A bag contains $8$ black, $5$ white, and $7$ green balls. Two balls are drawn without replacement. Calculate the probability that exactly one of the two drawn balls is green.

Step-by-step solution

1. Step 1: Total balls = $8+5+7=18$.\nStep 2: Total ways to choose any 2 balls = $\left(\genfrac{}{}{0ex}{}{18}{2}\right)=153$.\nStep 3: To have exactly one green ball:\n- Choose 1 green ball: $\left(\genfrac{}{}{0ex}{}{7}{1}\right)=7$ ways.\n- Choose 1 non‑green ball (black or white): $8+5=13$ balls, so $\left(\genfrac{}{}{0ex}{}{13}{1}\right)=13$ ways.\n- Total favourable unordered pairs = $7\times 13=91$.\nStep 4: Probability = $\frac{91}{153}$.
2. Step 5: Numerical value ≈ $0.595$.

Concept tested: Hypergeometric distribution – single draw without replacement

Published 11 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked