The Probability distribution of a discrete random variable X is given as :

 The mean of the distribution is :

$\frac{16}{9}$

The correct answer is Option (4) → $\frac{16}{9}$

Since, the sum of probabilities must be equal to 1.

$2k^2+k^2+3k^2+k=1$

$6k^2+k-1=0$

$6k^2+3k-2k-1=0$

$3k(2k+1)-1(2k+1)=0$

$(2k+1)(3k-1)=0$

$⇒k=\frac{1}{3}$ [Always positive]

∴ Mean of distribution = $0×2×(\frac{1}{3})^2+1×(\frac{1}{3})^2+2×3(\frac{1}{3})^2+(\frac{1}{3})×3$

$=\frac{0}{9}+\frac{1}{9}+\frac{6}{9}+\frac{3}{9}=\frac{6+1+9}{9}$

$=\frac{16}{9}$