A pair of dice is thrown 5 times. If getting a total of 7 is considered a success, then the probability of at least 3 success is :

$46\left(\frac{1}{6}\right)^4$

The correct answer is Option (1) → $46\left(\frac{1}{6}\right)^4$

Total thrown = 5

P(7 comes) = $\frac{1}{6}$

P(7 doesn't comes) = $\frac{5}{6}$

$={^5C}_3\frac{1}{63}×\frac{52}{62}+{^5C}_4\frac{1}{6}×\frac{5}{6}+{^5C}_5×\frac{1}{65}$

$=\frac{1}{64}\left(\frac{5×5×5×4}{2×6}+\frac{5×5}{6}+\frac{1}{6}\right)$

$=\frac{46}{64}$