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Question Number 47481 by limite last updated on 10/Nov/18

$$\mathrm{Three}\mathrm{six}-\mathrm{faced}\mathrm{dice}\mathrm{are}\mathrm{thrown}$$$$\mathrm{together}.\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{sum}$$$$\mathrm{of}\mathrm{the}\mathrm{numbers}\mathrm{appearing}\mathrm{on}\mathrm{the}\mathrm{dice}$$$$\mathrm{is}k(3⩽k⩽8)\mathrm{is}$$

Answered by ajfour last updated on 10/Nov/18

$$coeff.of{x}^3+coeff.of{x}^4+....$$$$...+coeff.of{x}^{8}intheexpansion$$$$of{(x+x^2+x^3+x^4+x^5+x^6)}^{3}beN$$$$P\left(k\right)=\frac{N}{216}.$$$${(x+x^2+x^3+x^4+x^5+x^6)}^3$$$$=x^3\frac{{(1-x^6)}^3}{{(1-x)}^3}$$$$=x^3(1-3x^6+...){(1-x)}^{-3}$$$$N=coeff.of{x}^{k}where3⩽k⩽8is$$$$=coeff.of{x}^{k-3}intheexpansion$$$$of{(1-x)}^{-3}$$$$=\mathrm{\Sigma }{}^{3+k-3-1}{C}_{3-1}=\mathrm{\Sigma }{}^{k-1}{C}_2$$$$={}^7{C}_2{+}^6{C}_2{+}^5{C}_2{+}^4{C}_2{+}^3{C}_2{+}^2{C}_2$$$$=21+15+10+6+3+1=56$$$$P\left(k\right)=\frac{56}{216}=\frac{7}{27}.$$

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