Question

Two lamps are to be chosen from pack of 12 lamps where four are defective and the rest are non defective. what is the probablity that
a.both are defective?
b.one is defective?
c.at most one is defective?

Answer 1

prob(defective) = 4/12 = 1/3
prob(not defective) = 2/3

a) prob(both defective)
= C(2,2) (1/3)^2
= 1/9

b) one defective:
= C(2,1)(1/3)(2/3)
= 2(2/9) = 4/9

c) at most 1 is defective
----> rule out both defective, which we did in a)
so Pro(of at most 1 defective) = 1 - 1/9 = 8/9

analysis:
we have the following cases
d = defective, g = good

gg = (2/3)(2/3) = 4/9
gd = (2/3)(1/3) = 2/9
dg = (1/3)(2/3) = 2/9
dd - (1/3)(1/3) = 1/9
total of prob's = 9/9 = 1

Answer 2

Two lamps are to be chosen from a pack of 12 lamps where four are defective and the rest are non defective. What is the probability that a both are defective? b One is defective? c at most one is defective?

Answer 3

A. 4 are defective and 8 are non-defective
P(2 are defective) =
C(4,2)/C(12, 2)
= 1/11

Answer 4

A. 4 are defective and 8 are non-defective
P(2 are defective) =
C(4,2)/C(12, 2)
= 1/11
B. One lamp is defective means the other one is non- defective
P(1- defective) =
(8, 1)(4, 1)/(12, 2)= 16/33
C.
P( at most onedefective)=
P(no defective) +
P(1- defective)
=(8, 2)(4, 2)+(8, 1)(4, 1)/(12, 2)
= 10/11