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There is only one ace of hearts. So, required probability = $\frac{{}^{51}C_4}{{}^{52}C_5} = \frac{5}{52}$

(ace is assumed to be picked and we need to pick any 4 from remaining 51)

Can also be solved like

Required probability = 1 - Probability that ace of heart won't be in hand

$= 1 -\frac{{}^{51}C_5}{{}^{52}C_5} \\= 1- \frac{47 }{52} \\= \frac{5}{52}$

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