Question Number 111447 by mohammad17 last updated on 03/Sep/20
Commented by mohammad17 last updated on 03/Sep/20
$${hep}\:{me}\:{sir} \\ $$
Answered by Aziztisffola last updated on 03/Sep/20
$$\mathrm{1}.\:\mathrm{X}\left(\Omega\right)=\left\{\mathrm{0};\mathrm{1};…;\mathrm{n}\right\} \\ $$$$\left.\mathrm{2}.\:\mathrm{a}\right)\:\mathrm{X}\rightsquigarrow\mathrm{B}\left(\mathrm{5};\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\:\mathrm{p}\left(\mathrm{X}=\mathrm{k}\right)=\begin{pmatrix}{\mathrm{5}}\\{\mathrm{k}}\end{pmatrix}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{k}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{5}−\mathrm{k}} \:\mathrm{k}=\mathrm{0};\mathrm{1}…\mathrm{5} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{p}\left(\mathrm{X}\geqslant\mathrm{3}\right)=\mathrm{p}\left(\mathrm{X}=\mathrm{3}\right)+\mathrm{p}\left(\mathrm{X}=\mathrm{4}\right)+\mathrm{p}\left(\mathrm{X}=\mathrm{5}\right) \\ $$$$\left.\:\mathrm{d}\right)\mathrm{E}\left(\mathrm{X}\right)=\mathrm{np}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\:\mathrm{V}\left(\mathrm{X}\right)=\mathrm{np}\left(\mathrm{1}−\mathrm{p}\right)=\frac{\mathrm{5}}{\mathrm{3}}×\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{10}}{\mathrm{9}} \\ $$$$ \\ $$
Commented by mohammad17 last updated on 03/Sep/20
$${thank}\:{you}\:{sir} \\ $$