A random variable X has the following probability distribution

Then value of P for $(0 < X < 5)$

\(\frac{4}{5}\)

P(X) = 1

∴ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1

∴ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1

9k + 10k² - 1 = 0

10k² + 9k - 1 = 0 ⇒ 10k² + 10k - k - 1 = 0

10k (k + 1) - 1(k + 1) = 0 ⇒ (k + 1)(10k - 1) = 0

∴ k +1 = 0

k = -1

k ≠ -1

∴ 10k - 1 = 0

$k = \frac{1}{10}$ ...(i)

Value of P(0 < x < 5) = 0 + k + 2k + 2k + 3k

$8k = 8×\frac{1}{10}=\frac{8}{10}=\frac{4}{5}$

Option 3 is correct.