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Question Number 55736 by byaw last updated on 03/Mar/19

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Mar/19

p(under wieight)=(1/(20))→p(u.w)=(1/(20))→a p(actual weight)=((19)/(20))→p(a.w)=((19)/(20))→b a+b=1 (b+a)^(72) =b^(72) +72c_1 b^(71) a^1 +72c_2 b^(70) a^2 +72c_3 b^(69) a^3 +... i) so less than four will be under weight =(b^(72) +72c_1 b^(71) a^1 +72c_2 b^(70) a^2 +72c_3 b^(69) a^3 ) ={(((19)/(20)))^(72) +72c_1 (((19)/(20)))^(71) ((1/(20)))^1 +72c_2 (((19)/(20)))^(70) ((1/(20)))^2 +72c_3 (((19)/(20)))^(69) ((1/(20)))^3 } ii)Two were under weight 72c_2 (((19)/(20)))((1/(20)))^2 pls check...

$${p}\left({under}\:{wieight}\right)=\frac{\mathrm{1}}{\mathrm{20}}\rightarrow{p}\left({u}.{w}\right)=\frac{\mathrm{1}}{\mathrm{20}}\rightarrow\boldsymbol{{a}} \\ $$$${p}\left({actual}\:{weight}\right)=\frac{\mathrm{19}}{\mathrm{20}}\rightarrow{p}\left({a}.{w}\right)=\frac{\mathrm{19}}{\mathrm{20}}\rightarrow{b} \\ $$$${a}+{b}=\mathrm{1} \\ $$$$\left({b}+{a}\right)^{\mathrm{72}} ={b}^{\mathrm{72}} +\mathrm{72}{c}_{\mathrm{1}} {b}^{\mathrm{71}} {a}^{\mathrm{1}} +\mathrm{72}{c}_{\mathrm{2}} {b}^{\mathrm{70}} {a}^{\mathrm{2}} +\mathrm{72}{c}_{\mathrm{3}} {b}^{\mathrm{69}} {a}^{\mathrm{3}} +... \\ $$$$\left.\:\:\:\:\:{i}\right)\:{so}\:{less}\:{than}\:{four}\:{will}\:{be}\:{under}\:{weight}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$=\left({b}^{\mathrm{72}} +\mathrm{72}{c}_{\mathrm{1}} {b}^{\mathrm{71}} {a}^{\mathrm{1}} +\mathrm{72}{c}_{\mathrm{2}} {b}^{\mathrm{70}} {a}^{\mathrm{2}} +\mathrm{72}{c}_{\mathrm{3}} {b}^{\mathrm{69}} {a}^{\mathrm{3}} \right) \\ $$$$=\left\{\left(\frac{\mathrm{19}}{\mathrm{20}}\right)^{\mathrm{72}} +\mathrm{72}{c}_{\mathrm{1}} \left(\frac{\mathrm{19}}{\mathrm{20}}\right)^{\mathrm{71}} \left(\frac{\mathrm{1}}{\mathrm{20}}\right)^{\mathrm{1}} +\mathrm{72}{c}_{\mathrm{2}} \left(\frac{\mathrm{19}}{\mathrm{20}}\right)^{\mathrm{70}} \left(\frac{\mathrm{1}}{\mathrm{20}}\right)^{\mathrm{2}} +\mathrm{72}{c}_{\mathrm{3}} \left(\frac{\mathrm{19}}{\mathrm{20}}\right)^{\mathrm{69}} \left(\frac{\mathrm{1}}{\mathrm{20}}\right)^{\mathrm{3}} \right\} \\ $$$$\left.{ii}\right){Two}\:{were}\:{under}\:{weight} \\ $$$$\mathrm{72}{c}_{\mathrm{2}} \left(\frac{\mathrm{19}}{\mathrm{20}}\right)\left(\frac{\mathrm{1}}{\mathrm{20}}\right)^{\mathrm{2}} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$$$ \\ $$

Commented by byaw last updated on 03/Mar/19

Thank you very much please help with question numbrer55735

$${Thank}\:{you}\:{very}\:{much} \\ $$$${please}\:{help}\:{with}\:{question}\:{numbrer}\mathrm{55735} \\ $$$$ \\ $$

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