Probability Distributions
The Free Throw Challenge
A basketball player makes 80% of free throws. What’s the probability she makes exactly 7 out of 10 attempts? At least 8?
This isn’t just a random question—it follows the Binomial Distribution, one of the most important probability distributions! Understanding distributions lets you model exams (how many questions will you get right?), defects in manufacturing, meteor strikes, and even radioactive decay!
What is a Probability Distribution?
A probability distribution describes how probabilities are distributed over the values of a random variable.
Why Study Standard Distributions?
Many real-world scenarios follow the same patterns:
- Coin tosses, exam questions → Binomial
- Rare events (earthquakes, typos) → Poisson
- Measurements (height, IQ) → Normal (not in JEE syllabus but good to know)
Instead of calculating from scratch each time, we use standard formulas!
Binomial Distribution
The Setup
Binomial experiment has these properties:
- Binary outcomes: Each trial has only 2 outcomes (success/failure, yes/no, heads/tails)
- Independent trials: Trials don’t affect each other
- Number fixed: Number of trials n is fixed in advance
- Same probability: Probability of success p is same for each trial
Mnemonic: BINS - Binary, Independent, Number fixed, Same probability
Examples:
- Tossing coin 10 times (n=10, p=0.5)
- 20 multiple choice questions with random guessing (n=20, p=0.25 if 4 choices)
- Testing 100 products (n=100, p=defect rate)
Not Binomial:
- Drawing cards without replacement (not independent)
- Variable number of trials
- Probability changes between trials
Binomial Distribution Formula
Let X = number of successes in n trials
where:
- $n$ = number of trials
- $r$ = number of successes (0, 1, 2, …, n)
- $p$ = probability of success on each trial
- $q = 1 - p$ = probability of failure
- $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ = number of ways to arrange r successes
Notation: $X \sim B(n, p)$ or $X \sim \text{Binomial}(n, p)$
Understanding the Formula
Three parts:
- $\binom{n}{r}$ - In how many ways can r successes occur among n trials?
- $p^r$ - Probability of r successes
- $q^{n-r}$ - Probability of (n-r) failures
Example: 3 heads in 5 tosses
- Ways to arrange HHH__ = $\binom{5}{3} = 10$
- Probability of specific sequence (e.g., HHHTT) = $(0.5)^3 (0.5)^2 = (0.5)^5$
- Total: $10 \times (0.5)^5 = 10/32 = 5/16$
Mean and Variance of Binomial
Mean (Expected Value):
$$E(X) = \mu = np$$Variance:
$$\text{Var}(X) = \sigma^2 = npq = np(1-p)$$Standard Deviation:
$$\sigma = \sqrt{npq}$$Intuition:
- If p=0.5 and n=100, expect 50 successes on average
- Variance is maximum when p=0.5 (most uncertain)
Properties of Binomial Distribution
Range: X can take values 0, 1, 2, …, n
Symmetry:
- If p = 0.5, distribution is symmetric
- If p < 0.5, skewed right (tail toward higher values)
- If p > 0.5, skewed left
Sum of probabilities:
$$\sum_{r=0}^{n} P(X = r) = 1$$(This comes from binomial theorem: $(p+q)^n = 1^n = 1$)
Mode (most likely value):
- If (n+1)p is integer: modes are (n+1)p and (n+1)p - 1
- Otherwise: mode = floor[(n+1)p]
Sum of binomials: If $X_1 \sim B(n_1, p)$ and $X_2 \sim B(n_2, p)$ are independent,
$$X_1 + X_2 \sim B(n_1 + n_2, p)$$
Poisson Distribution
The Setup
Poisson distribution models the number of rare events occurring in a fixed interval of time or space.
- Events occur randomly and independently
- Events occur at a constant average rate λ (lambda)
- Two events cannot occur at exactly the same instant
- The number of events in disjoint intervals are independent
Examples:
- Number of typos per page (λ = 2 per page)
- Number of calls to a help desk per hour (λ = 15 per hour)
- Number of meteors hitting Earth per year
- Number of students arriving at library per minute
- Number of goals in a football match
Not Poisson:
- If events are clustered (not random)
- If rate changes over time
- If events affect each other
Poisson Distribution Formula
Let X = number of events in the interval
where:
- $\lambda$ (lambda) = average number of events in the interval
- $e \approx 2.71828$ (Euler’s number)
- $r$ = actual number of events (0, 1, 2, 3, …)
Notation: $X \sim \text{Poisson}(\lambda)$ or $X \sim P(\lambda)$
Understanding the Formula
- $e^{-\lambda}$ - Normalizing constant (ensures probabilities sum to 1)
- $\lambda^r$ - Higher λ → more likely to get higher r
- $r!$ - Divides to account for arrangements
Mean and Variance of Poisson
Mean:
$$E(X) = \mu = \lambda$$Variance:
$$\text{Var}(X) = \sigma^2 = \lambda$$Standard Deviation:
$$\sigma = \sqrt{\lambda}$$Key property: Mean = Variance (unique to Poisson!)
Properties of Poisson Distribution
Range: X can take values 0, 1, 2, 3, … (infinite, but probabilities get very small)
Sum of probabilities:
$$\sum_{r=0}^{\infty} P(X = r) = 1$$Mode:
- If λ is integer: modes are λ and λ-1
- Otherwise: mode = floor[λ]
Additivity: If $X_1 \sim P(\lambda_1)$ and $X_2 \sim P(\lambda_2)$ are independent,
$$X_1 + X_2 \sim P(\lambda_1 + \lambda_2)$$Limiting case: Binomial(n, p) → Poisson(λ) as n → ∞, p → 0, with np = λ (When n is large, p is small, use Poisson to approximate Binomial!)
Poisson as Approximation to Binomial
When n is large (n > 20) and p is small (p < 0.05), Binomial is hard to compute. Use Poisson!
If $X \sim B(n, p)$ with large n and small p:
$$X \approx \text{Poisson}(\lambda = np)$$Rule of thumb: Use when n ≥ 20, p ≤ 0.05, and np < 5
Example: Defect rate is 0.2%. In batch of 1000, probability of exactly 3 defects?
- Exact: $\binom{1000}{3} (0.002)^3 (0.998)^{997}$ (very hard!)
- Poisson: λ = 1000 × 0.002 = 2, so $P(X=3) = \frac{e^{-2} \cdot 2^3}{3!}$ (easy!)
Interactive Visualization
Memory Tricks
Binomial
“BINS” for conditions: Binary, Independent, Number fixed, Same probability
Formula: “Choose-Power-Power”
- Choose r from n: $\binom{n}{r}$
- Power of p: $p^r$
- Power of q: $q^{n-r}$
Mean = n × p: “If you try n times with probability p, expect np successes”
Variance = npq: “Add q to mean formula and multiply”
Poisson
“Rare events, Poisson appears”
Formula: “e-Power-Factorial”
- e to the -λ: $e^{-\lambda}$
- Power of λ: $\lambda^r$
- Factorial: $r!$
Mean = Variance = λ: Only distribution where these are equal!
Lambda is everything: Just need one parameter λ
Comparison
- Binomial: n trials, count successes
- Poisson: time/space interval, count rare events
- Approximation: Big n, small p → Use Poisson (easier!)
Common Mistakes to Avoid
Using Binomial when trials aren’t independent: E.g., drawing without replacement
Confusing n and r: n = total trials, r = number of successes
Forgetting $\binom{n}{r}$: Can’t just do $p^r q^{n-r}$
Wrong probability: “At least r” means P(X ≥ r) = 1 - P(X < r) = 1 - P(X ≤ r-1)
Poisson with changing rate: λ must be constant over interval
Wrong λ: If given rate per hour, but want probability for 2 hours, λ doubles!
Calculation errors: $e^{-\lambda}$ is very small for large λ—use calculator/tables
Using Binomial formula when Poisson is easier: If n > 20 and p < 0.05, try Poisson
Solved Examples
Example 1: Basic Binomial (Fair Coin)
Problem: A fair coin is tossed 6 times. Find probability of: (a) Exactly 4 heads (b) At least 5 heads
Solution:
$X \sim B(6, 0.5)$, where X = number of heads
(a) Exactly 4 heads:
$$P(X = 4) = \binom{6}{4} (0.5)^4 (0.5)^2 = \frac{6!}{4!2!} \cdot (0.5)^6 = 15 \times \frac{1}{64} = \frac{15}{64}$$(b) At least 5 heads: P(X ≥ 5) = P(X = 5) + P(X = 6)
$$P(X = 5) = \binom{6}{5} (0.5)^6 = 6 \times \frac{1}{64} = \frac{6}{64}$$ $$P(X = 6) = \binom{6}{6} (0.5)^6 = 1 \times \frac{1}{64} = \frac{1}{64}$$ $$P(X \geq 5) = \frac{6 + 1}{64} = \frac{7}{64}$$Example 2: Binomial Mean and Variance
Problem: A biased coin shows heads with probability 0.6. It’s tossed 50 times. Find: (a) Expected number of heads (b) Variance (c) Probability of getting exactly 30 heads
Solution:
$X \sim B(50, 0.6)$
(a) Mean:
$$E(X) = np = 50 \times 0.6 = 30$$(b) Variance:
$$\text{Var}(X) = npq = 50 \times 0.6 \times 0.4 = 12$$(c) P(X = 30):
$$P(X = 30) = \binom{50}{30} (0.6)^{30} (0.4)^{20}$$(This requires calculator/software: P(X = 30) ≈ 0.1082)
Example 3: Binomial with “At Most” (JEE Pattern)
Problem: A student guesses answers on a 5-question true/false test. Find probability of: (a) Passing (at least 3 correct) (b) Failing (at most 2 correct)
Solution:
$X \sim B(5, 0.5)$, where X = number correct
(a) P(pass) = P(X ≥ 3):
$$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5)$$ $$P(X=3) = \binom{5}{3} (0.5)^5 = 10 \times \frac{1}{32} = \frac{10}{32}$$ $$P(X=4) = \binom{5}{4} (0.5)^5 = 5 \times \frac{1}{32} = \frac{5}{32}$$ $$P(X=5) = \binom{5}{5} (0.5)^5 = 1 \times \frac{1}{32} = \frac{1}{32}$$ $$P(X \geq 3) = \frac{10 + 5 + 1}{32} = \frac{16}{32} = \frac{1}{2}$$(b) P(fail) = P(X ≤ 2) = 1 - P(X ≥ 3) = 1 - 1/2 = 1/2
(Or calculate P(X=0) + P(X=1) + P(X=2) = 1/32 + 5/32 + 10/32 = 16/32 = 1/2)
Example 4: Basic Poisson
Problem: Average 3 typos per page. Find probability of: (a) Exactly 2 typos on a page (b) No typos on a page (c) At least 1 typo
Solution:
$X \sim \text{Poisson}(3)$
(a) P(X = 2):
$$P(X = 2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{e^{-3} \cdot 9}{2} = 4.5 e^{-3} \approx 4.5 \times 0.0498 = 0.224$$(b) P(X = 0):
$$P(X = 0) = \frac{e^{-3} \cdot 3^0}{0!} = e^{-3} \approx 0.0498$$(c) P(X ≥ 1) = 1 - P(X = 0):
$$P(X \geq 1) = 1 - e^{-3} \approx 1 - 0.0498 = 0.9502$$Example 5: Poisson with Rate Change
Problem: On average, 2 customers arrive per minute at a store. Find probability that: (a) Exactly 5 arrive in 2 minutes (b) No one arrives in 30 seconds
Solution:
(a) 2 minutes: λ = 2 per minute × 2 minutes = 4
$$P(X = 5) = \frac{e^{-4} \cdot 4^5}{5!} = \frac{e^{-4} \cdot 1024}{120} = \frac{1024}{120} e^{-4} \approx 8.53 \times 0.0183 = 0.156$$(b) 30 seconds: λ = 2 per minute × 0.5 minutes = 1
$$P(X = 0) = \frac{e^{-1} \cdot 1^0}{0!} = e^{-1} \approx 0.368$$Example 6: Poisson Approximation to Binomial
Problem: A factory produces 1000 items daily. Each has 0.2% defect probability. Find probability of: (a) Exactly 3 defective items (b) At least 2 defective items
Solution:
Exact: $X \sim B(1000, 0.002)$ (hard to calculate!)
Approximation: n = 1000 (large), p = 0.002 (small), λ = np = 2
Use $X \approx \text{Poisson}(2)$
(a) P(X = 3):
$$P(X = 3) = \frac{e^{-2} \cdot 2^3}{3!} = \frac{8 e^{-2}}{6} = \frac{4}{3} e^{-2} \approx \frac{4}{3} \times 0.1353 = 0.180$$(b) P(X ≥ 2) = 1 - P(X ≤ 1) = 1 - [P(X=0) + P(X=1)]:
$$P(X = 0) = e^{-2} \approx 0.1353$$ $$P(X = 1) = 2e^{-2} \approx 0.2707$$ $$P(X \geq 2) = 1 - (0.1353 + 0.2707) = 1 - 0.406 = 0.594$$Example 7: Combined Problem (JEE Advanced)
Problem: A die is rolled 180 times. Find the probability of getting a 6 exactly 30 times using: (a) Binomial distribution (b) Poisson approximation
Solution:
(a) Binomial: $X \sim B(180, 1/6)$
$$P(X = 30) = \binom{180}{30} \left(\frac{1}{6}\right)^{30} \left(\frac{5}{6}\right)^{150}$$(Extremely difficult to calculate by hand!)
(b) Poisson: λ = np = 180 × (1/6) = 30
Note: λ = 30 is large, so Poisson approximation not ideal here (p = 1/6 is not very small). But proceeding:
$$P(X = 30) \approx \frac{e^{-30} \cdot 30^{30}}{30!}$$(Still requires calculator/software)
Better approach: For such problems, use normal approximation (beyond JEE scope) or statistical tables.
Practice Problems
Level 1: Foundation (JEE Main)
A coin is tossed 4 times. Find P(exactly 2 heads).
Average 1.5 accidents per day. Find P(no accidents tomorrow) using Poisson.
For X ~ B(10, 0.3), find E(X) and Var(X).
Solutions
X ~ B(4, 0.5) P(X=2) = C(4,2) × (0.5)⁴ = 6/16 = 3/8
X ~ Poisson(1.5) P(X=0) = e⁻¹·⁵ ≈ 0.223
E(X) = 10 × 0.3 = 3 Var(X) = 10 × 0.3 × 0.7 = 2.1
Level 2: Intermediate (JEE Main/Advanced)
60% of students pass an exam. In a class of 20, find: (a) P(exactly 12 pass) (b) P(at least 15 pass)
On average, 4 emails per hour. Find P(more than 2 emails in next 30 minutes).
A machine produces 2% defective items. In batch of 100, find P(at most 1 defective) using Poisson approximation.
Solutions
X ~ B(20, 0.6) (a) P(X=12) = C(20,12) × (0.6)¹² × (0.4)⁸ ≈ 0.180 (b) P(X≥15) = Σ P(X=r) for r=15 to 20 (use calculator/table)
λ for 30 min = 4 × 0.5 = 2, X ~ Poisson(2) P(X>2) = 1 - P(X≤2) = 1 - [P(0) + P(1) + P(2)] = 1 - [e⁻² + 2e⁻² + 2e⁻²] = 1 - 5e⁻² ≈ 1 - 0.677 = 0.323
λ = np = 100 × 0.02 = 2, X ~ Poisson(2) P(X≤1) = P(0) + P(1) = e⁻² + 2e⁻² = 3e⁻² ≈ 0.406
Level 3: Advanced (JEE Advanced)
For X ~ B(n, p), prove that E(X) = np using the binomial theorem.
If X ~ Poisson(λ₁) and Y ~ Poisson(λ₂) are independent, prove X+Y ~ Poisson(λ₁+λ₂).
A fair die is rolled until a 6 appears. Find the probability that it takes exactly 5 rolls. (Hint: Geometric distribution, related to binomial)
Solutions
E(X) = Σ r × C(n,r) × pʳ × qⁿ⁻ʳ for r=0 to n = Σ r × [n!/(r!(n-r)!)] × pʳ × qⁿ⁻ʳ = np × Σ C(n-1, r-1) × pʳ⁻¹ × qⁿ⁻ʳ (after simplification) = np × (p+q)ⁿ⁻¹ = np × 1 = np ✓
P(X+Y = k) = Σ P(X=r)P(Y=k-r) for r=0 to k = Σ [e⁻λ¹λ₁ʳ/r!] × [e⁻λ²λ₂ᵏ⁻ʳ/(k-r)!] = e⁻⁽λ¹⁺λ²⁾ × (1/k!) × Σ C(k,r)λ₁ʳλ₂ᵏ⁻ʳ = e⁻⁽λ¹⁺λ²⁾ × (λ₁+λ₂)ᵏ/k! ✓ [Binomial theorem]
Need 4 non-6’s then a 6: (5/6)⁴ × (1/6) = 625/7776 ≈ 0.080
Real-Life Applications
Binomial Distribution
- Quality Control: Testing n items, each has probability p of being defective
- Medical Trials: n patients, probability p of recovery
- Sports: n free throws, probability p of making each
- Elections: Polling n people, probability p support candidate
- Exams: n questions, probability p of getting each right
Poisson Distribution
- Customer Service: Number of calls/customers in time period
- Traffic: Number of accidents on highway per month
- Publishing: Number of typos per page
- Natural Events: Earthquakes, meteor strikes per year
- Biology: Number of bacteria in sample, mutations per generation
- Network: Number of packets arriving at server
Connection to Other Topics
Related JEE Topics:
- Bernoulli Trials - Foundation for binomial distribution
- Random Variables - Distributions are special types of RVs
- Probability Basics - Sample spaces and events
- Permutations & Combinations - $\binom{n}{r}$ in binomial formula
- Binomial Theorem - $(p+q)^n$ expansion
Advanced Connections:
- Binomial → Normal (for large n, use normal approximation)
- Poisson → Exponential (time between Poisson events)
- Expectation → Decision making (expected profit/loss)
Distribution Comparison Table
| Feature | Binomial | Poisson |
|---|---|---|
| Type | Fixed n trials | Events in interval |
| Parameter(s) | n, p | λ |
| Range | 0, 1, …, n | 0, 1, 2, … (infinite) |
| Mean | np | λ |
| Variance | npq | λ |
| Formula | $\binom{n}{r}p^r q^{n-r}$ | $\frac{e^{-\lambda}\lambda^r}{r!}$ |
| When to use | Independent trials | Rare events |
| Examples | Coin tosses, exams | Accidents, typos |
| Approximation | Use Poisson if n large, p small | — |
Formula Quick Reference
Binomial Distribution
$$P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$$ $$E(X) = np$$ $$\text{Var}(X) = np(1-p)$$Poisson Distribution
$$P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!}$$ $$E(X) = \text{Var}(X) = \lambda$$Special Values
- $e \approx 2.71828$
- $0! = 1$
- $\binom{n}{0} = \binom{n}{n} = 1$
Quick Revision Checklist
Exam Tips
- Check conditions: Is it binomial or Poisson? Verify BINS for binomial
- Identify parameters: What are n, p (binomial) or λ (Poisson)?
- Use approximation: If n > 20 and p < 0.05, consider Poisson
- Complement: For “at least r”, use 1 - P(less than r)
- Calculator: Know how to use C(n,r), e^(-x) functions
- Tables: JEE may provide Poisson/Binomial tables—use them!
- Rate adjustment: If λ given per hour but want probability for 30 min, halve λ
Next Steps
Understanding how binomial distribution arises from repeated trials? Continue to:
- Bernoulli Trials - The building blocks of binomial distribution
Last updated: December 21, 2025 Master probability distributions at JEENotes Practice Portal