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
$$\frac{\mathrm{11}}{\mathrm{7776}}\:? \\ $$
Commented by benjo 1/2 santuyy last updated on 30/Dec/19
$${i}'{m}\:{got}\:\mathrm{341}/\mathrm{12}^{\mathrm{5}} \:{sir} \\ $$
Commented by benjo 1/2 santuyy last updated on 30/Dec/19
$${may}\:{be}\:{i}'{m}\:{wrong} \\ $$
Commented by benjo 1/2 santuyy last updated on 30/Dec/19
$${how}\:{your}\:{get}\:{it}\:{sir}\:? \\ $$
Commented by mr W last updated on 31/Dec/19
$${probability}\:{their}\:{birthdays}\:{are}\:{in} \\ $$$${the}\:{same}\:{month}:\:{C}_{\mathrm{1}} ^{\mathrm{12}} ×\left(\frac{\mathrm{1}}{\mathrm{12}}\right)^{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{248321}} \\ $$$${the}\:{same}\:\mathrm{2}\:{monthes}:\:{C}_{\mathrm{2}} ^{\mathrm{12}} ×\left(\frac{\mathrm{2}}{\mathrm{12}}\right)^{\mathrm{6}} =\frac{\mathrm{11}}{\mathrm{7776}} \\ $$$${the}\:{same}\:\mathrm{3}\:{monthes}:\:{C}_{\mathrm{3}} ^{\mathrm{12}} ×\left(\frac{\mathrm{3}}{\mathrm{12}}\right)^{\mathrm{6}} =\frac{\mathrm{55}}{\mathrm{1024}} \\ $$$${the}\:{same}\:\mathrm{4}\:{monthes}:\:{C}_{\mathrm{4}} ^{\mathrm{12}} ×\left(\frac{\mathrm{4}}{\mathrm{12}}\right)^{\mathrm{6}} =\frac{\mathrm{55}}{\mathrm{81}} \\ $$$$...... \\ $$$${the}\:{same}\:\mathrm{12}\:{monthes}:\:{C}_{\mathrm{12}} ^{\mathrm{12}} ×\left(\frac{\mathrm{12}}{\mathrm{12}}\right)^{\mathrm{6}} =\mathrm{1} \\ $$
Commented by mr W last updated on 31/Dec/19
$${to}\:{choose}\:{two}\:{monthes}\:{from}\:\mathrm{12} \\ $$$${there}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{12}} \:{possibilities}. \\ $$$$ \\ $$$${let}'{s}\:{take}\:{monthes}\:{January}\:{and} \\ $$$${February}.\:{for}\:{one}\:{person}\:{the} \\ $$$${probability}\:{that}\:{his}\:{birthday}\:{is}\:{in}\: \\ $$$${Jan}.\:{or}\:{Feb}.\:{is}\:\frac{\mathrm{2}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{6}}. \\ $$$${for}\:{six}\:{persons}\:{the}\:{probability} \\ $$$${that}\:{their}\:{birthdays}\:{are}\:{in}\:{Jan}.\:{or} \\ $$$${Feb}.\:{is}\:{then}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{6}} . \\ $$$${for}\:{any}\:{two}\:{monthes}: \\ $$$$\Rightarrow{p}={C}_{\mathrm{2}} ^{\mathrm{12}} ×\left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{6}} =\frac{\mathrm{11}}{\mathrm{7776}} \\ $$