Odpowiedź :
[tex]\displaystyle|\Omega|=\binom{1000}{3}=\dfrac{1000!}{3!997!}=\dfrac{998\cdot999\cdot1000}{2\cdot3}=166167000\\|A|=\binom{4}{3}=4\\\\P(A)=\dfrac{4}{166167000}=\dfrac{1}{41541750}[/tex]
[tex]\displaystyle|\Omega|=\binom{1000}{3}=\dfrac{1000!}{3!997!}=\dfrac{998\cdot999\cdot1000}{2\cdot3}=166167000\\|A|=\binom{4}{3}=4\\\\P(A)=\dfrac{4}{166167000}=\dfrac{1}{41541750}[/tex]