The mean and variance, respectively, of a binomial distribution for n independent trials with the probability of success as p, are

The correct option is 𝑛𝑝 , 𝑛𝑝(1 − 𝑝).

To understand why this option is correct, let us analyze the properties of a binomial distribution. A binomial distribution models the number of successes in a sequence of $n$ independent trials (or experiments), where each trial has a constant probability of success $p$ and a probability of failure $q=1-p$.

1. Finding the Mean:
The mean (or expected value) of a probability distribution represents the average outcome we expect over many trials. For a single independent trial (a Bernoulli trial), the expected value $E\left(X_i\right)$ is:
$E\left(X_i\right)=(1\times p)+(0\times (1-p\left)\right)=p$
Since a binomial random variable $X$ is the sum of $n$ such independent Bernoulli trials ($X=X_1+X_2+...+X_n$), we can use the linearity of expectation:
$E\left(X\right)=E\left(X_1\right)+E\left(X_2\right)+...+E\left(X_n\right)$
$E\left(X\right)=p+p+...+p=np$
Thus, the mean of the binomial distribution is $np$.

2. Finding the Variance:
The variance measures the spread of the distribution. For a single independent trial, the variance $\mathrm{Var}\left(X_i\right)$ is:
$\mathrm{Var}\left(X_i\right)=E\left(X_i^2\right)-{\left[E\right(X_i\left)\right]}^2$
Since $X_i$ can only take values of 0 or 1, $X_i^2=X_i$, which gives $E\left(X_i^2\right)=p$. Therefore:
$\mathrm{Var}\left(X_i\right)=p-p^2=p(1-p)$
Because the trials are independent, the variance of the sum of the trials is simply the sum of their individual variances:
$\mathrm{Var}\left(X\right)=\mathrm{Var}\left(X_1\right)+\mathrm{Var}\left(X_2\right)+...+\mathrm{Var}\left(X_n\right)$
$\mathrm{Var}\left(X\right)=p(1-p)+p(1-p)+...+p(1-p)$
$\mathrm{Var}\left(X\right)=np(1-p)$
Thus, the variance of the binomial distribution is $np(1-p)$.

Therefore, the mean and variance are respectively $np$ and $np(1-p)$, matching the correct option.