Question Number 224476 by Mingma last updated on 14/Sep/25
Answered by mr W last updated on 14/Sep/25
$$\left.\mathrm{9}\right) \\ $$$${when}\:{the}\:{coin}\:{is}\:{tossed}\:\mathrm{6}\:{times},\:{you} \\ $$$${might}\:{get} \\ $$$$\mathrm{6}\:{heads}\:{or} \\ $$$$\mathrm{5}\:{heads}\:{or} \\ $$$$\mathrm{4}\:{heads}\:{or} \\ $$$$\mathrm{3}\:{heads}\:{or} \\ $$$$\mathrm{2}\:{heads}\:{or} \\ $$$$\mathrm{1}\:{head}\:{or} \\ $$$$\mathrm{0}\:{head} \\ $$$${the}\:{probability}\:{that}\:{you}\:{are}\:{paid}\:\$\mathrm{2} \\ $$$${is}\:\frac{\mathrm{1}}{\mathrm{7}},\:{but}\:{the}\:{probalitity}\:{that}\:{you} \\ $$$${pay}\:\$\mathrm{1}\:{is}\:\frac{\mathrm{6}}{\mathrm{7}}.\: \\ $$$$\frac{\mathrm{2}}{\mathrm{7}}<\frac{\mathrm{6}}{\mathrm{7}}\:\Rightarrow\:{bad}\:{bet}\:{for}\:{you}! \\ $$
Commented by Mingma last updated on 15/Sep/25
Nice one, Prof!
Answered by mr W last updated on 14/Sep/25
$${bonus}\:{question} \\ $$$${say}\:{the}\:{number}\:{of}\:{balls}\:{in}\:{the}\:{box} \\ $$$${number}\:{i}\:{is}\:{n}_{{i}} \:\left({i}=\mathrm{1},\mathrm{2},…,\mathrm{8}\right) \\ $$$${we}\:{have}\:{n}_{{i}} \geqslant\mathrm{0},\:{n}_{\mathrm{1}} +{n}_{\mathrm{2}} +…+{n}_{\mathrm{8}} =\mathrm{40} \\ $$$${number}\:{of}\:{solutions}\:{is}\:{the}\:{coef}.\:{of} \\ $$$${x}^{\mathrm{40}} \:{in}\:{expansion}\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{8}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{7}} ^{{k}+\mathrm{7}} {x}^{{k}} , \\ $$$${i}.{e}.\:{C}_{\mathrm{7}} ^{\mathrm{47}} =\mathrm{62}\:\mathrm{891}\:\mathrm{499} \\ $$$${generally}\:{to}\:{distribute}\:\boldsymbol{{n}}\:{identical}\: \\ $$$${objects}\:{into}\:\boldsymbol{{r}}\:{distinct}\:{boxes}\:{there} \\ $$$${are}\:\boldsymbol{{C}}_{\boldsymbol{{r}}−\mathrm{1}} ^{\boldsymbol{{n}}+\boldsymbol{{r}}−\mathrm{1}} \:{ways}. \\ $$$${this}\:{can}\:{be}\:{also}\:{proved}\:\:{using}\: \\ $$$${star}\:\&\:{bar}\:{approach}: \\ $$
Commented by mr W last updated on 14/Sep/25
