- A$\frac{2}{5}$
- B$\frac{1}{2}$
- C$\frac{4}{5}$
- ✓$\frac{1}{5}$
The probability of getting neither green nor red ball is equal to the probability of getting blue balls.
Number of blue balls $= 2$
Total number of balls $= 4 + 4 + 2 = 10$
Therefore
Probability of getting neither green nor red ball $=\frac{2}{10}=\frac{1}{5}$
Hence, the correct option is $(d).$