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Question Number 33724 by mondodotto@gmail.com last updated on 22/Apr/18

$$\mathbf{if}\mathbf{three}\mathbf{persons}\mathbf{selected}\mathbf{at}\mathbf{random}$$$$\mathbf{are}\mathbf{stopped}\mathbf{on}\mathbf{a}\mathbf{street},\mathbf{what}\mathbf{is}$$$$\mathbf{the}\mathbf{probability}\mathbf{that}$$$$\left(\mathbf{i}\right)\mathbf{all}\mathbf{were}\mathbf{born}\mathbf{on}\mathbf{a}\mathbf{friday}?$$$$\left(\mathbf{ii}\right)\mathbf{two}\mathbf{were}\mathbf{born}\mathbf{on}\mathbf{a}\mathbf{friday}\mathbf{and}$$$$\mathbf{the}\mathbf{others}\mathbf{on}\mathbf{a}\mathbf{thursday}?$$$$\left(\mathbf{iii}\right)\mathbf{none}\mathbf{was}\mathbf{born}\mathbf{on}\mathbf{a}\mathbf{friday}$$$$$$

Commented by mondodotto@gmail.com last updated on 22/Apr/18

$$\mathrm{it}\mathrm{is}\mathrm{urgent}\mathrm{please}\mathrm{help}\mathrm{this}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Apr/18

$$52weeks=1yearso52frida$$$$\left(i\right)52C_3/365C_3$$$$\left(ii\right)\left(52C_2\times 52C_1\right)/365C_3$$$$\left(iio\right)1-\left(52C_3/365C_3\right)$$$$$$$$$$

Answered by MJS last updated on 22/Apr/18

$$\mathrm{week}\mathrm{has}7\mathrm{days}$$$$\left(\mathrm{i}\right)\frac{1}{7^3}=\frac{1}{343}$$$$\left(\mathrm{ii}\right)\frac{1}{7^3}\times \left(\begin{array}{c}3\\ 2\end{array}\right)=\frac{3}{343}$$$$\left(\mathrm{iii}\right){\left(\frac{6}{7}\right)}^3=\frac{216}{343}$$

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