Use Binomial Theorem
[tex]P(x=k) = (nCk) p^k (1-p)^{n-k}[/tex]
Where k is number that are defective, n is total selected (3), and p is probability that a transistor is defective (3/17).
a) All are defective means k = 3
[tex]P = (3C3) (\frac{3}{17})^3 (\frac{14}{17})^0 = (\frac{3}{17})^3 = 0.0055[/tex]
b) None are defective means k = 0
[tex]P = (3C0) (\frac{3}{17})^0 (\frac{14}{17})^3 = (\frac{14}{17})^3 = 0.5585[/tex]
