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Question Number 133821 by bramlexs22 last updated on 24/Feb/21
$$\mathrm{There}\:\mathrm{are}\:\mathrm{5}\:\mathrm{positive}\:\mathrm{numbers}\: \\ $$$$\mathrm{and}\:\mathrm{6}\:\mathrm{negative}\:\mathrm{numbers}.\:\mathrm{Three}\: \\ $$$$\mathrm{numbers}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{at}\:\mathrm{random} \\ $$$$\mathrm{and}\:\mathrm{multiplied}\:.\:\mathrm{The}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{product}\:\mathrm{being}\:\mathrm{a}\:\mathrm{negative} \\ $$$$\mathrm{number}\:\mathrm{is}\: \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{17}}{\mathrm{33}}\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{11}}{\mathrm{34}}\:\left(\mathrm{c}\right)\:\frac{\mathrm{16}}{\mathrm{33}}\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{16}}{\mathrm{35}} \\ $$
Answered by john_santu last updated on 24/Feb/21
$${Required}\:{probability}\:=\:\frac{{C}_{\mathrm{2}} ^{\:\mathrm{5}} ×{C}_{\mathrm{1}} ^{\:\mathrm{6}} \:+\:{C}_{\mathrm{3}} ^{\:\mathrm{6}} }{{C}_{\mathrm{3}} ^{\:^{\mathrm{11}} } }\:=\:\frac{\mathrm{16}}{\mathrm{33}} \\ $$
Commented by bramlexs22 last updated on 24/Feb/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by malwan last updated on 24/Feb/21
$$\left({i}\right)\:\mathrm{1}\:−{ve}\:{and}\:\mathrm{2}\:+{ve} \\ $$$$\left({ii}\right)\:\mathrm{3}\:−{ve} \\ $$$$\therefore\:{p}\:=\:\frac{{C}_{\mathrm{1}} ^{\:\mathrm{6}} ×{C}_{\mathrm{2}} ^{\:\mathrm{5}} \:+\:{C}_{\:\mathrm{3}} ^{\:\mathrm{6}} }{{C}_{\mathrm{3}} ^{\:\mathrm{11}} }\:\:=\:\frac{\mathrm{16}}{\mathrm{33}} \\ $$
Commented by bramlexs22 last updated on 24/Feb/21
$$\mathrm{thank}\:\mathrm{you} \\ $$