Question Number 152940 by nadovic last updated on 03/Sep/21
$$\:\mathrm{In}\:\mathrm{bottle}\:\mathrm{manufacturing}\:\mathrm{company},\:\mathrm{it} \\ $$$$\mathrm{was}\:\mathrm{observed}\:\mathrm{that}\:\mathrm{5\%}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bottles} \\ $$$$\mathrm{manufactured}\:\mathrm{were}\:\mathrm{defective}.\:\mathrm{In}\:\mathrm{a}\: \\ $$$$\mathrm{random}\:\mathrm{sample}\:\mathrm{of}\:\mathrm{150}\:\mathrm{bottles},\:\mathrm{find}\: \\ $$$$\mathrm{probability}\:\mathrm{that}\: \\ $$$$\:\left({a}\right)\:\mathrm{exactly}\:\mathrm{3}, \\ $$$$\:\left({b}\right)\:\mathrm{between}\:\mathrm{3}\:\mathrm{and}\:\mathrm{6}, \\ $$$$\:\left({c}\right)\:\mathrm{at}\:\mathrm{most}\:\mathrm{4}, \\ $$$$\:\mathrm{manufactured}\:\mathrm{bottles}\:\mathrm{are}\:\mathrm{defective}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{Take}\:\:{e}\:=\:\mathrm{2}.\mathrm{718}\right] \\ $$
Answered by physicstutes last updated on 03/Sep/21
$$\mathrm{here}\:\mathrm{we}'\mathrm{re}\:\mathrm{gonna}\:\mathrm{use}\:\mathrm{the}\:\mathrm{binomial} \\ $$$$\mathrm{distribution}:\:{p}\left({X}={x}\right)=\:^{{n}} {C}_{{x}} {p}^{{x}} {q}^{{n}−{x}} \\ $$$${X}\sim{B}\left(\mathrm{150},\mathrm{0}.\mathrm{05}\right) \\ $$$$\left(\mathrm{a}\right)\:{p}\left({X}=\mathrm{5}\right)=\:^{\mathrm{150}} {C}_{\mathrm{5}} \left(\mathrm{0}.\mathrm{05}\right)^{\mathrm{5}} \left(\mathrm{0}.\mathrm{95}\right)^{\mathrm{150}−\mathrm{5}} \\ $$$$\left[\boldsymbol{\mathrm{note}}:\:{p}+{q}=\mathrm{1}\right] \\ $$$$\left(\mathrm{b}\right)\:{p}\left(\mathrm{3}<{x}<\mathrm{6}\right)=\:{p}\left({X}=\mathrm{4}\right)+{p}\left({X}=\mathrm{5}\right) \\ $$$$\left(\mathrm{c}\right)\:{p}\left({X}\leqslant\mathrm{4}\right)=\:{p}\left({X}=\mathrm{0}\right)+{p}\left({X}=\mathrm{1}\right)+{p}\left({X}=\mathrm{2}\right) \\ $$$$+{p}\left({X}=\mathrm{3}\right)+{p}\left({X}=\mathrm{4}\right) \\ $$
Commented by nadovic last updated on 04/Sep/21
$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{use}\:{e}\:=\:\mathrm{2}.\mathrm{718}\:{please}\:??? \\ $$