Is it correct to say: Using the binomial distribution for draws made without replacement?

Binomial needs independent trials with the same p each time. Drawing without replacement changes p after every draw, so the binomial formula does not apply.

Is it correct to say: Mixing up p and q?

Define 'success' as exactly what X counts (a defective bulb, a six, a hit). p is the probability of that event; q = 1 − p.

Is it correct to say: Getting the boundary wrong in 'more than' or 'at least' questions?

P(X > 2) = 1 − P(X ≤ 2), whereas P(X ≥ 2) = 1 − P(X ≤ 1). Write the inequality first, then the complement.

Is it correct to say: Taking the variance as np, or the standard deviation as npq?

Mean = np, variance = npq, standard deviation = √(npq).

Is it correct to say: Leaving out the factor ⁿCₓ?

pˣqⁿ⁻ˣ is the probability of one particular order of successes and failures. Multiply by ⁿCₓ to count all the orders.

Is it correct to say: Rounding intermediate values too early?

Powers such as (0.9)⁵ magnify rounding errors. Keep exact values or at least 4–5 decimal places until the final answer.

📄 Formula sheet (PDF)

Binomial Distribution

Bernoulli trials, the binomial distribution B(n, p), its probability formula, and the mean, variance and standard deviation of a binomial variable — Maharashtra HSC Maths Part II Ch 8

📐 14 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Begin every answer with 'Let X = number of …; then X ~ B(n, p)' and state n, p and q explicitly.
→Translate the words into an inequality before calculating: 'at least 3' is X ≥ 3, 'at most 3' is X ≤ 3, 'more than 3' is X > 3, i.e. X ≥ 4.
→For 'at least one', use 1 − qⁿ instead of adding n separate terms.
→If the mean and variance are given, divide the variance by the mean to get q straight away, then p = 1 − q and n = mean/p.
→Keep fractions (or powers such as (3/4)¹⁶) until the last step; round only if a decimal answer is asked for.

📐 Formulas(14)

Bernoulli Trial

P (success) = p, P (failure) = q, p + q = 1

Symbol	Meaning
$p$	Probability of success in one trial
$q$	Probability of failure in one trial, 1 − p

Bernoulli Distribution (one trial)

P (X = 1) = p, P (X = 0) = q; E (X) = p, Var (X) = pq

Binomial Distribution Notation

X \sim B (n, p), X = 0, 1, 2, \dots, n

Binomial Probability Formula★ Board fav

P (X = x) =^{n} C_{x} p^{x} q^{n - x}, x = 0, 1, \dots, n

Symbol	Meaning
$n$	Number of trials
$x$	Number of successes (0 to n)
$p, q$	Probabilities of success and failure in one trial

Probabilities as Terms of a Binomial Expansion

\sum_{x = 0}^{n}^{n} C_{x} p^{x} q^{n - x} = (q + p)^{n} = 1

No Successes and All Successes

P (X = 0) = q^{n}, P (X = n) = p^{n}

At Least One Success★ Board fav

P (X \geq 1) = 1 - P (X = 0) = 1 - q^{n}

Cumulative Probabilities

P (X \leq r) = \sum_{x = 0}^{r} P (X = x), P (X \geq r) = 1 - P (X \leq r - 1)

Fair Coin Case

p = q = \frac{1}{2} \Rightarrow P (X = x) = \frac{^{n} C _{x}}{2 ^{n}}

Mean of a Binomial Distribution★ Board fav

E (X) = μ = n p

Variance of a Binomial Distribution★ Board fav

Var (X) = σ^{2} = n pq

Standard Deviation

σ = n pq

Variance is Less Than the Mean

\frac{Var ( X )}{E ( X )} = \frac{n pq}{n p} = q < 1 \Rightarrow Var (X) < E (X)

Finding n and p from Mean and Variance

q = \frac{n pq}{n p}, p = 1 - q, n = \frac{n p}{p}

✏️ Solved Examples

1Solved Exampleeasy2 steps

A fair coin is tossed 6 times. Find the probability of getting exactly 4 heads.

Let X = number of heads. Then X ~ B(6, 1/2), with p = q = 1/2.

n = 6, p = q = \frac{1}{2}

2Solved Exampleboard4 steps

The probability that a bulb produced by a factory is defective is 0.1. Find the probability that in a sample of 5 bulbs, at most one is defective.

Let X = number of defective bulbs. Then X ~ B(5, 0.1), with q = 0.9.

n = 5, p = 0.1, q = 0.9

3Solved Exampleboard4 steps

The mean and variance of a binomial distribution are 4 and 3 respectively. Find n, p and P(X ≥ 1).

Write the two given facts.

n p = 4, n pq = 3

4Solved ExampleHOTS4 steps

The probability that a person hits a target in one shot is 1/4. Find the least number of shots needed so that the probability of hitting the target at least once is greater than 2/3.

Let X = number of hits in n shots. Then X ~ B(n, 1/4), with q = 3/4.

P (X \geq 1) = 1 - (\frac{3}{4})^{n}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Using the binomial distribution for draws made without replacement
✓Binomial needs independent trials with the same p each time. Drawing without replacement changes p after every draw, so the binomial formula does not apply.
2
Mixing up p and q
✓Define 'success' as exactly what X counts (a defective bulb, a six, a hit). p is the probability of that event; q = 1 − p.
3
Getting the boundary wrong in 'more than' or 'at least' questions
✓P(X > 2) = 1 − P(X ≤ 2), whereas P(X ≥ 2) = 1 − P(X ≤ 1). Write the inequality first, then the complement.
4
Taking the variance as np, or the standard deviation as npq
✓Mean = np, variance = npq, standard deviation = √(npq).
5
Leaving out the factor ⁿCₓ
✓pˣqⁿ⁻ˣ is the probability of one particular order of successes and failures. Multiply by ⁿCₓ to count all the orders.
6
Rounding intermediate values too early
✓Powers such as (0.9)⁵ magnify rounding errors. Keep exact values or at least 4–5 decimal places until the final answer.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
A die is thrown 4 times. Getting a six is a success. Find the probability of exactly 2 successes.
Q2
If X ~ B(10, 0.4), find the mean, variance and standard deviation of X.
Q3
Can a binomial distribution have mean 6 and variance 8? Give a reason.
Q4
Five fair coins are tossed together. Find the probability of getting at least 4 heads.
Q5
The probability that a student passes a test is 0.8. Find the probability that at least 5 out of 6 students pass.
Q6
The mean and variance of a binomial distribution are 3 and 2 respectively. Find n and p.

📝 Notes

Binomial Distribution

This chapter applies the ideas of random variables from the previous chapter (Probability Distributions) to one very common situation: counting the number of successes in a fixed number of repeated, independent trials.

Is it binomial?

Check all four conditions before using the formula:

A fixed number of trials, $n$ .
Each trial has only two outcomes — success or failure.
The trials are independent.
The probability of success, $p$ , is the same in every trial.

Tossing coins, throwing dice, testing items with replacement and repeated shots at a target all qualify. Drawing cards or balls without replacement does not.

Reading the question

Most of the work is translating words into an event on $X$ . Write it as an inequality, then decide whether direct addition or the complement is shorter. For "at least one", the complement $1 - q^n$ is almost always the right move. For "how many trials are needed" questions, set up $1 - q^n > k$ and test whole numbers until the inequality first holds.

Mean and variance shortcuts

For $X \sim B(n, p)$ there is no need to build a probability table: $E(X) = np$ and $\text{Var}(X) = npq$ . Because $q < 1$ , the variance is always less than the mean — a quick check on any given or computed values. When both are given, $q = \frac{\text{variance}}{\text{mean}}$ unlocks $p$ and $n$ in two lines.