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Question Number 76774 by benjo 1/2 santuyy last updated on 30/Dec/19

$$\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{birth}\mathrm{days}\mathrm{of}$$$$\mathrm{six}\mathrm{different}\mathrm{persons}\mathrm{will}\mathrm{fall}\mathrm{in}\mathrm{exactly}$$$$\mathrm{two}\mathrm{calendar}\mathrm{months}\mathrm{is}$$

Commented by mr W last updated on 30/Dec/19

$$\frac{11}{7776}?$$

Commented by benjo 1/2 santuyy last updated on 30/Dec/19

$$i^{\prime }mgot341/{12}^{5}sir$$

Commented by benjo 1/2 santuyy last updated on 30/Dec/19

$$maybe{i}^{\prime }mwrong$$

Commented by benjo 1/2 santuyy last updated on 30/Dec/19

$$howyourgetitsir?$$

Commented by mr W last updated on 31/Dec/19

$$probabilitytheirbirthdaysarein$$$$thesamemonth:C_1^{12}\times {\left(\frac{1}{12}\right)}^6=\frac{1}{248321}$$$$thesame2monthes:C_2^{12}\times {\left(\frac{2}{12}\right)}^6=\frac{11}{7776}$$$$thesame3monthes:C_3^{12}\times {\left(\frac{3}{12}\right)}^6=\frac{55}{1024}$$$$thesame4monthes:C_4^{12}\times {\left(\frac{4}{12}\right)}^6=\frac{55}{81}$$$$......$$$$thesame12monthes:C_{12}^{12}\times {\left(\frac{12}{12}\right)}^6=1$$

Commented by mr W last updated on 31/Dec/19

$$tochoosetwomonthesfrom12$$$$thereare{C}_2^{12}possibilities.$$$$$$$${let}^{\prime }stakemonthesJanuaryand$$$$February.foronepersonthe$$$$probabilitythathisbirthdayisin$$$$Jan.orFeb.is\frac{2}{12}=\frac{1}{6}.$$$$forsixpersonstheprobability$$$$thattheirbirthdaysareinJan.or$$$$Feb.isthen{\left(\frac{1}{2}\right)}^6.$$$$foranytwomonthes:$$$$\Rightarrow p=C_2^{12}\times {\left(\frac{1}{6}\right)}^6=\frac{11}{7776}$$

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