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Option C

There are 4 aces in a deck

Probability of 1st card drawn to be ace = $\frac{\binom{4}{1}}{\binom{52}{1}} = \frac{4}{52}$

Now, Remaining aces $= 4 – 1 = 3$ and Remaining cards $= 52 – 1 = 51$

So, Probability of 2nd card drawn to be ace = $\frac{\binom{3}{1}}{\binom{51}{1}} = \frac{3}{51}$

Hence, Total Probability that top and bottom cards of a randomly shuffled deck are both aces is $=\mathbf{\frac{4}{52}\times \frac{3}{51}}$

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