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 answer:

p =  1.2097 %

Step-by-step explanation:

$$\begin{gathered} a=(28\mathrm{/}32)\cdot (4\mathrm{/}31)\cdot (3\mathrm{/}30)={\displaystyle \frac{7}{620}}\doteq 0.0113 \\ b=(4\mathrm{/}32)\cdot (3\mathrm{/}31)\cdot (2\mathrm{/}30)={\displaystyle \frac{1}{1240}}\doteq 0.0008 \\ p=100\cdot (a+b)=100\cdot (0.0113+0.0008)=1.2097\mathrm{\%} \end{gathered}$$

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