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Question Number 60635 by naka3546 last updated on 23/May/19

Answered by MJS last updated on 23/May/19

we have 6^6 =46656 possible results 33= 666663 ⇒ 6 permutations 666654 ⇒ 30 permutations 666555 ⇒ 20 permutations ⇒ probability of sum=33 is ((56)/(46656))=(7/(5832))

$$\mathrm{we}\:\mathrm{have}\:\mathrm{6}^{\mathrm{6}} =\mathrm{46656}\:\mathrm{possible}\:\mathrm{results} \\ $$$$\mathrm{33}= \\ $$$$\mathrm{666663}\:\Rightarrow\:\mathrm{6}\:\mathrm{permutations} \\ $$$$\mathrm{666654}\:\Rightarrow\:\mathrm{30}\:\mathrm{permutations} \\ $$$$\mathrm{666555}\:\Rightarrow\:\mathrm{20}\:\mathrm{permutations} \\ $$$$\Rightarrow\:\mathrm{probability}\:\mathrm{of}\:\mathrm{sum}=\mathrm{33}\:\mathrm{is}\:\frac{\mathrm{56}}{\mathrm{46656}}=\frac{\mathrm{7}}{\mathrm{5832}} \\ $$

Answered by mr W last updated on 23/May/19

(x+x^2 +x^3 +...+x^6 )^6 =((x^6 (1−x^6 )^6 )/((1−x)^6 )) =x^6 (Σ_(k=0) ^6 (−1)^k C_k ^6 x^(6k) )(Σ_(h=0) ^∞ C_5 ^(h+5) x^h ) coef. of x^(33) =C_0 ^6 C_5 ^(32) −C_1 ^6 C_5 ^(26) +C_2 ^6 C_5 ^(20) −C_3 ^6 C_5 ^(14) +C_4 ^6 C_5 ^8 =56 ⇒there are 56 possibilities that the sum is 33. total possibilities=6^6 ⇒p=((56)/6^6 )≈0.12%

$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...+{x}^{\mathrm{6}} \right)^{\mathrm{6}} =\frac{{x}^{\mathrm{6}} \left(\mathrm{1}−{x}^{\mathrm{6}} \right)^{\mathrm{6}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{6}} } \\ $$$$={x}^{\mathrm{6}} \left(\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{{k}} ^{\mathrm{6}} {x}^{\mathrm{6}{k}} \right)\left(\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{5}} ^{{h}+\mathrm{5}} {x}^{{h}} \right) \\ $$$${coef}.\:{of}\:{x}^{\mathrm{33}} ={C}_{\mathrm{0}} ^{\mathrm{6}} {C}_{\mathrm{5}} ^{\mathrm{32}} −{C}_{\mathrm{1}} ^{\mathrm{6}} {C}_{\mathrm{5}} ^{\mathrm{26}} +{C}_{\mathrm{2}} ^{\mathrm{6}} {C}_{\mathrm{5}} ^{\mathrm{20}} −{C}_{\mathrm{3}} ^{\mathrm{6}} {C}_{\mathrm{5}} ^{\mathrm{14}} +{C}_{\mathrm{4}} ^{\mathrm{6}} {C}_{\mathrm{5}} ^{\mathrm{8}} =\mathrm{56} \\ $$$$\Rightarrow{there}\:{are}\:\mathrm{56}\:{possibilities}\:{that}\:{the} \\ $$$${sum}\:{is}\:\mathrm{33}. \\ $$$${total}\:{possibilities}=\mathrm{6}^{\mathrm{6}} \\ $$$$\Rightarrow{p}=\frac{\mathrm{56}}{\mathrm{6}^{\mathrm{6}} }\approx\mathrm{0}.\mathrm{12\%} \\ $$

Commented by mr W last updated on 24/May/19

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