Respuesta :

konrad509

[tex] \displaystyle
|\Omega|=\binom{52}{5}=\dfrac{52!}{5!47!}=\dfrac{48\cdot49\cdot50\cdot51\cdot52}{2\cdot3\cdot4\cdot5}=2598960\\
|A|=\binom{13}{5}\cdot4=\dfrac{13!}{5!8!}\cdot4=\dfrac{9\cdot10\cdot11\cdot12\cdot13}{2\cdot3\cdot4\cdot5}\cdot4=5148\\\\
P(A)=\dfrac{5148}{2598960}=\dfrac{33}{16660}\approx0.2\% [/tex]