CUET Preparation Today
| Three dice are thrown randomly. The probability of coming \(3\) in at least one dice is |
\(\frac{180}{216}\) \(\frac{91}{216}\) \(\frac{5}{216}\) \(\frac{125}{216}\) |
| \(\frac{91}{216}\) |
| Probability of coming \(3=1-(\)probability of not coming \(3)\) |
