Two aces

From a 32 card box we randomly pick 1 card and then 2 more cards. What is the probability that last two drawn cards are aces?

Correct result:

p =  1.21 %

Solution:

a=(28/32)⋅ (4/31)⋅ (3/30)=7620≐0.0113 b=(4/32)⋅ (3/31)⋅ (2/30)=11240≐0.0008 p=100⋅ (a+b)=100⋅ (0.0113+0.0008)=7562=1.21%a=(28/32) \cdot \ (4/31) \cdot \ (3/30)=\dfrac{ 7 }{ 620 } \doteq 0.0113 \ \\ b=(4/32) \cdot \ (3/31) \cdot \ (2/30)=\dfrac{ 1 }{ 1240 } \doteq 0.0008 \ \\ p=100 \cdot \ (a+b)=100 \cdot \ (0.0113+0.0008)=\dfrac{ 75 }{ 62 }=1.21 \%

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