Since the
drawing is done with replacement, the trials are Bernoulli trials Clearly, X has the
binomial distribution with n = 10 and 
10
1
100
10
p Therefore
q =
9
1
−p10
=
Now
P(at least one defective egg) = P(X ≥ 1) = 1 – P (X = 0)
=
10
10
0
9
1
C
⎛10
⎞
−
⎜
⎟
⎝
⎠  = 
10
910
1
10
−
EXERCISE 13 5
1