Q.If the mean and variance of a binomial distribution are $4/3$ and $8/9$ resp. find $P(x=1)$

Variance of the Bernoulli Distribution

First, the intuition

Imagine a single coin flip — but not a fair coin. Let’s say the coin lands Heads with probability $p$ and Tails with probability $1-p$. We code Heads as 1 and Tails as 0. That’s a Bernoulli trial: a single experiment with exactly two outcomes.

Now ask: how "spread out" is this outcome? The answer is not obvious because there are only two possible values, 0 and 1. The spread depends entirely on $p$.

If $p = 0$ or $p = 1$, the outcome is certain — no spread at all. If $p = 0.5$, the outcome is maximally uncertain: half the time you get 0, half the time 1. That’s when the spread is largest.

Variance measures exactly this spread — the average squared distance of the outcome from its mean.

The mean first

The mean (expected value) of a Bernoulli random variable $X$ is:

So the "centre" of the distribution is at $p$. For a fair coin, the mean is 0.5 — a number that never actually occurs, but that’s fine; it’s the long-run average.

Now variance

Variance is defined as:

That is: take each possible outcome, subtract the mean, square the result, then average those squared deviations (weighted by their probabilities).

For Bernoulli:

• When $X = 1$: deviation is $1 - p$, squared is $(1-p)^2$, weight is $p$.
• When $X = 0$: deviation is $0 - p = -p$, squared is $p^2$, weight is $1-p$.

So:

Factor $p(1-p)$:

That’s it. One clean expression.

What the formula tells you

• When $p = 0$ or $p = 1$, variance is $0$ — no uncertainty.
• When $p = 0.5$, variance is $0.5 \times 0.5 = 0.25$, the maximum possible.
• The variance is symmetric: swapping $p$ and $1-p$ gives the same number. That makes sense — calling Heads "success" vs "failure" shouldn’t change the spread. …