Hi,

  

p(correct) = 1/5 = .2, p(incorrect) = 4/5 = .8,  n = 4

P(x = 4) = (.2)^4 = .0016

B. Using TI for cumulative P:  P(x ≥ 3) =  1 - binomcdf(n, p, largest x-value)

 P(x ≥ 3) = 1 – P(x ≤ 2) = 1 - .9728 = .0272

or using Formula

 P(x ≥ 3) = P(x=3) + P(x =4) = 4(.2)^3(.8)^1 + .0016 =  .0256 + .0016 = .0272 

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