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Question Number 182360 by Rasheed.Sindhi last updated on 08/Dec/22
In equation ax^2 +bx+c=0, a,b,c are randomly selected from integers; what is the probability that roots will be real?
$$\mathrm{In}\:\mathrm{equation}\:\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}=\mathrm{0},\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are} \\ $$$$\mathrm{randomly}\:\mathrm{selected}\:\mathrm{from}\:\mathrm{integers}; \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{roots} \\ $$$$\mathrm{will}\:\mathrm{be}\:\mathrm{real}? \\ $$
Commented by mr W last updated on 08/Dec/22
i′m not sure if there is a solution. but i think there is a solution when the question is e.g. like a,b,c ∈R, and 0≤a,b,c≤10. in this case we have p=((5+6 ln 2)/(36))≈25.4%, see below. please check.
$${i}'{m}\:{not}\:{sure}\:{if}\:{there}\:{is}\:{a}\:{solution}. \\ $$$${but}\:{i}\:{think}\:{there}\:{is}\:{a}\:{solution}\:{when} \\ $$$${the}\:{question}\:{is}\:{e}.{g}.\:{like}\:{a},{b},{c}\:\in{R},\:{and}\: \\ $$$$\mathrm{0}\leqslant{a},{b},{c}\leqslant\mathrm{10}.\:{in}\:{this}\:{case}\:{we}\:{have} \\ $$$${p}=\frac{\mathrm{5}+\mathrm{6}\:\mathrm{ln}\:\mathrm{2}}{\mathrm{36}}\approx\mathrm{25}.\mathrm{4\%},\:{see}\:{below}.\:{please} \\ $$$${check}. \\ $$
Commented by mr W last updated on 08/Dec/22
this is a nice question. i hope you have a solution for a,b,c as integers.
$${this}\:{is}\:{a}\:{nice}\:{question}.\:{i}\:{hope}\:{you} \\ $$$${have}\:{a}\:{solution}\:{for}\:{a},{b},{c}\:{as}\:{integers}. \\ $$
Commented by Rasheed.Sindhi last updated on 09/Dec/22
No sir,I know nothing about answer/solution.Actually it′s beyond my capacity! But only question raised in my mind and I threw it for discussion.
$${No}\:\boldsymbol{{sir}},{I}\:\:{know}\:{nothing}\:{about} \\ $$$${answer}/{solution}.{Actually}\:{it}'{s}\:{beyond} \\ $$$${my}\:{capacity}!\:{But}\:{only}\:{question} \\ $$$${raised}\:{in}\:{my}\:{mind}\:{and}\:{I}\:{threw} \\ $$$${it}\:{for}\:{discussion}. \\ $$
Commented by mr W last updated on 09/Dec/22
i see, thanks! i think we can′t answer the question, if we don′t restrain the range of the values. that′s the same as we can′t determine the probability of drawing a black ball from a bag, if we don′t know how many balls are in the bag or if the number of balls in the bag is infinite. an other issue is, when a,b,c are integers in a given range, say −10≤a≤20, 10≤b≤50, −20≤c≤0 we can solve it, but we have to check all 31×41×21=26691 discrete points one after one to see which points fulfill b^2 ≥4ac and get the number of such points, say n_1 . then we can calculate the probability p=(n_1 /(26691)). this is an easy but tough and stupid process. but when a,b,c are not discrete integer values, but continuous real values in a given range, we can apply functions and their integrals to find the probability as i showed below. any way, thank you for creating a very interesting question!
$${i}\:{see},\:{thanks}! \\ $$$${i}\:{think}\:{we}\:{can}'{t}\:{answer}\:{the}\:{question}, \\ $$$${if}\:{we}\:{don}'{t}\:{restrain}\:{the}\:{range}\:{of}\:{the} \\ $$$${values}.\:{that}'{s}\:{the}\:{same}\:{as}\:{we}\:{can}'{t} \\ $$$${determine}\:{the}\:{probability}\:{of}\:{drawing} \\ $$$${a}\:{black}\:{ball}\:{from}\:{a}\:{bag},\:{if}\:{we}\:{don}'{t}\: \\ $$$${know}\:{how}\:{many}\:{balls}\:{are}\:{in}\:{the}\:{bag} \\ $$$${or}\:{if}\:{the}\:{number}\:{of}\:{balls}\:{in}\:{the}\:{bag}\:{is} \\ $$$${infinite}.\:{an}\:{other}\:{issue}\:{is},\:{when}\: \\ $$$${a},{b},{c}\:{are}\:{integers}\:{in}\:{a}\:{given}\:{range}, \\ $$$${say}\:−\mathrm{10}\leqslant{a}\leqslant\mathrm{20},\:\mathrm{10}\leqslant{b}\leqslant\mathrm{50},\:−\mathrm{20}\leqslant{c}\leqslant\mathrm{0} \\ $$$${we}\:{can}\:{solve}\:{it},\:{but}\:{we}\:{have}\:{to}\:{check}\: \\ $$$${all}\:\mathrm{31}×\mathrm{41}×\mathrm{21}=\mathrm{26691}\:{discrete}\:{points}\: \\ $$$${one}\:{after}\:{one}\:{to}\:{see}\:{which}\:{points}\: \\ $$$${fulfill}\:\:{b}^{\mathrm{2}} \geqslant\mathrm{4}{ac}\:{and}\:{get}\:{the}\:{number}\: \\ $$$${of}\:{such}\:{points},\:{say}\:{n}_{\mathrm{1}} .\:{then}\:{we}\:{can}\: \\ $$$${calculate}\:{the}\:{probability}\:{p}=\frac{{n}_{\mathrm{1}} }{\mathrm{26691}}.\: \\ $$$${this}\:{is}\:{an}\:{easy}\:{but}\:{tough}\:{and}\:{stupid}\: \\ $$$${process}. \\ $$$${but}\:{when}\:{a},{b},{c}\:{are}\:{not}\:{discrete}\:{integer} \\ $$$${values},\:{but}\:{continuous}\:{real}\:{values}\:{in} \\ $$$${a}\:{given}\:{range},\:{we}\:{can}\:{apply}\:{functions} \\ $$$${and}\:{their}\:{integrals}\:{to}\:{find}\:{the} \\ $$$${probability}\:{as}\:{i}\:{showed}\:{below}. \\ $$$${any}\:{way},\:{thank}\:{you}\:{for}\:{creating}\:{a} \\ $$$${very}\:{interesting}\:{question}! \\ $$
Commented by Rasheed.Sindhi last updated on 09/Dec/22
Grateful sir, for your very deep disscussion and for your very precious time!
$$\mathbb{G}\boldsymbol{\mathrm{rateful}}\:\boldsymbol{\mathrm{sir}},\:\mathrm{for}\:\mathrm{your}\:\boldsymbol{\mathrm{very}}\:\boldsymbol{\mathrm{deep}}\:\mathrm{disscussion} \\ $$$$\mathrm{and}\:\mathrm{for}\:\mathrm{your}\:\boldsymbol{\mathrm{very}}\:\boldsymbol{\mathrm{precious}}\:\mathrm{time}! \\ $$
Answered by mr W last updated on 09/Dec/22
say the question is: a,b,c ∈R and 0≤a,b,c≤10. we know the equation at^2 +bt+c=0 has real roots, when b^2 ≥4ac. let a=x, c=y, b=z. each randomly selected values for a,b,c represent a point inside the cube 10×10×10. the equation has real roots, when the point lies inside the zone b^2 ≥4ac, i.e. z^2 ≥4xy. that′s the hatched areas in the diagram. the probability that the roots are real is then p=((volume of the zone z^2 ≥4xy)/(volume of the cube))
$${say}\:{the}\:{question}\:{is}: \\ $$$${a},{b},{c}\:\in{R}\:{and}\:\mathrm{0}\leqslant{a},{b},{c}\leqslant\mathrm{10}. \\ $$$${we}\:{know}\:{the}\:{equation}\:{at}^{\mathrm{2}} +{bt}+{c}=\mathrm{0} \\ $$$${has}\:{real}\:{roots},\:{when}\:{b}^{\mathrm{2}} \geqslant\mathrm{4}{ac}. \\ $$$${let}\:{a}={x},\:{c}={y},\:{b}={z}. \\ $$$${each}\:{randomly}\:{selected}\:{values}\:{for} \\ $$$${a},{b},{c}\:{represent}\:{a}\:{point}\:{inside}\:{the}\:{cube} \\ $$$$\mathrm{10}×\mathrm{10}×\mathrm{10}.\:{the}\:{equation}\:{has}\:{real} \\ $$$${roots},\:{when}\:{the}\:{point}\:{lies}\:{inside}\:{the} \\ $$$${zone}\:{b}^{\mathrm{2}} \geqslant\mathrm{4}{ac},\:{i}.{e}.\:{z}^{\mathrm{2}} \geqslant\mathrm{4}{xy}.\:{that}'{s}\:{the} \\ $$$${hatched}\:{areas}\:{in}\:{the}\:{diagram}. \\ $$$${the}\:{probability}\:{that}\:{the}\:{roots}\:{are} \\ $$$${real}\:{is}\:{then} \\ $$$${p}=\frac{{volume}\:{of}\:{the}\:{zone}\:{z}^{\mathrm{2}} \geqslant\mathrm{4}{xy}}{{volume}\:{of}\:{the}\:{cube}} \\ $$
Commented by mr W last updated on 08/Dec/22
Commented by mr W last updated on 08/Dec/22
4xy≤z^2 y=(z^2 /(4x)) at y=10: x_1 =(z^2 /(40)) A(z)=(z^2 /(40))×10+∫_x_1 ^(10) (z^2 /(4x))dx A(z)=(z^2 /4)+(z^2 /4)×ln ((400)/z^2 ) V_(zone) =∫_0 ^(10) A(z)dz V_(zone) =∫_0 ^(10) ((z^2 /4)+(z^2 /4)×ln ((400)/z^2 ))dz V_(zone) =(1/4)(((10^3 )/3)+∫_0 ^(10) z^2 ×ln ((400)/z^2 )dz) V_(zone) =((1000)/4)((5/9)+((2 ln 2)/3))=((1000(5+6 ln 2))/(36)) V_(cube) =10^3 =1000 ⇒p=(V_(zone) /V_(cube) )=((5+6 ln 2)/(36))≈25.4%
$$\mathrm{4}{xy}\leqslant{z}^{\mathrm{2}} \\ $$$${y}=\frac{{z}^{\mathrm{2}} }{\mathrm{4}{x}} \\ $$$${at}\:{y}=\mathrm{10}:\:{x}_{\mathrm{1}} =\frac{{z}^{\mathrm{2}} }{\mathrm{40}} \\ $$$${A}\left({z}\right)=\frac{{z}^{\mathrm{2}} }{\mathrm{40}}×\mathrm{10}+\int_{{x}_{\mathrm{1}} } ^{\mathrm{10}} \frac{{z}^{\mathrm{2}} }{\mathrm{4}{x}}{dx} \\ $$$${A}\left({z}\right)=\frac{{z}^{\mathrm{2}} }{\mathrm{4}}+\frac{{z}^{\mathrm{2}} }{\mathrm{4}}×\mathrm{ln}\:\frac{\mathrm{400}}{{z}^{\mathrm{2}} } \\ $$$${V}_{{zone}} =\int_{\mathrm{0}} ^{\mathrm{10}} {A}\left({z}\right){dz} \\ $$$${V}_{{zone}} =\int_{\mathrm{0}} ^{\mathrm{10}} \left(\frac{{z}^{\mathrm{2}} }{\mathrm{4}}+\frac{{z}^{\mathrm{2}} }{\mathrm{4}}×\mathrm{ln}\:\frac{\mathrm{400}}{{z}^{\mathrm{2}} }\right){dz} \\ $$$${V}_{{zone}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{10}^{\mathrm{3}} }{\mathrm{3}}+\int_{\mathrm{0}} ^{\mathrm{10}} {z}^{\mathrm{2}} ×\mathrm{ln}\:\frac{\mathrm{400}}{{z}^{\mathrm{2}} }{dz}\right) \\ $$$${V}_{{zone}} =\frac{\mathrm{1000}}{\mathrm{4}}\left(\frac{\mathrm{5}}{\mathrm{9}}+\frac{\mathrm{2}\:\mathrm{ln}\:\mathrm{2}}{\mathrm{3}}\right)=\frac{\mathrm{1000}\left(\mathrm{5}+\mathrm{6}\:\mathrm{ln}\:\mathrm{2}\right)}{\mathrm{36}} \\ $$$${V}_{{cube}} =\mathrm{10}^{\mathrm{3}} =\mathrm{1000} \\ $$$$\Rightarrow{p}=\frac{{V}_{{zone}} }{{V}_{{cube}} }=\frac{\mathrm{5}+\mathrm{6}\:\mathrm{ln}\:\mathrm{2}}{\mathrm{36}}\approx\mathrm{25}.\mathrm{4\%} \\ $$
Commented by mr W last updated on 09/Dec/22
i hope somebody could comfirm my method and result, or disprove it.
$${i}\:{hope}\:{somebody}\:{could}\:{comfirm}\:{my} \\ $$$${method}\:{and}\:{result},\:{or}\:{disprove}\:{it}. \\ $$

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