Variance of a Bernoulli trial with p=0.3

medium CBSE JEE-MAIN • 3 min read
Tags Expected Value And Variance

Question

A Bernoulli trial has probability of success p=0.3p = 0.3. Find the variance of the Bernoulli random variable XX.

Solution — Step by Step

A Bernoulli trial has exactly two outcomes: success (with probability pp) and failure (with probability q=1−pq = 1 - p).

The Bernoulli random variable XX takes:

  • X=1X = 1 with probability p=0.3p = 0.3 (success)
  • X=0X = 0 with probability q=1−0.3=0.7q = 1 - 0.3 = 0.7 (failure)
 E(X)=∑x⋅P(X=x)=1⋅p+0⋅q=p=0.3E(X) = \sum x \cdot P(X = x) = 1 \cdot p + 0 \cdot q = p = 0.3 E(X2)=∑x2⋅P(X=x)=12⋅p+02⋅q=p=0.3E(X^2) = \sum x^2 \cdot P(X = x) = 1^2 \cdot p + 0^2 \cdot q = p = 0.3

(Note: for a Bernoulli variable, X2=XX^2 = X since XX only takes values 0 and 1.)

 Var(X)=E(X2)−[E(X)]2=p−p2=p(1−p)=pq\text{Var}(X) = E(X^2) - [E(X)]^2 = p - p^2 = p(1-p) = pq

Substituting p=0.3p = 0.3, q=0.7q = 0.7:

Var(X)=0.3×0.7=0.21\text{Var}(X) = 0.3 \times 0.7 = \boxed{0.21}

Why This Works

The variance formula Var(X)=pq\text{Var}(X) = pq has a nice intuitive interpretation. When p=0p = 0 or p=1p = 1, the outcome is certain (no randomness), so variance = 0. Maximum variance occurs at p=0.5p = 0.5 (most uncertainty), giving Var=0.25\text{Var} = 0.25.

For p=0.3p = 0.3: Var=0.21\text{Var} = 0.21, which is less than 0.25 — reflecting that success (30%) is less likely than failure (70%), making the distribution somewhat more predictable.

The standard deviation is σ=pq=0.21≈0.458\sigma = \sqrt{pq} = \sqrt{0.21} \approx 0.458.

Alternative Method — Direct Definition

Var(X)=E[(X−μ)2]=p(1−p)2+q(0−p)2\text{Var}(X) = E\left[(X - \mu)^2\right] = p(1-p)^2 + q(0-p)^2 =p⋅q2+q⋅p2=pq(q+p)=pq⋅1=pq= p \cdot q^2 + q \cdot p^2 = pq(q + p) = pq \cdot 1 = pq

Same result, confirming Var(X)=pq=0.21\text{Var}(X) = pq = 0.21.

For a Binomial distribution B(n,p)B(n, p) (which is nn independent Bernoulli trials): Mean = npnp, Variance = npqnpq. These are simply the Bernoulli mean and variance multiplied by nn. If n=1n = 1, you get the Bernoulli case. This scaling property is useful for JEE statistics problems.

Common Mistake

Students sometimes compute Var(X)=p2+q2\text{Var}(X) = p^2 + q^2 or just p2p^2 — both are wrong. The correct formula is Var(X)=p−p2=pq\text{Var}(X) = p - p^2 = pq. Remember: variance uses E(X2)−[E(X)]2E(X^2) - [E(X)]^2, not just [E(X)]2[E(X)]^2. Always use the two-step approach: find E(X)E(X) and E(X2)E(X^2) separately, then subtract.

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