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Question Number 76774 by benjo 1/2 santuyy last updated on 30/Dec/19
The probability that the birth days of  six different persons will fall in exactly  two calendar months is
$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{birth}\:\mathrm{days}\:\mathrm{of} \\ $$$$\mathrm{six}\:\mathrm{different}\:\mathrm{persons}\:\mathrm{will}\:\mathrm{fall}\:\mathrm{in}\:\mathrm{exactly} \\ $$$$\mathrm{two}\:\mathrm{calendar}\:\mathrm{months}\:\mathrm{is} \\ $$
Commented by mr W last updated on 30/Dec/19
((11)/(7776)) ?
$$\frac{\mathrm{11}}{\mathrm{7776}}\:? \\ $$
Commented by benjo 1/2 santuyy last updated on 30/Dec/19
i′m got 341/12^5  sir
$${i}'{m}\:{got}\:\mathrm{341}/\mathrm{12}^{\mathrm{5}} \:{sir} \\ $$
Commented by benjo 1/2 santuyy last updated on 30/Dec/19
may be i′m wrong
$${may}\:{be}\:{i}'{m}\:{wrong} \\ $$
Commented by benjo 1/2 santuyy last updated on 30/Dec/19
how your get it sir ?
$${how}\:{your}\:{get}\:{it}\:{sir}\:? \\ $$
Commented by mr W last updated on 31/Dec/19
probability their birthdays are in  the same month: C_1 ^(12) ×((1/(12)))^6 =(1/(248321))  the same 2 monthes: C_2 ^(12) ×((2/(12)))^6 =((11)/(7776))  the same 3 monthes: C_3 ^(12) ×((3/(12)))^6 =((55)/(1024))  the same 4 monthes: C_4 ^(12) ×((4/(12)))^6 =((55)/(81))  ......  the same 12 monthes: C_(12) ^(12) ×(((12)/(12)))^6 =1
$${probability}\:{their}\:{birthdays}\:{are}\:{in} \\ $$$${the}\:{same}\:{month}:\:{C}_{\mathrm{1}} ^{\mathrm{12}} ×\left(\frac{\mathrm{1}}{\mathrm{12}}\right)^{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{248321}} \\ $$$${the}\:{same}\:\mathrm{2}\:{monthes}:\:{C}_{\mathrm{2}} ^{\mathrm{12}} ×\left(\frac{\mathrm{2}}{\mathrm{12}}\right)^{\mathrm{6}} =\frac{\mathrm{11}}{\mathrm{7776}} \\ $$$${the}\:{same}\:\mathrm{3}\:{monthes}:\:{C}_{\mathrm{3}} ^{\mathrm{12}} ×\left(\frac{\mathrm{3}}{\mathrm{12}}\right)^{\mathrm{6}} =\frac{\mathrm{55}}{\mathrm{1024}} \\ $$$${the}\:{same}\:\mathrm{4}\:{monthes}:\:{C}_{\mathrm{4}} ^{\mathrm{12}} ×\left(\frac{\mathrm{4}}{\mathrm{12}}\right)^{\mathrm{6}} =\frac{\mathrm{55}}{\mathrm{81}} \\ $$$$…… \\ $$$${the}\:{same}\:\mathrm{12}\:{monthes}:\:{C}_{\mathrm{12}} ^{\mathrm{12}} ×\left(\frac{\mathrm{12}}{\mathrm{12}}\right)^{\mathrm{6}} =\mathrm{1} \\ $$
Commented by mr W last updated on 31/Dec/19
to choose two monthes from 12  there are C_2 ^(12)  possibilities.    let′s take monthes January and  February. for one person the  probability that his birthday is in   Jan. or Feb. is (2/(12))=(1/6).  for six persons the probability  that their birthdays are in Jan. or  Feb. is then ((1/2))^6 .  for any two monthes:  ⇒p=C_2 ^(12) ×((1/6))^6 =((11)/(7776))
$${to}\:{choose}\:{two}\:{monthes}\:{from}\:\mathrm{12} \\ $$$${there}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{12}} \:{possibilities}. \\ $$$$ \\ $$$${let}'{s}\:{take}\:{monthes}\:{January}\:{and} \\ $$$${February}.\:{for}\:{one}\:{person}\:{the} \\ $$$${probability}\:{that}\:{his}\:{birthday}\:{is}\:{in}\: \\ $$$${Jan}.\:{or}\:{Feb}.\:{is}\:\frac{\mathrm{2}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{6}}. \\ $$$${for}\:{six}\:{persons}\:{the}\:{probability} \\ $$$${that}\:{their}\:{birthdays}\:{are}\:{in}\:{Jan}.\:{or} \\ $$$${Feb}.\:{is}\:{then}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{6}} . \\ $$$${for}\:{any}\:{two}\:{monthes}: \\ $$$$\Rightarrow{p}={C}_{\mathrm{2}} ^{\mathrm{12}} ×\left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{6}} =\frac{\mathrm{11}}{\mathrm{7776}} \\ $$

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