Mean and Variance of Binomial Distribution:
- The mean of binomial distribution is given by μ = np, where n is the number of trials and p is the probability of success in each trial.
- The variance of binomial distribution is given by σ^2 = np(1-p).

Finding the Mode:
- The mode is the value that occurs most frequently in the distribution.
- For a binomial distribution, the mode is the value of x that corresponds to the highest point on the probability mass function (PMF).

Given that the mean is 4 and variance is 3:
- μ = np = 4
- σ^2 = np(1-p) = 3

Substituting μ in the first equation:
- np = 4
- p = 4/n

Substituting p in the second equation:
- σ^2 = np(1-p) = 4(1-4/n)
- σ^2 = 4 - 16/n
- n = 16/σ^2 + 4

Substituting n in the equation for p:
- p = 4/n = 4/(16/σ^2 + 4) = σ^2/(4+σ^2)

Finding the Mode using PMF:
- The PMF of binomial distribution is given by P(x) = (nCx) * p^x * (1-p)^(n-x), where nCx is the number of combinations of n things taken x at a time.
- The mode occurs at the value of x that maximizes P(x).

Substituting the values of n and p:
- P(x) = [(16/σ^2 + 4)Cx] * (σ^2/(16/σ^2 + 4))^x * (1-σ^2/(16/σ^2 + 4))^(16/σ^2 + 4-x)

To maximize P(x), we can take the derivative of P(x) with respect to x and set it equal to 0:
- dP(x)/dx = [(16/σ^2 + 4)C(x-1)] * (σ^2/(16/σ^2 + 4))^(x-1) * (1-σ^2/(16/σ^2 + 4))^(16/σ^2 + 4-x) * [σ^2/(16/σ^2 + 4) - (16/σ^2 + 4-x)σ^2/(16/σ^2 + 4)] = 0
- Simplifying, we get:
- x = (16/σ^2 + 4)p = 4p

Substituting p = 4/n, we get:
- x = (16σ^2 + 16)/σ^2n = (4σ^2 + 4)/σ^2

Since x is an integer, the closest value to (4σ^2 + 4)/σ^2 is 4. Therefore, the mode is 4.

Hence, the correct answer is option A) 4.