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Question Number 81926 by zainal tanjung last updated on 16/Feb/20

$$3\mathrm{integers}\mathrm{are}\mathrm{chosen}\mathrm{at}\mathrm{random}\mathrm{from}$$$$\mathrm{the}\mathrm{first}20\mathrm{integers}.\mathrm{The}\mathrm{probability}$$$$\mathrm{that}\mathrm{their}\mathrm{ptoduct}\mathrm{is}\mathrm{even},\mathrm{is}$$

Commented by mr W last updated on 16/Feb/20

$$p=1-\frac{C_3^{10}}{C_3^{20}}=1-\frac{120}{1140}=89.5\mathrm{\%}$$$$or$$$$p=\frac{C_3^{10}+C_2^{10}{C}_1^{10}+C_1^{10}{C}_2^{10}}{C_3^{20}}=\frac{1020}{1140}=89.5\mathrm{\%}$$

Answered by MJS last updated on 16/Feb/20

$$\mathrm{the}\mathrm{product}\mathrm{is}\mathrm{uneven}\mathrm{when}\mathrm{no}\mathrm{even}\mathrm{number}$$$$\mathrm{is}\mathrm{among}\mathrm{the}\mathrm{chosen}\mathrm{numbers}$$$$\mathrm{the}\mathrm{probability}\mathrm{to}\mathrm{get}\mathrm{an}\mathrm{uneven}\mathrm{number}\mathrm{is}$$$$\mathrm{for}\mathrm{the}{1}^{\mathrm{st}}\mathrm{number}\frac{10}{20}=\frac{1}{2}$$$$\mathrm{for}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{number}\frac{9}{19}$$$$\mathrm{for}\mathrm{the}{3}^{\mathrm{rd}}\mathrm{number}\frac{8}{18}=\frac{4}{9}$$$$3\mathrm{uneven}\mathrm{numbers}\frac{1}{2}\times \frac{9}{19}\times \frac{4}{9}=\frac{2}{19}\approx 10.53\mathrm{\%}$$$$\Rightarrow \mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{product}\mathrm{is}\mathrm{even}$$$$\mathrm{is}\frac{17}{19}\approx 89.47\mathrm{\%}$$

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