Question Number 144801 by mohammad17 last updated on 29/Jun/21
$${if}\:{you}\:{know}\:{that}\:{the}\:{probability}\:{of}\:{a}\:{picture} \\ $$$${appearing}\:{when}\:{acoin}\:{is}\:{tossed}\:{is}\:\mathrm{2}/\mathrm{5} \\ $$$${then}\:{the}\:{probability}\:{of}\:{getting}\:{writings}\: \\ $$$${when}\:{this}\:{coin}\:{is}\:{tossed}\:\mathrm{6}\:{times}\:? \\ $$
Commented by mohammad17 last updated on 29/Jun/21
$${help}\:{me}\:{sir}\:\:? \\ $$
Commented by mohammad17 last updated on 29/Jun/21
$$????????? \\ $$
Answered by gsk2684 last updated on 29/Jun/21
$$\mathrm{X}\approx\mathrm{B}\left(\mathrm{n},\mathrm{p}\right) \\ $$$$\mathrm{p}=\:\mathrm{prob}'\:\mathrm{of}\:\mathrm{getting}\:\mathrm{writings}\: \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{coin}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{q}=\mathrm{prob}'\:\mathrm{of}\:\mathrm{getting}\:\mathrm{picture}\: \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{coin}\:=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\mathrm{then}\:\mathrm{p}\left(\mathrm{X}\geqslant\mathrm{1}\right)=\mathrm{1}−\mathrm{p}\left(\mathrm{X}=\mathrm{0}\right) \\ $$$$=\mathrm{1}−\mathrm{6}_{\mathrm{C}_{\mathrm{0}} } \mathrm{q}^{\mathrm{6}} \mathrm{p}^{\mathrm{0}} =\mathrm{1}−\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{\mathrm{6}} \\ $$
Commented by mohammad17 last updated on 29/Jun/21
$${sir}\:{i}\:{want}\:{the}\:{result}\: \\ $$
Commented by SLVR last updated on 29/Jun/21
$${very}…{easy}..=\mathrm{15561}/\mathrm{15625} \\ $$