Odpowiedź :
[tex]|\Omega|=n+4\\|A|=4\\P(A)=\dfrac{4}{n+4}=\dfrac{2}{3}\\\\\dfrac{4}{n+4}=\dfrac{2}{3}\\2(n+4)=12\\2n+8=12\\2n=4\\n=2[/tex]
[tex]4 = \frac{2}{3} x/*3\\12 = 2x\\x = 6\\6 - 4 =2 \\odp : 2 kule[/tex]
[tex]|\Omega|=n+4\\|A|=4\\P(A)=\dfrac{4}{n+4}=\dfrac{2}{3}\\\\\dfrac{4}{n+4}=\dfrac{2}{3}\\2(n+4)=12\\2n+8=12\\2n=4\\n=2[/tex]
[tex]4 = \frac{2}{3} x/*3\\12 = 2x\\x = 6\\6 - 4 =2 \\odp : 2 kule[/tex]