Question Number 133821 by bramlexs22 last updated on 24/Feb/21
![There are 5 positive numbers and 6 negative numbers. Three numbers are chosen at random and multiplied . The probability that the product being a negative number is (a) ((17)/(33)) (b) ((11)/(34)) (c) ((16)/(33)) (d) ((16)/(35))](https://www.tinkutara.com/question/Q133821.png)
$$\mathrm{There}\:\mathrm{are}\:\mathrm{5}\:\mathrm{positive}\:\mathrm{numbers}\: \\ $$$$\mathrm{and}\:\mathrm{6}\:\mathrm{negative}\:\mathrm{numbers}.\:\mathrm{Three}\: \\ $$$$\mathrm{numbers}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{at}\:\mathrm{random} \\ $$$$\mathrm{and}\:\mathrm{multiplied}\:.\:\mathrm{The}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{product}\:\mathrm{being}\:\mathrm{a}\:\mathrm{negative} \\ $$$$\mathrm{number}\:\mathrm{is}\: \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{17}}{\mathrm{33}}\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{11}}{\mathrm{34}}\:\left(\mathrm{c}\right)\:\frac{\mathrm{16}}{\mathrm{33}}\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{16}}{\mathrm{35}} \\ $$
Answered by john_santu last updated on 24/Feb/21
![Required probability = ((C_2 ^( 5) ×C_1 ^( 6) + C_3 ^( 6) )/C_3 ^^(11) ) = ((16)/(33))](https://www.tinkutara.com/question/Q133822.png)
$${Required}\:{probability}\:=\:\frac{{C}_{\mathrm{2}} ^{\:\mathrm{5}} ×{C}_{\mathrm{1}} ^{\:\mathrm{6}} \:+\:{C}_{\mathrm{3}} ^{\:\mathrm{6}} }{{C}_{\mathrm{3}} ^{\:^{\mathrm{11}} } }\:=\:\frac{\mathrm{16}}{\mathrm{33}} \\ $$
Commented by bramlexs22 last updated on 24/Feb/21
![thank you](https://www.tinkutara.com/question/Q133828.png)
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by malwan last updated on 24/Feb/21
![(i) 1 −ve and 2 +ve (ii) 3 −ve ∴ p = ((C_1 ^( 6) ×C_2 ^( 5) + C_( 3) ^( 6) )/C_3 ^( 11) ) = ((16)/(33))](https://www.tinkutara.com/question/Q133824.png)
$$\left({i}\right)\:\mathrm{1}\:−{ve}\:{and}\:\mathrm{2}\:+{ve} \\ $$$$\left({ii}\right)\:\mathrm{3}\:−{ve} \\ $$$$\therefore\:{p}\:=\:\frac{{C}_{\mathrm{1}} ^{\:\mathrm{6}} ×{C}_{\mathrm{2}} ^{\:\mathrm{5}} \:+\:{C}_{\:\mathrm{3}} ^{\:\mathrm{6}} }{{C}_{\mathrm{3}} ^{\:\mathrm{11}} }\:\:=\:\frac{\mathrm{16}}{\mathrm{33}} \\ $$
Commented by bramlexs22 last updated on 24/Feb/21
![thank you](https://www.tinkutara.com/question/Q133827.png)
$$\mathrm{thank}\:\mathrm{you} \\ $$