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Question Number 152940 by nadovic last updated on 03/Sep/21

$$\mathrm{In}\mathrm{bottle}\mathrm{manufacturing}\mathrm{company},\mathrm{it}$$$$\mathrm{was}\mathrm{observed}\mathrm{that}5\mathrm{\%}\mathrm{of}\mathrm{the}\mathrm{bottles}$$$$\mathrm{manufactured}\mathrm{were}\mathrm{defective}.\mathrm{In}\mathrm{a}$$$$\mathrm{random}\mathrm{sample}\mathrm{of}150\mathrm{bottles},\mathrm{find}$$$$\mathrm{probability}\mathrm{that}$$$$\left(a\right)\mathrm{exactly}3,$$$$\left(b\right)\mathrm{between}3\mathrm{and}6,$$$$\left(c\right)\mathrm{at}\mathrm{most}4,$$$$\mathrm{manufactured}\mathrm{bottles}\mathrm{are}\mathrm{defective}.$$$$[\mathrm{Take}e=2.718]$$

Answered by physicstutes last updated on 03/Sep/21

$$\mathrm{here}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{gonna}\mathrm{use}\mathrm{the}\mathrm{binomial}$$$$\mathrm{distribution}:p(X=x)={}^n{C}_x{p}^x{q}^{n-x}$$$$X\sim B(150,0.05)$$$$\left(\mathrm{a}\right)p(X=5)={}^{150}{C}_5{(0.05)}^5{(0.95)}^{150-5}$$$$[\mathbf{note}:p+q=1]$$$$\left(\mathrm{b}\right)p(3<x<6)=p(X=4)+p(X=5)$$$$\left(\mathrm{c}\right)p(X⩽4)=p(X=0)+p(X=1)+p(X=2)$$$$+p(X=3)+p(X=4)$$

Commented by nadovic last updated on 04/Sep/21

$$\mathrm{How}\mathrm{do}\mathrm{you}\mathrm{use}e=2.718please???$$

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