in a shuffled pack of 52 cards, when two cards are picked randomly, what is the probability that one will be and ace?

P(exactly one ace)= 2* 4/52 * 48/51 = 0.14479638 approx

P (one or 2 aces) = the above + 4/52*3/51 = 0.14932 approx

Same as Alan, how about that !

From the four aces, we want to choose any one of them = C(4,1)

Then, we want to choose any card from the other 12 ranks   = 12*C(4,1)

And the total number of possible "hands" is just C(52,2)

So we have

C(4,1) * 12C(4,1) / C(52,2)  ≈  .217  ≈ 21.7%

I'm not sure I agree with Chris here.  If the question is asking what is the probability of getting exactly one ace, then my reasoning would be:

prob of 1st card being an ace = 4/52

prob of 2nd card not being an ace = 48/51

so prob of "ace" followed by "not ace" = (4/52)*(48/51)

prob of 1st card not being an ace = 48/52

prob of 2nd card being an ace = 4/51

so prob of "not ace" followed by "ace" = (48/52)*(4/51)

Overall prob of exactly one ace = (4/52)*(48/51) + (48/52)*(4/51)

$$\left({\frac{{\mathtt{4}}}{{\mathtt{52}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{48}}}{{\mathtt{51}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{48}}}{{\mathtt{52}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{51}}}}\right) = {\frac{{\mathtt{32}}}{{\mathtt{221}}}} = {\mathtt{0.144\: \!796\: \!380\: \!090\: \!497\: \!7}}$$

If the question is what is the probability of at least one ace then:

prob of 1st card not an ace = 48/52

prob of 2nd card not an ace = 47/51

prob of at least one ace = 1 - (48/52)*(47/51)

$${\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{48}}}{{\mathtt{52}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{47}}}{{\mathtt{51}}}}\right) = {\frac{{\mathtt{33}}}{{\mathtt{221}}}} = {\mathtt{0.149\: \!321\: \!266\: \!968\: \!325\: \!8}}$$

Ah.....I see my error here....we're picking cards one at a time....!!!

OK...I assumed "dealt"

Sorry!!