How do I solve $4p^{3}(1-p) = [p^{4} + 4p^{3}(1-p)][1 - p^{4} - (1-p)^{4}]$?
$begingroup$
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
asked Jan 11 at 22:06
user1337user1337
41710
$endgroup$
add a comment |
$begingroup$
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
asked Jan 11 at 22:06
user1337user1337
41710
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 11 at 22:06
user1337user1337
41710
$endgroup$
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
asked Jan 11 at 22:06
user1337user1337
41710
asked Jan 11 at 22:06
user1337user1337
41710
asked Jan 11 at 22:06
user1337user1337
41710
asked Jan 11 at 22:06
user1337user1337
41710
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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?
answered Jan 11 at 22:10
David G. StorkDavid G. Stork
10.8k31432
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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?
answered Jan 11 at 22:10
David G. StorkDavid G. Stork
10.8k31432
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
answered Jan 11 at 22:10
David G. StorkDavid G. Stork
10.8k31432
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
answered Jan 11 at 22:10
David G. StorkDavid G. Stork
10.8k31432
$endgroup$
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?
answered Jan 11 at 22:10
David G. StorkDavid G. Stork
10.8k31432
answered Jan 11 at 22:10
David G. StorkDavid G. Stork
10.8k31432
answered Jan 11 at 22:10
David G. StorkDavid G. Stork
10.8k31432
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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