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Question Number 117339 by john santu last updated on 11/Oct/20
$${a}\:{coin}\:{tossed}\:\mathrm{8}\:{times}.\:{Probability} \\ $$$${of}\:{appearing}\:{head}\:{at}\:{least}\:\mathrm{3}\:{times}\:{is} \\ $$$$\_\_ \\ $$
Answered by bemath last updated on 11/Oct/20
$$\mathrm{by}\:\mathrm{Binomial}\:\mathrm{distribution}\: \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{p}=\frac{\mathrm{1}}{\mathrm{2}}\:,\:\mathrm{n}\:=\:\mathrm{8}\:,\:\mathrm{X}\geqslant\mathrm{3} \\ $$$$\mathrm{p}\left(\mathrm{X}\geqslant\mathrm{3}\right)\:=\:\underset{\mathrm{n}=\mathrm{3}} {\overset{\mathrm{8}} {\sum}}\mathrm{b}\left(\mathrm{x},\mathrm{8},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\:\mathrm{1}−\left\{\mathrm{p}\left(\mathrm{X}=\mathrm{0}\right)+\mathrm{p}\left(\mathrm{X}=\mathrm{1}\right)+\mathrm{p}\left(\mathrm{X}=\mathrm{2}\right)\right\} \\ $$$$=\mathrm{1}−\left\{\mathrm{C}_{\mathrm{0}} ^{\mathrm{8}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{8}} +\mathrm{C}_{\mathrm{1}} ^{\mathrm{8}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{8}} +\mathrm{C}_{\mathrm{2}} ^{\mathrm{8}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{8}} \right\} \\ $$$$=\mathrm{1}−\left\{\frac{\mathrm{1}}{\mathrm{256}}+\frac{\mathrm{8}}{\mathrm{256}}+\frac{\mathrm{28}}{\mathrm{256}}\right\} \\ $$$$=\mathrm{1}−\frac{\mathrm{37}}{\mathrm{256}}\:=\:\mathrm{0}.\mathrm{855469} \\ $$