Question Number 207372 by sniper237 last updated on 12/May/24
$$\mathrm{2}\:{students}\:{are}\:{passing}\: \\ $$$${a}\:{test}\:{of}\:\:{n}\:{questions}\:{with} \\ $$$${the}\:{same}\:{chance}\:{to}\:{find}\:{each}\:{one} \\ $$$${Show}\:\:{the}\:{chance}\:{that}\:{they}\:{both} \\ $$$$\:{don}'{t}\:{find}\:{a}\:{same}\:{question}\:{is}\:\:\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} \\ $$
Commented by A5T last updated on 13/May/24
$${If}\:{the}\:{chance}\:{of}\:{finding}\left({getting}\:?\right)\:{one}\:{question}\: \\ $$$${is}\:\frac{{x}}{{y}}\leqslant\mathrm{1};{and}\:{the}\:{students}\:{be}\:{A}\:{and}\:{B}.\:{When}\:{n}=\mathrm{1}; \\ $$$$\:{there}\:{are}\:{three}\:{cases}: \\ $$$$\left({i}\right)\:{A}\:{gets}\:{the}\:{question}\:{and}\:{B}\:{misses}. \\ $$$$\left({ii}\right)\:{B}\:{gets}\:{the}\:{question}\:{and}\:{A}\:{misses} \\ $$$$\left({iii}\right)\:{A}\:{and}\:{B}\:{miss}\:{the}\:{question} \\ $$$$\Rightarrow{Probability}=\frac{{x}}{{y}}\left(\mathrm{1}−\frac{{x}}{{y}}\right)+\left(\mathrm{1}−\frac{{x}}{{y}}\right)\left(\frac{{x}}{{y}}\right)+\frac{\left({y}−{x}\right)^{\mathrm{2}} }{{y}^{\mathrm{2}} } \\ $$$$=\frac{{x}\left({y}−{x}\right)}{{y}^{\mathrm{2}} }+\frac{\left({y}−{x}\right)^{\mathrm{2}} }{{y}^{\mathrm{2}} }=\frac{\left({y}−{x}\right)\left({y}−{x}+{x}\right)}{{y}^{\mathrm{2}} }=\frac{{y}−{x}}{{y}} \\ $$$${So},\:{the}\:{probability}\:{is}\:{dependent}\:{on}\:\frac{{x}}{{y}}. \\ $$$$\:{For}\:{example},\:{if}\:\frac{{x}}{{y}}=\frac{\mathrm{1}}{\mathrm{4}};\:{then}\:{probability}=\frac{\mathrm{3}}{\mathrm{4}}. \\ $$$${But}\:{if}\:\frac{{x}}{{y}}=\mathrm{1};\:{then}\:{probability}=\mathrm{0} \\ $$
Commented by sniper237 last updated on 13/May/24
$${Hint}:\:{The}\:{results}\:{of}\:{that}\:{experience}\: \\ $$$${are}\:\:{sets}\:\left\{\left({A},\:{B}\right)/\:{A},{B}\subset\left\{\mathrm{1},....,{n}\right\}\right\} \\ $$$$\:{of}\:{questions}\:{correctly}\:\:{answered} \\ $$$${And}\:{the}\:{event}\:{here}\:{are}\:{all}\:{those} \\ $$$${such}\:{as}\:{A}\cap{B}=\varnothing \\ $$
Commented by A5T last updated on 13/May/24
$${It}\:{is}\:{dependent}\:{on}\:\:\frac{{x}}{{y}}\left({the}\:{probability}\:{of}\:{them}\right. \\ $$$$\left.{getting}\:{the}\:{correct}\:{answer}\right).\:{So},{unless}\:\frac{{x}}{{y}}=\frac{\mathrm{1}}{\mathrm{4}}, \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} \:{failed}\:{to}\:{hold}\:{for}\:{n}=\mathrm{1}. \\ $$
Commented by sniper237 last updated on 13/May/24
$${Read}\:{the}\:{post}\:{again}.\:{you}\:{don}'{t}\:{need}\:{that}\: \\ $$
Commented by A5T last updated on 13/May/24
$${Do}\:{you}\:{have}\:{a}\:{solution}? \\ $$
Commented by A5T last updated on 13/May/24
$$\left({i}\right)\:{A}\:{gets}\:{the}\:{question}\:{and}\:{B}\:{misses}. \\ $$$$\left({ii}\right)\:{B}\:{gets}\:{the}\:{question}\:{and}\:{A}\:{misses} \\ $$$$\left({iii}\right)\:{A}\:{and}\:{B}\:{miss}\:{the}\:{question} \\ $$$$\left({iv}\right)\:{A}\:{and}\:{B}\:{get}\:{the}\:{question} \\ $$$$ \\ $$$${Then}\:{there}\:{are}\:\frac{\mathrm{3}}{\mathrm{4}}\:{ways}\:{for}\:{A}\:{and}\:{B}\:{not}\:{to}\:{get} \\ $$$${the}\:{same}\:{question}? \\ $$$${So},\:\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} \:{for}\:{n}\:{questions}? \\ $$