Question 851626
a company purchasing department is busy looking at the quality of computers bought by the company. 
they found that 15% of all new computers screens have to be replaced within 6 months P(screens bad) = 0.15
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27% of all new computers must have their screens or mouse replaced within 6 months. P(screens OR mouse bad) = 0.27
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given that a computer's mouse was replaced, 10% must also have their screens replaced within 6 months:: P(screens bad | mouse bad) = 0.10
Note: P(screens AND mouse bad) = 0.10*P(mouse bad)
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a) what is the probability that a computer's mouse must be replaced within 6 months?
Note: Since P(A OR B) = P(A)+P(B)-P(A AND B),
P(m) = P(m OR s)-P(s)+P(m AND s)

P(mouse bad) = P(screens OR mouse) + P(screens AND mouse)-P(screen)
P(mouse bad) = 0.27+0.10(mouse bad)-0.15
0.90P(mouse bad) = 0.12 
P(mouse bad) = 2/15
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b) what is the probability that given a computer's screen was replaced, the same computer's mouse will also be replaced within 6 months?
P(mouse bad | screen bad) = P(mouse AND screen bad)/P(screen bad) 
= (0.10)(2/15)/0.15 = 0.089
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Cheers,
Stan H.
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