Question 1083927
I) with replacement

x: number of defective in 4 picking.

Range of x ={0,1,2,3,4}

Every picking is independent from the rest.

p= P(pick one defective) = number of defective/total items = 5/25 =1/5.

x is binomial (n= 4, p=1/5).

P(x=0) = 4C0(p^0)(1-p)^(4-0) = 256/625.
P(x=1) = 4C1(p^1)(1-p)^(4-1) = 256/625.
P(x=2) = 4C2(p^2)(1-p)^(4-2) = 96/625.
P(x=3) = 4C3(p^3)(1-p)^(4-3) = 16/625.
P(x=4) = 4C4(p^4)(1-p)^(4-4) = 1/625.

II) without replacement

y: number of defective in 4 picking.

Range of y ={0,1,2,3,4}

y is hyper-geometric with N=25, n=4, d=5.

P(y=0) = (5C0)((25-5)C(4-0))/(25C4) = 969/2530.
P(y=1) = (5C1)((25-5)C(4-1))/(25C4) = 114/253
P(y=2) = (5C2)((25-5)C(4-2))/(25C4) = 38/253
P(y=3) = (5C3)((25-5)C(4-3))/(25C4) = 4/253
P(y=4) = (5C4)((25-5)C(4-4))/(25C4) = 1/2530

check for yourself that, in both results, the sum of every probabilities is equal to one.

@natolino_