A random variable X has the following probability distribution:

Determine: $P(0<X<3)$

$\frac{3}{10}$

The correct answer is Option (1) → $\frac{3}{10}$

We know that $Σp_i = 1$

$⇒ 0+ k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k=1$

$⇒ 10k^2 + 9k-1=0⇒ (10k-1) (k + 1) = 0$

$⇒ k=\frac{1}{10},-1$ but $k$ cannot be negative

$⇒ k=\frac{1}{10}$

$P(0 <X < 3) = P(1) + P(2) = k + 2k = 3k = 3 ×\frac{1}{10}=\frac{3}{10}$.