Q.If a fair coin is tossed 9 times, find the probability of

Binomial Distribution: From Intuition to Precision

Imagine you flip a fair coin 10 times. You want to know the probability of getting exactly 6 heads. This is the kind of question the binomial distribution answers — it models the number of "successes" in a fixed number of independent trials, where each trial has only two outcomes.

The Intuition

Think of a single trial. You roll a die and call "getting a 4" a success. That's one trial. Now roll the die 5 times. The number of times you get a 4 could be 0, 1, 2, 3, 4, or 5. Each roll is independent — the result of one roll doesn't affect the next. The probability of success (getting a 4) stays the same every time: $p = \frac{1}{6}$.

The binomial distribution tells you exactly how likely each possible count of successes is, given:

• a fixed number of trials $n$,
• a constant success probability $p$ per trial,
• independent trials.

The Key Conditions

For a situation to be modelled by a binomial distribution, all four must hold:

1. Fixed number of trials ($n$). You decide in advance how many times you'll repeat the experiment.
2. Two outcomes per trial — "success" and "failure". These are just labels; success is whatever outcome you're counting.
3. Constant probability of success ($p$) on every trial.
4. Independent trials. The outcome of one trial does not influence another.

The Precise Statement

Let $X$ be the number of successes in $n$ independent trials, each with success probability $p$. Then $X$ follows a binomial distribution with parameters $n$ and $p$, written:

The probability of getting exactly $k$ successes (where $k = 0, 1, 2, \dots, n$) is:

Breaking Down the Formula

Three pieces multiply together:

• $p^k$ — the probability that the $k$ successes happen. Since each success has probability $p$, and there are $k$ of them, this factor is $p$ multiplied by itself $k$ times.
• $(1-p)^{n-k}$ — the probability that the remaining $n-k$ trials are failures. Each failure has probability $1-p$.
• $\binom{n}{k}$ — the number of ways to choose which $k$ of the $n$ trials are the successes. The successes could be the first $k$ trials, or the last $k$, or any arrangement. This binomial coefficient counts all possible arrangements.

A Worked Example

Problem: A multiple-choice test has 10 questions, each with 4 options. You guess every answer. What is the probability you get exactly 3 correct?

Solution:

• $n = 10$ (10 trials)
• $p = \frac{1}{4}$ (probability of guessing correctly on one question)
• $k = 3$ (we want exactly 3 successes)

Compute $\binom{10}{3} = 120$, so:

$\left(\frac{3}{4}\right)^7 = \frac{2187}{16384}$, so:

Simplify: $120/64 = 15/8$, so:

So there's about a 25% chance of getting exactly 3 correct by pure guessing.

Mean and Variance

For a binomial distribution, two important summary measures are:

• Mean (expected number of successes): $\mu = np$
• Variance: $\sigma^2 = np(1-p)$

The mean makes intuitive sense: if you flip a coin 100 times with $p = 0.5$, you expect about 50 heads. The variance tells you how spread out the distribution is — it's largest when $p = 0.5$ (maximum uncertainty) and smallest when $p$ is near 0 or 1 (outcome is almost certain).

When Not to Use It

The binomial distribution fails if any condition is violated. Common exam traps:

• Without replacement from a finite population — use hypergeometric.
• Trials continue until a success — use geometric distribution.
• Counting events in a fixed interval — use Poisson distribution.

The binomial distribution is your tool when you have a fixed number of independent, identical trials, each with two outcomes. Master this, and you've unlocked a core piece of probability.

Binomial distribution is a core topic of the NCERT Class 12 Mathematics chapter on Probability, and searches like "binomial distribution formula and examples" or "binomial distribution class 12 important questions" reflect how frequently it appears in CBSE board papers and JEE Main statistics questions. The mean-variance relationship and the contrast with Poisson and hypergeometric distributions covered above are exactly the distinctions competitive exams like to test.