A consignment of 12 refurbished laptops contains 3 defective units. If 4 laptops are randomly chosen for inspection , what is the probability that at least 2 of them will be defective
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Step-by-step explanation:
Explanation:
The probability that a laptop is defective is
(
6
22
)
The probability of a laptop not being defective is
(
16
22
)
The probability that at least one laptop is defective is given by:
P(1 defective)+P(2 defective)+P(3 defective)
, as this probability is cumulative. Let
X
be the number of laptops found to be defective.
P
(
X
=
1
)
= (3 choose 1)
(
6
22
)
1
×
(
16
22
)
2
=
0.43275
P
(
X
=
2
)
= (3 choose 2)
(
6
22
)
2
×
(
16
22
)
1
=
0.16228
P
(
X
=
3
)
=(3 choose 3)
(
6
22
)
3
=
0.02028
(Sum up all of the probabilities)
=
0.61531
≈
0.615
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