A box contains 6 blue balls, y white balls, and (y + 4) black balls. The probability of selecting a white ball at random is 1/4. If two balls are drawn simultaneously, what is the probability that both balls are black?

Total balls in the box =
→ 6 blue + y white + (y + 4) black = (2y + 10)

Given:
Probability of picking a white ball =
→ white balls / total balls = 1/4
⇒ y / (2y + 10) = 1/4

Cross-multiply:
⇒ 4y = 2y + 10
⇒ 4y − 2y = 10
⇒ y = 5

So,

Blue balls = 6

White balls = y = 5

Black balls = y + 4 = 9

Total balls = 6 + 5 + 9 = 20

Probability of both black balls in 1 draw:

Two balls are drawn simultaneously without replacement.

Favorable ways = selecting 2 black balls out of 9
→ = ⁹C₂ = (9 × 8)/2 = 36

Total ways = selecting any 2 balls from 20
→ = ²⁰C₂ = (20 × 19)/2 = 190

Probability = ⁹C₂ / ²⁰C₂ = 36 / 190 = 18 / 95

Thus, the correct answer is 18/95.