Answer:
Option C. 1/221
Step-by-step explanation:
Total cards in a standard deck of cards = 52
Number of ace in a deck of cards = 4
So probability to draw an ace ![P_{1}=\frac{4}{52}](https://tex.z-dn.net/?f=P_%7B1%7D%3D%5Cfrac%7B4%7D%7B52%7D)
Since the cards were drawn without replacement means remaining cards in deck of cards now = 51
Number of aces remaining = 3
So probability of drawing another ace ![P_{2}=\frac{3}{51}](https://tex.z-dn.net/?f=P_%7B2%7D%3D%5Cfrac%7B3%7D%7B51%7D)
Now probability to draw both the aces
= ![P_{1}\times P_{2}](https://tex.z-dn.net/?f=P_%7B1%7D%5Ctimes%20P_%7B2%7D)
![\frac{4}{52}\times \frac{3}{51}=\frac{1}{13}\times \frac{1}{17}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B52%7D%5Ctimes%20%5Cfrac%7B3%7D%7B51%7D%3D%5Cfrac%7B1%7D%7B13%7D%5Ctimes%20%5Cfrac%7B1%7D%7B17%7D)
= ![\frac{1}{221}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B221%7D)
Option C. is the answer.