Question Number 90258 by jagoll last updated on 22/Apr/20
Commented by john santu last updated on 22/Apr/20
$$\left({a}\right){P}\left({X}\leqslant\mathrm{3}\right)\:=\:\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{3}} {\sum}}{C}_{{k}} ^{\mathrm{25}} \left(\mathrm{0}.\mathrm{05}\right)^{{k}} \:\left(\mathrm{0}.\mathrm{95}\right)^{\mathrm{3}−{k}} \\ $$$$\left({b}\right){P}\left({X}\geqslant\mathrm{4}\right)\:=\:\mathrm{1}−\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{3}} {\sum}}{C}_{{k}} ^{\mathrm{25}} \:\left(\mathrm{0}.\mathrm{05}\right)^{{k}} \:\left(\mathrm{0}.\mathrm{95}\right)^{\mathrm{3}−{k}} \\ $$$$\left({c}\right)\:{P}\left(\mathrm{1}\leqslant{X}\leqslant\mathrm{3}\right)\:=\:\underset{{k}\:=\:\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{C}_{{k}} ^{\mathrm{25}} \:\left(\mathrm{0}.\mathrm{05}\right)^{{k}} \:\left(\mathrm{0}.\mathrm{95}\right)^{\mathrm{3}−{k}} \\ $$