) In a video game factory, machines A, B and C manufacture 20%, 30% and 50% of the total output, respectively. Also 5%, 3% and 7% of their outputs are defective, respectively.
A video game is selected at random, calculate the probability that
a) Video game is defective
b) The defective video game is produced by machine A
c) The defective video game is produced by machine B
d) The defective video game is produced by machine C
Answers
Explanation:
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Answer:
a)0.135
b)0.074
c)0.066
d)0.259
Explanations:
Video game manufactured by machine A = 20%
Video game manufactured by machine B = 30%
Video game manufactured by machine C = 50%
Defective outputs by A = 5%
Defective outputs by B = 3%
Defective outputs by C = 7%
E₁ : Video game produced by A
E₂ : Video game produced by B
E₃ : Video game produced by C
D : Video game is defective
P(E₁) = 20/100 = 2/10 = 1/5
P(E₂) = 30/100 = 3/10
P(E₃) = 50/100 = 5/10 = 1/2
a) Probability of defective video game P(D) is given by,
P(D) = P(E₁) *P(D/E₁) + P(E₂)*P(D/E₂) + P(E₃)*P(D/E₃)
= 1/5*1/20 + 3/10*3/100 + 1/2*7/100
= 1/100 + 9/100 + 7/100
= 27/100 = 0.135.
b) Probability of defective video game produced by machine A is given by,
P(E₁/D) = P(E₁)*P(D/E₁) / P(D)
= 1/5*5/100/ 27/200
= 2/27 = 0.074.
c) Probability of defective video game produced by machine B is given by,
P(E₂/D) = P(E₂)*P(D/E₂) / P(D)
= 3/10*3/100 / 27/200
= 1/15 = 0.066.
d) Probability of defective video game produced by machine C is given by,
P(E₃/D) = P(E₃)*P(D/E₃) / P(D)
= 1/2*7/100 / 27/200
= 7/27 = 0.259.