Asked by Estifanos
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?
a.both are defective?
b.one is defective?
c.at most one is defective?
Answers
Answered by
Reiny
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
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
Answered by
Lakachew Bimerew Teferi
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?
Answered by
Mehari
A. 4 are defective and 8 are non-defective
P(2 are defective) =
C(4,2)/C(12, 2)
= 1/11
P(2 are defective) =
C(4,2)/C(12, 2)
= 1/11
Answered by
Mehari
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
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
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