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Question-171993

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Question Number 171993 by Tawa11 last updated on 23/Jun/22
Commented by Tawa11 last updated on 23/Jun/22
A soldier fires shots from his rifle and he is expected to hit a target 5 out of 6 shots. If the soldier fires five shots, find the probability that he will (a) miss each time (b) hit the target thrice (c) hit the target at most twice
$$\mathrm{A}\:\mathrm{soldier}\:\mathrm{fires}\:\mathrm{shots}\:\mathrm{from}\:\mathrm{his}\:\mathrm{rifle}\:\mathrm{and}\:\mathrm{he}\:\mathrm{is}\:\mathrm{expected}\:\mathrm{to}\:\mathrm{hit}\:\mathrm{a}\:\mathrm{target}\:\:\mathrm{5}\:\:\mathrm{out}\:\mathrm{of}\:\:\mathrm{6} \\ $$$$\mathrm{shots}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{soldier}\:\mathrm{fires}\:\mathrm{five}\:\mathrm{shots},\:\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{he}\:\mathrm{will} \\ $$$$\left(\mathrm{a}\right)\:\:\mathrm{miss}\:\mathrm{each}\:\mathrm{time} \\ $$$$\left(\mathrm{b}\right)\:\:\mathrm{hit}\:\mathrm{the}\:\mathrm{target}\:\mathrm{thrice} \\ $$$$\left(\mathrm{c}\right)\:\:\mathrm{hit}\:\mathrm{the}\:\mathrm{target}\:\mathrm{at}\:\mathrm{most}\:\mathrm{twice} \\ $$
Commented by mr W last updated on 25/Jun/22
for a shot the probability that he hits is p=(5/6). the probability that he misses is q=1−p=(1/6). (a) p(0)=C_0 ^5 ((5/6))^0 ((1/6))^5 =(1/(7776)) (b) p(3)=C_3 ^5 ((5/6))^3 ((1/6))^2 =((625)/(3888))=16.1% (c) p(1)=C_1 ^5 ((5/6))^1 ((1/6))^4 =((25)/(7776)) p(2)=C_2 ^5 ((5/6))^2 ((1/6))^3 =((125)/(3888)) p(≤2)=p(0)+p(1)+p(2) =(1/(7776))+((25)/(7776))+((125)/(3888))=((23)/(648))=3.5%
$${for}\:{a}\:{shot}\:{the}\:{probability}\:{that}\:{he}\:{hits} \\ $$$${is}\:{p}=\frac{\mathrm{5}}{\mathrm{6}}.\:{the}\:{probability}\:{that}\:{he}\:{misses} \\ $$$${is}\:{q}=\mathrm{1}−{p}=\frac{\mathrm{1}}{\mathrm{6}}. \\ $$$$\left({a}\right) \\ $$$${p}\left(\mathrm{0}\right)={C}_{\mathrm{0}} ^{\mathrm{5}} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{7776}} \\ $$$$\left({b}\right) \\ $$$${p}\left(\mathrm{3}\right)={C}_{\mathrm{3}} ^{\mathrm{5}} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} =\frac{\mathrm{625}}{\mathrm{3888}}=\mathrm{16}.\mathrm{1\%} \\ $$$$\left({c}\right) \\ $$$${p}\left(\mathrm{1}\right)={C}_{\mathrm{1}} ^{\mathrm{5}} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{4}} =\frac{\mathrm{25}}{\mathrm{7776}} \\ $$$${p}\left(\mathrm{2}\right)={C}_{\mathrm{2}} ^{\mathrm{5}} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{3}} =\frac{\mathrm{125}}{\mathrm{3888}} \\ $$$$ \\ $$$${p}\left(\leqslant\mathrm{2}\right)={p}\left(\mathrm{0}\right)+{p}\left(\mathrm{1}\right)+{p}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{7776}}+\frac{\mathrm{25}}{\mathrm{7776}}+\frac{\mathrm{125}}{\mathrm{3888}}=\frac{\mathrm{23}}{\mathrm{648}}=\mathrm{3}.\mathrm{5\%} \\ $$
Commented by Tawa11 last updated on 25/Jun/22
God bless you sir. I appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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