Answer:
Given,
p = 26% = 0.26
sample n = 3
Consider,
Binomial distriuution P(X) = nCx*p^x*q^(n-x)
nCx = n!/(n-x)!*x!
a)
P(X < 2) = P(0) + P(1)
= 3C0*0.26^0*(1-0.26)^3 + 3C1*0.26^1*(1-0.26)^2
= 0.405 + 0.427
= 0.832
b)
P(X >= 2) = 1 - P(X <= 1)
= 1 - [P(0) + P(1)]
= 1 - [3C0*0.26^0*(1-0.26)^3 + 3C1*0.26^1*(1-0.26)^2]
= 1 - 0.832
= 0.168
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