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Question Number 133821 by bramlexs22 last updated on 24/Feb/21

$$\mathrm{There}\mathrm{are}5\mathrm{positive}\mathrm{numbers}$$$$\mathrm{and}6\mathrm{negative}\mathrm{numbers}.\mathrm{Three}$$$$\mathrm{numbers}\mathrm{are}\mathrm{chosen}\mathrm{at}\mathrm{random}$$$$\mathrm{and}\mathrm{multiplied}.\mathrm{The}\mathrm{probability}$$$$\mathrm{that}\mathrm{the}\mathrm{product}\mathrm{being}\mathrm{a}\mathrm{negative}$$$$\mathrm{number}\mathrm{is}$$$$\left(\mathrm{a}\right)\frac{17}{33}\left(\mathrm{b}\right)\frac{11}{34}\left(\mathrm{c}\right)\frac{16}{33}\left(\mathrm{d}\right)\frac{16}{35}$$

Answered by john_santu last updated on 24/Feb/21

$$Requiredprobability=\frac{C_2^5\times C_1^6+C_3^6}{C_3^{{}^{11}}}=\frac{16}{33}$$

Commented by bramlexs22 last updated on 24/Feb/21

$$\mathrm{thank}\mathrm{you}$$

Answered by malwan last updated on 24/Feb/21

$$\left(i\right)1-veand2+ve$$$$\left(ii\right)3-ve$$$$\therefore p=\frac{C_1^6\times C_2^5+C_3^6}{C_3^{11}}=\frac{16}{33}$$

Commented by bramlexs22 last updated on 24/Feb/21

$$\mathrm{thank}\mathrm{you}$$

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