Answer:
P(X=0) = 0.7158
P(X=1) = 0.2684
P(X=2) =0.0158
Step-by-step explanation:
Number of defective laptops = 3
Number of normal laptops = 17
If the school chooses 2 laptops from this lot, the number of defectives (X) could be 0, 1, or 2.
For zero defectives:
[tex]P(X=0) = \frac{17}{20} *\frac{16}{19}\\P(X=0) = 0.7158[/tex]
For one defective:
[tex]P(X=1) = \frac{17}{20} *\frac{3}{19}+\frac{3}{20} *\frac{17}{19}\\P(X=1) = 0.2684[/tex]
For two defectives:
[tex]P(X=2) = \frac{3}{20} *\frac{2}{19}\\P(X=2) = 0.0158[/tex]
The probability distribution for the defective laptops of the shipment for the school
X 0 1 2
P(X) 0.716 0.268 0.0157
According to the question
A shipment of 20 similar laptop computers to a retail outlet contains
Given that , In that lot 3 laptops are defective.
If a school makes a random purchase of 2 of these computers,
we have to find the probability distribution for the number of defective laptop computers
Total number of laptops = 20
Total number of defective laptops = 3
Total number of non defective laptops = 20-3 = 17
Given that school choose two laptops out of this lot
so let P(X) represents the probability of choosing X defective laptops
The value of X can be 0, 1 or 2
Case 1 X = 0
The probability of choosing zero (X=0) defective laptop
[tex]\rm P(0) = \dfrac{17}{20}\times \dfrac{16}{19}=0.716\\[/tex]
The probability of choosing one (X=1 ) defective laptop
[tex]\rm P(1) = \dfrac{3}{20}\times \dfrac{17}{19}+\dfrac{17}{20} \times\dfrac{3}{19} =0.268\\[/tex]
The probability of choosing two (X=2 ) defective laptops
[tex]\rm P(2) = \dfrac{3}{20}\times \dfrac{2}{19}=0.0157\\[/tex]
So the probability distribution for the defective laptops of the shipment for the school
X 0 1 2
P(X) 0.716 0.268 0.0157
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https://brainly.com/question/11234923
