If X∼Binomial(n=8,p=0.25)X \sim Binomial(n=8, p=0.25)X∼Binomial(n=8,p=0.25), what is the probability of observing exactly 111 success, P(X=1)P(X=1)P(X=1)?
8⋅(0.25)⋅(0.75)78 \cdot (0.25) \cdot (0.75)^78⋅(0.25)⋅(0.75)7
8⋅(0.25)7⋅(0.75)8 \cdot (0.25)^7 \cdot (0.75)8⋅(0.25)7⋅(0.75)
(0.25)⋅(0.75)8(0.25) \cdot (0.75)^8(0.25)⋅(0.75)8
8⋅(0.25)2⋅(0.75)68 \cdot (0.25)^2 \cdot (0.75)^68⋅(0.25)2⋅(0.75)6