How do I solve $4p^{3}(1-p) = [p^{4} + 4p^{3}(1-p)][1 - p^{4} - (1-p)^{4}]$?












0












$begingroup$


Suppose any child is male with probability $p$ or female with probability $1 − p$, independently of other children. In a family with four children, let $A$ be the event that there is at most one girl, and
$B$ the event that there are children of both sexes. Show that there is a value of $p$, with $0 < p < 1/2$, such that $A$ and $B$ are independent.



MY ATTEMPT



In the first place, we get $textbf{P}(A) = p^{4} + 4p^{3}(1-p)$ and $textbf{P}(B) = 1 - p^{4} - (1-p)^{4}$. Finally, we have $textbf{P}(Acap B) = 4p^{3}(1-p)$. According to the definition of independency, we obtain the following equation to solve



begin{align*}
4p^{3}(1-p) = [p^{4} + 4p^{3}(1-p)][1 - p^{4} - (1-p)^{4}]
end{align*}



Wolfram gives $1$ and $0.41133$ as the reals solution, but I suspect there is another approach to the problem. Could someone provide me an answer or an alternative way to tackle this exercise? I will be grateful anyway. Thanks in advance.






probability polynomials probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not sure what you mean, but $$ 4,{p}^{3} left( 1-p right) - left( {p}^{4}+4,{p}^{3} left( 1-p right) right) left( 1-{p}^{4}- left( 1-p right) ^{4} right) = -2,{p}^{3} left( 3,{p}^{3}-4,{p}^{2}+6,p-2 right) left( p-1 right) ^{2} $$ so actually it is 8th degree and I don't see how you turn the main equation of degree 3 into 2?
    $endgroup$
    – Diger
    Jan 11 at 22:47










  • $begingroup$
    @Diger You are right, I got a sign wrong - so easy to do.
    $endgroup$
    – Mark Bennet
    Jan 11 at 22:59










  • $begingroup$
    @Diger You still get the cubic factor on which you can use intermediate value - the idea of finding a root in a range is an indication that this might be a method to use.
    $endgroup$
    – Mark Bennet
    Jan 11 at 23:00










  • $begingroup$
    As in $f(p)=3p^3-4p^2+6p-2$ and $f(0)=-2$ and $f(1/2)=3/8$ and now use IVT?
    $endgroup$
    – Diger
    Jan 11 at 23:11


















0












$begingroup$


Suppose any child is male with probability $p$ or female with probability $1 − p$, independently of other children. In a family with four children, let $A$ be the event that there is at most one girl, and
$B$ the event that there are children of both sexes. Show that there is a value of $p$, with $0 < p < 1/2$, such that $A$ and $B$ are independent.



MY ATTEMPT



In the first place, we get $textbf{P}(A) = p^{4} + 4p^{3}(1-p)$ and $textbf{P}(B) = 1 - p^{4} - (1-p)^{4}$. Finally, we have $textbf{P}(Acap B) = 4p^{3}(1-p)$. According to the definition of independency, we obtain the following equation to solve



begin{align*}
4p^{3}(1-p) = [p^{4} + 4p^{3}(1-p)][1 - p^{4} - (1-p)^{4}]
end{align*}



Wolfram gives $1$ and $0.41133$ as the reals solution, but I suspect there is another approach to the problem. Could someone provide me an answer or an alternative way to tackle this exercise? I will be grateful anyway. Thanks in advance.






probability polynomials probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not sure what you mean, but $$ 4,{p}^{3} left( 1-p right) - left( {p}^{4}+4,{p}^{3} left( 1-p right) right) left( 1-{p}^{4}- left( 1-p right) ^{4} right) = -2,{p}^{3} left( 3,{p}^{3}-4,{p}^{2}+6,p-2 right) left( p-1 right) ^{2} $$ so actually it is 8th degree and I don't see how you turn the main equation of degree 3 into 2?
    $endgroup$
    – Diger
    Jan 11 at 22:47










  • $begingroup$
    @Diger You are right, I got a sign wrong - so easy to do.
    $endgroup$
    – Mark Bennet
    Jan 11 at 22:59










  • $begingroup$
    @Diger You still get the cubic factor on which you can use intermediate value - the idea of finding a root in a range is an indication that this might be a method to use.
    $endgroup$
    – Mark Bennet
    Jan 11 at 23:00










  • $begingroup$
    As in $f(p)=3p^3-4p^2+6p-2$ and $f(0)=-2$ and $f(1/2)=3/8$ and now use IVT?
    $endgroup$
    – Diger
    Jan 11 at 23:11
















0












0








0





$begingroup$


Suppose any child is male with probability $p$ or female with probability $1 − p$, independently of other children. In a family with four children, let $A$ be the event that there is at most one girl, and
$B$ the event that there are children of both sexes. Show that there is a value of $p$, with $0 < p < 1/2$, such that $A$ and $B$ are independent.



MY ATTEMPT



In the first place, we get $textbf{P}(A) = p^{4} + 4p^{3}(1-p)$ and $textbf{P}(B) = 1 - p^{4} - (1-p)^{4}$. Finally, we have $textbf{P}(Acap B) = 4p^{3}(1-p)$. According to the definition of independency, we obtain the following equation to solve



begin{align*}
4p^{3}(1-p) = [p^{4} + 4p^{3}(1-p)][1 - p^{4} - (1-p)^{4}]
end{align*}



Wolfram gives $1$ and $0.41133$ as the reals solution, but I suspect there is another approach to the problem. Could someone provide me an answer or an alternative way to tackle this exercise? I will be grateful anyway. Thanks in advance.






probability polynomials probability-distributions







share|cite|improve this question









$endgroup$




Suppose any child is male with probability $p$ or female with probability $1 − p$, independently of other children. In a family with four children, let $A$ be the event that there is at most one girl, and
$B$ the event that there are children of both sexes. Show that there is a value of $p$, with $0 < p < 1/2$, such that $A$ and $B$ are independent.



MY ATTEMPT



In the first place, we get $textbf{P}(A) = p^{4} + 4p^{3}(1-p)$ and $textbf{P}(B) = 1 - p^{4} - (1-p)^{4}$. Finally, we have $textbf{P}(Acap B) = 4p^{3}(1-p)$. According to the definition of independency, we obtain the following equation to solve



begin{align*}
4p^{3}(1-p) = [p^{4} + 4p^{3}(1-p)][1 - p^{4} - (1-p)^{4}]
end{align*}



Wolfram gives $1$ and $0.41133$ as the reals solution, but I suspect there is another approach to the problem. Could someone provide me an answer or an alternative way to tackle this exercise? I will be grateful anyway. Thanks in advance.






probability polynomials probability-distributions




probability polynomials probability-distributions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 11 at 22:06









user1337user1337

41710




41710












  • $begingroup$
    Not sure what you mean, but $$ 4,{p}^{3} left( 1-p right) - left( {p}^{4}+4,{p}^{3} left( 1-p right) right) left( 1-{p}^{4}- left( 1-p right) ^{4} right) = -2,{p}^{3} left( 3,{p}^{3}-4,{p}^{2}+6,p-2 right) left( p-1 right) ^{2} $$ so actually it is 8th degree and I don't see how you turn the main equation of degree 3 into 2?
    $endgroup$
    – Diger
    Jan 11 at 22:47










  • $begingroup$
    @Diger You are right, I got a sign wrong - so easy to do.
    $endgroup$
    – Mark Bennet
    Jan 11 at 22:59










  • $begingroup$
    @Diger You still get the cubic factor on which you can use intermediate value - the idea of finding a root in a range is an indication that this might be a method to use.
    $endgroup$
    – Mark Bennet
    Jan 11 at 23:00










  • $begingroup$
    As in $f(p)=3p^3-4p^2+6p-2$ and $f(0)=-2$ and $f(1/2)=3/8$ and now use IVT?
    $endgroup$
    – Diger
    Jan 11 at 23:11




















  • $begingroup$
    Not sure what you mean, but $$ 4,{p}^{3} left( 1-p right) - left( {p}^{4}+4,{p}^{3} left( 1-p right) right) left( 1-{p}^{4}- left( 1-p right) ^{4} right) = -2,{p}^{3} left( 3,{p}^{3}-4,{p}^{2}+6,p-2 right) left( p-1 right) ^{2} $$ so actually it is 8th degree and I don't see how you turn the main equation of degree 3 into 2?
    $endgroup$
    – Diger
    Jan 11 at 22:47










  • $begingroup$
    @Diger You are right, I got a sign wrong - so easy to do.
    $endgroup$
    – Mark Bennet
    Jan 11 at 22:59










  • $begingroup$
    @Diger You still get the cubic factor on which you can use intermediate value - the idea of finding a root in a range is an indication that this might be a method to use.
    $endgroup$
    – Mark Bennet
    Jan 11 at 23:00










  • $begingroup$
    As in $f(p)=3p^3-4p^2+6p-2$ and $f(0)=-2$ and $f(1/2)=3/8$ and now use IVT?
    $endgroup$
    – Diger
    Jan 11 at 23:11


















$begingroup$
Not sure what you mean, but $$ 4,{p}^{3} left( 1-p right) - left( {p}^{4}+4,{p}^{3} left( 1-p right) right) left( 1-{p}^{4}- left( 1-p right) ^{4} right) = -2,{p}^{3} left( 3,{p}^{3}-4,{p}^{2}+6,p-2 right) left( p-1 right) ^{2} $$ so actually it is 8th degree and I don't see how you turn the main equation of degree 3 into 2?
$endgroup$
– Diger
Jan 11 at 22:47




$begingroup$
Not sure what you mean, but $$ 4,{p}^{3} left( 1-p right) - left( {p}^{4}+4,{p}^{3} left( 1-p right) right) left( 1-{p}^{4}- left( 1-p right) ^{4} right) = -2,{p}^{3} left( 3,{p}^{3}-4,{p}^{2}+6,p-2 right) left( p-1 right) ^{2} $$ so actually it is 8th degree and I don't see how you turn the main equation of degree 3 into 2?
$endgroup$
– Diger
Jan 11 at 22:47












$begingroup$
@Diger You are right, I got a sign wrong - so easy to do.
$endgroup$
– Mark Bennet
Jan 11 at 22:59




$begingroup$
@Diger You are right, I got a sign wrong - so easy to do.
$endgroup$
– Mark Bennet
Jan 11 at 22:59












$begingroup$
@Diger You still get the cubic factor on which you can use intermediate value - the idea of finding a root in a range is an indication that this might be a method to use.
$endgroup$
– Mark Bennet
Jan 11 at 23:00




$begingroup$
@Diger You still get the cubic factor on which you can use intermediate value - the idea of finding a root in a range is an indication that this might be a method to use.
$endgroup$
– Mark Bennet
Jan 11 at 23:00












$begingroup$
As in $f(p)=3p^3-4p^2+6p-2$ and $f(0)=-2$ and $f(1/2)=3/8$ and now use IVT?
$endgroup$
– Diger
Jan 11 at 23:11






$begingroup$
As in $f(p)=3p^3-4p^2+6p-2$ and $f(0)=-2$ and $f(1/2)=3/8$ and now use IVT?
$endgroup$
– Diger
Jan 11 at 23:11












1 Answer
1






active

oldest

votes


















0












$begingroup$

Use computer algebra to get all the answers:



$$left{0,0,0,1,1,frac{1}{9} left(4-frac{38}{sqrt[3]{9 sqrt{681}-17}}+sqrt[3]{9 sqrt{681}-17}right),frac{4}{9}+frac{19 left(1+i sqrt{3}right)}{9 sqrt[3]{9 sqrt{681}-17}}-frac{1}{18}
left(1-i sqrt{3}right) sqrt[3]{9 sqrt{681}-17},frac{4}{9}+frac{19 left(1-i sqrt{3}right)}{9 sqrt[3]{9
sqrt{681}-17}}-frac{1}{18} left(1+i sqrt{3}right) sqrt[3]{9 sqrt{681}-17}right}$$



Why does this not suffice?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the contribution, David. But I am afraid the author expected a "clever" solution.
    $endgroup$
    – user1337
    Jan 11 at 22:12










  • $begingroup$
    Alas, I don't see a "clever" solution... and wonder why one is needed—or even *wanted*—for such a problem.
    $endgroup$
    – David G. Stork
    Jan 11 at 22:13












  • $begingroup$
    If it helps, the problem comes from a probability textbook.
    $endgroup$
    – user1337
    Jan 11 at 22:14











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070389%2fhow-do-i-solve-4p31-p-p4-4p31-p1-p4-1-p4%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Use computer algebra to get all the answers:



$$left{0,0,0,1,1,frac{1}{9} left(4-frac{38}{sqrt[3]{9 sqrt{681}-17}}+sqrt[3]{9 sqrt{681}-17}right),frac{4}{9}+frac{19 left(1+i sqrt{3}right)}{9 sqrt[3]{9 sqrt{681}-17}}-frac{1}{18}
left(1-i sqrt{3}right) sqrt[3]{9 sqrt{681}-17},frac{4}{9}+frac{19 left(1-i sqrt{3}right)}{9 sqrt[3]{9
sqrt{681}-17}}-frac{1}{18} left(1+i sqrt{3}right) sqrt[3]{9 sqrt{681}-17}right}$$



Why does this not suffice?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the contribution, David. But I am afraid the author expected a "clever" solution.
    $endgroup$
    – user1337
    Jan 11 at 22:12










  • $begingroup$
    Alas, I don't see a "clever" solution... and wonder why one is needed—or even *wanted*—for such a problem.
    $endgroup$
    – David G. Stork
    Jan 11 at 22:13












  • $begingroup$
    If it helps, the problem comes from a probability textbook.
    $endgroup$
    – user1337
    Jan 11 at 22:14
















0












$begingroup$

Use computer algebra to get all the answers:



$$left{0,0,0,1,1,frac{1}{9} left(4-frac{38}{sqrt[3]{9 sqrt{681}-17}}+sqrt[3]{9 sqrt{681}-17}right),frac{4}{9}+frac{19 left(1+i sqrt{3}right)}{9 sqrt[3]{9 sqrt{681}-17}}-frac{1}{18}
left(1-i sqrt{3}right) sqrt[3]{9 sqrt{681}-17},frac{4}{9}+frac{19 left(1-i sqrt{3}right)}{9 sqrt[3]{9
sqrt{681}-17}}-frac{1}{18} left(1+i sqrt{3}right) sqrt[3]{9 sqrt{681}-17}right}$$



Why does this not suffice?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the contribution, David. But I am afraid the author expected a "clever" solution.
    $endgroup$
    – user1337
    Jan 11 at 22:12










  • $begingroup$
    Alas, I don't see a "clever" solution... and wonder why one is needed—or even *wanted*—for such a problem.
    $endgroup$
    – David G. Stork
    Jan 11 at 22:13












  • $begingroup$
    If it helps, the problem comes from a probability textbook.
    $endgroup$
    – user1337
    Jan 11 at 22:14














0












0








0





$begingroup$

Use computer algebra to get all the answers:



$$left{0,0,0,1,1,frac{1}{9} left(4-frac{38}{sqrt[3]{9 sqrt{681}-17}}+sqrt[3]{9 sqrt{681}-17}right),frac{4}{9}+frac{19 left(1+i sqrt{3}right)}{9 sqrt[3]{9 sqrt{681}-17}}-frac{1}{18}
left(1-i sqrt{3}right) sqrt[3]{9 sqrt{681}-17},frac{4}{9}+frac{19 left(1-i sqrt{3}right)}{9 sqrt[3]{9
sqrt{681}-17}}-frac{1}{18} left(1+i sqrt{3}right) sqrt[3]{9 sqrt{681}-17}right}$$



Why does this not suffice?






share|cite|improve this answer









$endgroup$



Use computer algebra to get all the answers:



$$left{0,0,0,1,1,frac{1}{9} left(4-frac{38}{sqrt[3]{9 sqrt{681}-17}}+sqrt[3]{9 sqrt{681}-17}right),frac{4}{9}+frac{19 left(1+i sqrt{3}right)}{9 sqrt[3]{9 sqrt{681}-17}}-frac{1}{18}
left(1-i sqrt{3}right) sqrt[3]{9 sqrt{681}-17},frac{4}{9}+frac{19 left(1-i sqrt{3}right)}{9 sqrt[3]{9
sqrt{681}-17}}-frac{1}{18} left(1+i sqrt{3}right) sqrt[3]{9 sqrt{681}-17}right}$$



Why does this not suffice?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 22:10









David G. StorkDavid G. Stork

10.8k31432




10.8k31432












  • $begingroup$
    Thanks for the contribution, David. But I am afraid the author expected a "clever" solution.
    $endgroup$
    – user1337
    Jan 11 at 22:12










  • $begingroup$
    Alas, I don't see a "clever" solution... and wonder why one is needed—or even *wanted*—for such a problem.
    $endgroup$
    – David G. Stork
    Jan 11 at 22:13












  • $begingroup$
    If it helps, the problem comes from a probability textbook.
    $endgroup$
    – user1337
    Jan 11 at 22:14


















  • $begingroup$
    Thanks for the contribution, David. But I am afraid the author expected a "clever" solution.
    $endgroup$
    – user1337
    Jan 11 at 22:12










  • $begingroup$
    Alas, I don't see a "clever" solution... and wonder why one is needed—or even *wanted*—for such a problem.
    $endgroup$
    – David G. Stork
    Jan 11 at 22:13












  • $begingroup$
    If it helps, the problem comes from a probability textbook.
    $endgroup$
    – user1337
    Jan 11 at 22:14
















$begingroup$
Thanks for the contribution, David. But I am afraid the author expected a "clever" solution.
$endgroup$
– user1337
Jan 11 at 22:12




$begingroup$
Thanks for the contribution, David. But I am afraid the author expected a "clever" solution.
$endgroup$
– user1337
Jan 11 at 22:12












$begingroup$
Alas, I don't see a "clever" solution... and wonder why one is needed—or even *wanted*—for such a problem.
$endgroup$
– David G. Stork
Jan 11 at 22:13






$begingroup$
Alas, I don't see a "clever" solution... and wonder why one is needed—or even *wanted*—for such a problem.
$endgroup$
– David G. Stork
Jan 11 at 22:13














$begingroup$
If it helps, the problem comes from a probability textbook.
$endgroup$
– user1337
Jan 11 at 22:14




$begingroup$
If it helps, the problem comes from a probability textbook.
$endgroup$
– user1337
Jan 11 at 22:14


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070389%2fhow-do-i-solve-4p31-p-p4-4p31-p1-p4-1-p4%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
Read more

The Binding of Isaac: Rebirth/Afterbirth

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
Read more