Respuesta :

5 numbers chosen from 25 can be done in 25C5 = 25!/(5!*20!) = 53,130 ways. The probability of choosing the correct set of 5 numbers is
  1/53,130

[tex] \displaystyle
|\Omega|=\binom{25}{5}=\dfrac{25!}{5!20!}=\dfrac{21\cdot22\cdot23\cdot24\cdot25}{2\cdot3\cdot4\cdot5}=53130\\
|A|=1\\\\
P(A)=\dfrac{1}{53130}\approx0.002\% [/tex]

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