \frac{5}{16}$$ $$P(\text{2 in last 5}) = \binom{5}{2} (0.5)^5 = 10 \times \frac{1}{32} = \frac{10}{32} = \frac{5}{16}$$ $$P(\text{both}) = \frac{5}{16} \times \frac{5}{16} = \frac{25}{256}$$

Example 6: Negative Binomial (Advanced)

Problem: A basketball player has 60% free throw success rate. Find probability that her 3rd successful shot is on her 5th attempt.

Solution:

Need exactly 2 successes in first 4 attempts, then success on 5th:

$$P = \binom{4}{2} (0.6)^2 (0.4)^2 \times 0.6$$ $$= 6 \times 0.36 \times 0.16 \times 0.6 = 6 \times 0.03456 = 0.20736$$

Alternative (negative binomial formula):

$$P = \binom{4}{2} (0.6)^3 (0.4)^2 = 6 \times 0.216 \times 0.16 = 0.20736$$

Answer: About 20.7%

Practice Problems

Level 1: Foundation (JEE Main)

  1. A fair coin is tossed 4 times. Find P(exactly HTHT in that order).

  2. A die is rolled 3 times. Find P(getting three 6’s).

  3. For 5 Bernoulli trials with p = 0.3, find P(exactly 2 successes).

Solutions

  1. P(HTHT) = (1/2)⁴ = 1/16

  2. P(666) = (1/6)³ = 1/216

  3. P(X=2) = C(5,2) × (0.3)² × (0.7)³ = 10 × 0.09 × 0.343 = 0.3087

Level 2: Intermediate (JEE Main/Advanced)

  1. A student answers 10 multiple choice questions (4 options each) by guessing. Find: (a) P(exactly 5 correct) (b) P(at least 8 correct)

  2. A basketball player makes 70% of free throws. She takes 6 shots. Find P(makes first 3, misses next 2, makes last 1).

  3. Rolling a die repeatedly, find P(first 1 or 2 appears on 4th roll).

Solutions

  1. p = 1/4, n = 10 (a) P(X=5) = C(10,5) × (1/4)⁵ × (3/4)⁵ ≈ 0.0584 (b) P(X≥8) = P(X=8) + P(X=9) + P(X=10) ≈ 0.0004

  2. P = (0.7)³ × (0.3)² × 0.7 = (0.7)⁴ × (0.3)² = 0.2401 × 0.09 = 0.0216

  3. P(success) = 2/6 = 1/3, need 3 failures then success P = (2/3)³ × (1/3) = 8/81

Level 3: Advanced (JEE Advanced)

  1. In n Bernoulli trials with probability p, show that P(even number of successes) = $\frac{1 + (1-2p)^n}{2}$.

  2. A coin is tossed until 2 heads appear. Find the probability this takes exactly 5 tosses.

  3. Prove that for Bernoulli trials, the mode (most likely number of successes) is $\lfloor (n+1)p \rfloor$.

Solutions

  1. P(even) = Σ C(n,2k) p²ᵏ q^(n-2k) for k=0,1,2,… Using binomial theorem: (p+q)ⁿ = 1 and (p-q)ⁿ = (1-2p)ⁿ Add: 2×P(even) = 1 + (1-2p)ⁿ → P(even) = [1 + (1-2p)ⁿ]/2 ✓

  2. Need exactly 1 head in first 4 tosses, then head on 5th: P = C(4,1) × (1/2)⁴ × (1/2) = 4 × (1/2)⁵ = 4/32 = 1/8

  3. P(X=k) is maximum when C(n,k)pᵏq^(n-k) is maximum. Consider ratio: P(X=k)/P(X=k-1) = [(n-k+1)/k] × [p/q] P(X=k) > P(X=k-1) when (n-k+1)p > kq → k < (n+1)p Maximum when k = floor[(n+1)p] ✓

Real-Life Applications

1. Quality Control

  • Test n items independently
  • Each pass/fail with probability p
  • Calculate defect probabilities

2. Sports Analytics

  • Free throws, penalty kicks (independent attempts)
  • Batting average (hits vs at-bats)
  • Win streaks analysis

3. Medical Testing

  • n patients receive treatment
  • Each recovers with probability p
  • Predict success rates

4. Marketing

  • n customers contacted
  • Each buys with probability p
  • Sales forecasting

5. Reliability Engineering

  • n components in system
  • Each fails with probability q
  • System reliability calculation

6. Games and Gambling

  • Repeated plays of game
  • Calculate winning probabilities
  • Expected value analysis

Connection to Other Topics

Cross-Topic Links

Related JEE Topics:

Key Connections:

  • Single Bernoulli trial → Bernoulli distribution
  • n Bernoulli trials → Binomial distribution
  • Wait for success → Geometric distribution
  • Wait for r successes → Negative binomial distribution

Formula Quick Reference

Essential Formulas

Single Bernoulli Trial

  • P(success) = p, P(failure) = q = 1 - p
  • E(X) = p, Var(X) = pq

Specific Sequence (n trials)

  • r successes, (n-r) failures: $p^r q^{n-r}$

Any Order (n trials, r successes)

  • $P(X = r) = \binom{n}{r} p^r q^{n-r}$ (Binomial)

Geometric Distribution

  • First success on trial k: $(1-p)^{k-1} p$
  • Expected trials: $E = 1/p$

Negative Binomial

  • r-th success on trial k: $\binom{k-1}{r-1} p^r (1-p)^{k-r}$

Quick Revision Checklist

Quick Revision
✓ Bernoulli trial: Two outcomes (success/failure), fixed p ✓ TFII conditions: Two outcomes, Fixed p, Independent, Identical ✓ Specific sequence: Just multiply → $p^r q^{n-r}$ ✓ Any order: Choose positions + multiply → $\binom{n}{r} p^r q^{n-r}$ ✓ Independence: Crucial! Can’t use Bernoulli without it ✓ Geometric: Wait for first success, E = 1/p ✓ n Bernoulli trials = Binomial distribution ✓ At least r: Use complement often easier

Exam Tips

  1. Check independence: First verify trials are independent
  2. Specific vs general: Does order matter? If yes, no $\binom{n}{r}$
  3. “At least”: Consider using complement (1 - P(less than))
  4. Geometric shortcut: For “first success”, use $(1-p)^{k-1}p$ directly
  5. Split problems: “First m trials and last n-m trials” → multiply probabilities
  6. Verify p + q = 1: Common check to avoid errors
  7. Calculator: Know C(n,r) function on your calculator

Summary

Bernoulli trials are the building blocks of many probability problems:

  • One trial → Bernoulli distribution
  • Count successes in n trials → Binomial distribution
  • Wait for first success → Geometric distribution
  • Wait for r-th success → Negative binomial

Master Bernoulli trials, and you’ve mastered the foundation of discrete probability!

Next Steps

You’ve completed the Statistics & Probability chapter! Review and consolidate:

Last updated: December 22, 2025 Perfect your Bernoulli trials at JEENotes Practice Portal