Answer
$0.0202$
$0.0256$
$0.9991$
Work Step by Step
$p=0.119, n=15$
the probability that exactly one-third of them belong to a trade union
$P(X=5)=_{15}C_50.119^50.881^{10}=0.0202$
At least one-third
$P(X\geq5)=1-P(X\leq4)=1-_{15}C_00.119^00.881^{15}-_{15}C_10.119^10.881^{14}-_{15}C_20.119^20.881^{13}-_{15}C_30.119^30.881^{12}-_{15}C_40.119^40.881^{11}=0.0256$
What is the probability that at least 9 did not belong
$P(X\leq6)=P(X\leq4)+P(X=5)+P(X=6)=0.9991$
