Question Number 134751 by mathocean1 last updated on 06/Mar/21
$$\:{In}\:{my}\:{house},\:{there}\:{is}\:\mathrm{250}\:{laptops}:\: \\ $$$$\mathrm{40}\:{are}\:{new},\:\mathrm{100}\:{are}\:{recent}\:{and}\:{the}\: \\ $$$${others}\:{are}\:{old}.\:{A}\:{statistic}\:{showed}\:{that} \\ $$$$\mathrm{4\%}\:{of}\:{new}\:{laptops}\:{are}\:{faulty},\:\mathrm{12\%}\:{of} \\ $$$${recent}\:{ones}\:{are}\:{faulty}\:{and}\:\mathrm{25\%}\:{of}\:{old} \\ $$$${ones}\:{are}\:{faulty}. \\ $$$${Calculate}\:{the}\:{probability}\:{that}\:\mathrm{1}\:{laptop}\:{be} \\ $$$${new}\:,\:{knowing}\:{that}\:{it}\:{is}\:{faulty}. \\ $$
Answered by EDWIN88 last updated on 07/Mar/21
$$\mathrm{p}\left(\mathrm{A}\right)=\frac{\mathrm{0}.\mathrm{04}}{\mathrm{0}.\mathrm{04}+\mathrm{0}.\mathrm{12}+\mathrm{0}.\mathrm{25}}\:=\:\frac{\mathrm{0}.\mathrm{04}}{\mathrm{0}.\mathrm{41}}=\:\frac{\mathrm{4}}{\mathrm{41}} \\ $$$$ \\ $$