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Question Number 171993 by Tawa11 last updated on 23/Jun/22

Commented by Tawa11 last updated on 23/Jun/22

$$\mathrm{A}\mathrm{soldier}\mathrm{fires}\mathrm{shots}\mathrm{from}\mathrm{his}\mathrm{rifle}\mathrm{and}\mathrm{he}\mathrm{is}\mathrm{expected}\mathrm{to}\mathrm{hit}\mathrm{a}\mathrm{target}5\mathrm{out}\mathrm{of}6$$$$\mathrm{shots}.\mathrm{If}\mathrm{the}\mathrm{soldier}\mathrm{fires}\mathrm{five}\mathrm{shots},\mathrm{find}\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{he}\mathrm{will}$$$$\left(\mathrm{a}\right)\mathrm{miss}\mathrm{each}\mathrm{time}$$$$\left(\mathrm{b}\right)\mathrm{hit}\mathrm{the}\mathrm{target}\mathrm{thrice}$$$$\left(\mathrm{c}\right)\mathrm{hit}\mathrm{the}\mathrm{target}\mathrm{at}\mathrm{most}\mathrm{twice}$$

Commented by mr W last updated on 25/Jun/22

$$forashottheprobabilitythathehits$$$$isp=\frac{5}{6}.theprobabilitythathemisses$$$$isq=1-p=\frac{1}{6}.$$$$\left(a\right)$$$$p\left(0\right)=C_0^5{\left(\frac{5}{6}\right)}^0{\left(\frac{1}{6}\right)}^5=\frac{1}{7776}$$$$\left(b\right)$$$$p\left(3\right)=C_3^5{\left(\frac{5}{6}\right)}^3{\left(\frac{1}{6}\right)}^2=\frac{625}{3888}=16.1\mathrm{\%}$$$$\left(c\right)$$$$p\left(1\right)=C_1^5{\left(\frac{5}{6}\right)}^1{\left(\frac{1}{6}\right)}^4=\frac{25}{7776}$$$$p\left(2\right)=C_2^5{\left(\frac{5}{6}\right)}^2{\left(\frac{1}{6}\right)}^3=\frac{125}{3888}$$$$$$$$p(⩽2)=p\left(0\right)+p\left(1\right)+p\left(2\right)$$$$=\frac{1}{7776}+\frac{25}{7776}+\frac{125}{3888}=\frac{23}{648}=3.5\mathrm{\%}$$

Commented by Tawa11 last updated on 25/Jun/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

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