A pack of cards c

As total number of cards = 16

2 cards can be drawn in ¹⁶C₂ ways = 120 ways

As we need to find the probability for the event E such that at least one of them is an ace.

We can solve this problem using negation.

We will find the probability for event that both cards are ace (E’)

Automatically 1 – P(E’) will give P(E)

For finding P(E’) we will select both cards out of 4 ace cards

∴ P(E’) =

∴ P(E) = 1 – 1/20 = 19/20

Hence,

P(E) = 19/20

As our answer matches only with option (c)

∴ Option (c) is the only correct choice.