If the random variable \(X\) is such that&nbsp;\(X \sim B(n, p)\) then
\[\mathrm{E}(X) = np \quad \text{and} \quad \mathrm{Var}(X) = npq = np(1 - p)\]
where&nbsp;\(q = 1 - p\)
Recall that \[
\mathrm{E}(X) = \sum_{\rm{all } x} x P(X =x), 
\] and \[
\mathrm{Var}(X) = \mathrm{E}(X^2) - [(\mathrm{E}(X)]^2. 
\]
With \(P(X = x) = {}^n C_{x} p^x (1-p)^{n-x }\), where&nbsp;\(x = 0, \ldots, n\), the derivations for the expected value and variance of a binomial random variable are non-trivial.
The definition of expectation is: \[
\mathrm{E}(X) = \sum_{k=0}^{n} k \cdot P(X=k).
\] Substituting the Binomial probability distribution formula: \[
\mathrm{E}(X) = \sum_{k=0}^{n} k \binom{n}{k} p^k q^{n-k}.
\]
Since&nbsp;\( k=0 \) makes the term zero, we start from \( k=1 \): \[
\mathrm{E}(X) = \sum_{k=1}^{n} k \frac{n!}{k!(n-k)!} p^k q^{n-k}.
\] Using the identity \( k \binom{n}{k} = n \binom{n-1}{k-1} \), \[
\mathrm{E}(X) = \sum_{k=1}^{n} n \binom{n-1}{k-1} p^k q^{n-k}.
\]
Factorinf out \( np \), \[
\mathrm{E}(X) = np \sum_{k=1}^{n} \binom{n-1}{k-1} p^{k-1} q^{(n-1)-(k-1)}.
\] Let&nbsp;\( j = k-1 \) and \( m = n-1 \), \[
\mathrm{E}(X) = np \sum_{j=0}^{m} \binom{m}{j} p^j q^{m-j}.
\] By the Binomial Theorem, the sum is \( (p+q)^m = 1^m = 1 \), and so \[
\mathrm{E}(X) = np.
\]
The derivation of \(\mathrm{Var(X)\) is also complex, and so we omit it.

If the random variable \(X\) is such that \(X \sim B(n, p)\) then

\[\mathrm{E}(X) = np \quad \text{and} \quad \mathrm{Var}(X) = npq = np(1 - p)\]

Recall that 
\[
\mathrm{E}(X) = \sum_{\rm{all } x} x P(X =x), 
\]
and
\[
\mathrm{Var}(X) = \mathrm{E}(X^2) - [(\mathrm{E}(X)]^2. 
\]

With \(P(X = x) = {}^n C_{x} p^x (1-p)^{n-x }\), where \(x = 0, \ldots, n\), the derivations for the expected value and variance of a binomial random variable are non-trivial.

The definition of expectation is:
\[
\mathrm{E}(X) = \sum_{k=0}^{n} k \cdot P(X=k).
\]

Substituting the Binomial probability distribution formula:
\[
\mathrm{E}(X) = \sum_{k=0}^{n} k \binom{n}{k} p^k q^{n-k}.
\]

Since \( k=0  \) makes the term zero, we start from \( k=1 \):
\[
\mathrm{E}(X) = \sum_{k=1}^{n} k \frac{n!}{k!(n-k)!} p^k q^{n-k}.
\]

Using the identity \( k \binom{n}{k} = n \binom{n-1}{k-1} \), 
\[
\mathrm{E}(X) = \sum_{k=1}^{n} n \binom{n-1}{k-1} p^k q^{n-k}.
\]

Factorinf out \( np \), 
\[
\mathrm{E}(X) = np \sum_{k=1}^{n} \binom{n-1}{k-1} p^{k-1} q^{(n-1)-(k-1)}.
\]

Let \( j = k-1 \) and \(  m = n-1 \),
\[
\mathrm{E}(X) = np \sum_{j=0}^{m} \binom{m}{j} p^j q^{m-j}.
\]

By the Binomial Theorem, the sum is \( (p+q)^m = 1^m = 1 \), and so
\[
\mathrm{E}(X) = np.
\]

Expectation of a binomial random variable

Expectation of a binomial random variable