Question 890948
You need to calculate the probability of first drawing an ace then not an ace and add that to the probability of not drawing an ace and then drawing an ace.
{{{P[1]=(4/52)(48/52)=(1/13)(12/13)=12/169}}}
{{{P[2]=(48/52)(4/52)=12/169}}}
{{{P=P[1]+P[2]=(12+12)/169}}}
{{{P=24/169}}}