Q.(a) A die with numbers 1 to 6 is biased such that $P(2) = \dfrac{3}{10}$ and the probability of other numbers is equal. Find the mean of the number of times the number 2 appears on the die, if the die is thrown twice.

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: …

Event Independence

Two events are independent when the occurrence of one does not change the probability of the other. Toss a coin and roll a die: the coin landing heads tells you nothing about whether the die shows a six. Contrast this with drawing cards without replacement, where the first draw does change the odds for the second — those events are dependent.

From Conditional Probability to a Clean Test

"Knowing $B$ doesn't change $A$" means $P(A \mid B) = P(A)$. Substituting the definition $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ and clearing the fraction gives the symmetric form used in practice:

Events $A$ and $B$ are independent exactly when the probability of both occurring equals the product of their individual probabilities. This version is preferred because it needs no non-zero condition and treats $A$ and $B$ alike.

A Quick Check

Roll a fair die. Let $A = \{2,4,6\}$ (even) and $B = \{4,5,6\}$ (greater than 3). Then $P(A) = P(B) = \tfrac{1}{2}$, and $A \cap B = \{4,6\}$ so $P(A \cap B) = \tfrac{1}{3}$. Since $\tfrac{1}{3} \neq \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4}$, these events are not independent.

Three or More Events

Events $A, B, C$ are mutually independent only if all four conditions hold: the three pairwise products and

Pairwise independence alone is not enough to guarantee mutual independence. …