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"2" wypadnie 0 lub 2 lub 4 razy.
Schemat Bernoulliego.
n=5
k=0 v k=2 v k=4
p=1/3
q=2/3
P(A)=5!/5! *(2/3)⁵+5!/(2!*3!)*(1/3)²*(2/3)³+5!/4!*(1/3)⁴*2/3=
32/243+10*1/9*8/27+5*1/81 *2/3=122/243