Expected values of squares
up vote
2
down vote
favorite
Question A fair coin is tossed three times. Let Y be the random variable that denotes the square of the number of heads. For example, in the outcome HTH, there are two heads and Y = 4. What is E[Y]?
My answer:
possible outcomes to toss a coin three times : 0, 1, 2, 3
possible outcomes of Y : 0, 1, 4, 9
E[Y] = (1/6 * 0) + (1/6 * 1) + (1/6 * 4) + (1/6 * 9)
Is it ok? Thanks!
probability discrete-mathematics
asked Dec 4 at 19:50
John
345
add a comment |
up vote
2
down vote
favorite
Question A fair coin is tossed three times. Let Y be the random variable that denotes the square of the number of heads. For example, in the outcome HTH, there are two heads and Y = 4. What is E[Y]?
My answer:
possible outcomes to toss a coin three times : 0, 1, 2, 3
possible outcomes of Y : 0, 1, 4, 9
E[Y] = (1/6 * 0) + (1/6 * 1) + (1/6 * 4) + (1/6 * 9)
Is it ok? Thanks!
probability discrete-mathematics
asked Dec 4 at 19:50
John
345
Where do the factors of $frac 16$ come from? For example, there is only one way to get $Y=3$ so the coefficient of $9$ should be $frac 18$.
– lulu
Dec 4 at 19:57
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Question A fair coin is tossed three times. Let Y be the random variable that denotes the square of the number of heads. For example, in the outcome HTH, there are two heads and Y = 4. What is E[Y]?
My answer:
possible outcomes to toss a coin three times : 0, 1, 2, 3
possible outcomes of Y : 0, 1, 4, 9
E[Y] = (1/6 * 0) + (1/6 * 1) + (1/6 * 4) + (1/6 * 9)
Is it ok? Thanks!
probability discrete-mathematics
asked Dec 4 at 19:50
John
345
Question A fair coin is tossed three times. Let Y be the random variable that denotes the square of the number of heads. For example, in the outcome HTH, there are two heads and Y = 4. What is E[Y]?
My answer:
possible outcomes to toss a coin three times : 0, 1, 2, 3
possible outcomes of Y : 0, 1, 4, 9
E[Y] = (1/6 * 0) + (1/6 * 1) + (1/6 * 4) + (1/6 * 9)
Is it ok? Thanks!
probability discrete-mathematics
probability discrete-mathematics
asked Dec 4 at 19:50
John
345
asked Dec 4 at 19:50
John
345
asked Dec 4 at 19:50
John
345
asked Dec 4 at 19:50
John
345
asked Dec 4 at 19:50
John
345
345
Where do the factors of $frac 16$ come from? For example, there is only one way to get $Y=3$ so the coefficient of $9$ should be $frac 18$.
– lulu
Dec 4 at 19:57
add a comment |
Where do the factors of $frac 16$ come from? For example, there is only one way to get $Y=3$ so the coefficient of $9$ should be $frac 18$.
– lulu
Dec 4 at 19:57
Where do the factors of $frac 16$ come from? For example, there is only one way to get $Y=3$ so the coefficient of $9$ should be $frac 18$.
– lulu
Dec 4 at 19:57
Where do the factors of $frac 16$ come from? For example, there is only one way to get $Y=3$ so the coefficient of $9$ should be $frac 18$.
– lulu
Dec 4 at 19:57
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{h: HEAD. t: TOSSES}$.
begin{align}
&bbox[10px,#ffd]{sum_{h = 0}^{t}{t choose h}
pars{1 over 2}^{t - h}pars{1 over 2}^{h}h^{2}} =
left.{1 over 2^{t}},pars{x,partiald{}{x}}^{2}sum_{h = 0}^{t}
{t choose h}x^{h},rightvert_{ x = 1}
\[5mm] = &
left.{1 over 2^{t}},pars{x,partiald{}{x}}^{2}pars{1 + x}^{t},rightvert_{ x = 1} =
left.{1 over 2^{t}},x,partiald{}{x}txpars{1 + x}^{t - 1}
,rightvert_{ x = 1}
\[5mm] = &
{t over 2^{t}},xbracks{%
pars{1 + x}^{t - 1} + xpars{t - 1}pars{1 + x}^{t - 2}}
_{ x = 1}
\[5mm] = &
{t over 2^{t}}bracks{2^{t - 1} + pars{t - 1}2^{t - 2}}
=
{1 over 2},t + {1 over 4},tpars{t - 1} =
bbx{tpars{t + 1} over 4}
end{align}
answered Dec 4 at 20:21
Felix Marin
66.8k7107139
add a comment |
up vote
0
down vote
In general for $n$ tosses
$$
mathbf{E} = sum_{i=0}^n{n choose i}left(frac1{2}right)^{!!n} i^2
$$
answered Dec 4 at 20:09
karakfa
1,943811
I think simplifies to $frac{n(n+1)}{4}$
– karakfa
Dec 4 at 20:19
add a comment |
up vote
0
down vote
In this particular case the probabilities of 0, 1, 2, and 3 heads are 1/8, 3/8, 3/8, 1/8 and so
$$ E(Y) = (1/8) times 0^2 + (3/8) times 1^2 + (3/8) times 2^2 + (1/8) times 3^2 = 3. $$
In general for $n$ tosses, you have $E(Y) = E(X^2)$ where $X$ is a binomial(n, 1/2) random variable. Furthermore $E(X^2) = E(X)^2 + Var(X)$. It's well-known that $E(X) = n/2, Var(X) = n/4$, and so you have $E(X)^2 = (n/2)^2 + n/4 = n(n+1)/4$. Felix Marin already proved this by a different route.
answered Dec 4 at 20:47
Michael Lugo
17.9k33576
add a comment |
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',
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026066%2fexpected-values-of-squares%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{h: HEAD. t: TOSSES}$.
begin{align}
&bbox[10px,#ffd]{sum_{h = 0}^{t}{t choose h}
pars{1 over 2}^{t - h}pars{1 over 2}^{h}h^{2}} =
left.{1 over 2^{t}},pars{x,partiald{}{x}}^{2}sum_{h = 0}^{t}
{t choose h}x^{h},rightvert_{ x = 1}
\[5mm] = &
left.{1 over 2^{t}},pars{x,partiald{}{x}}^{2}pars{1 + x}^{t},rightvert_{ x = 1} =
left.{1 over 2^{t}},x,partiald{}{x}txpars{1 + x}^{t - 1}
,rightvert_{ x = 1}
\[5mm] = &
{t over 2^{t}},xbracks{%
pars{1 + x}^{t - 1} + xpars{t - 1}pars{1 + x}^{t - 2}}
_{ x = 1}
\[5mm] = &
{t over 2^{t}}bracks{2^{t - 1} + pars{t - 1}2^{t - 2}}
=
{1 over 2},t + {1 over 4},tpars{t - 1} =
bbx{tpars{t + 1} over 4}
end{align}
answered Dec 4 at 20:21
Felix Marin
66.8k7107139
add a comment |
up vote
1
down vote
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{h: HEAD. t: TOSSES}$.
begin{align}
&bbox[10px,#ffd]{sum_{h = 0}^{t}{t choose h}
pars{1 over 2}^{t - h}pars{1 over 2}^{h}h^{2}} =
left.{1 over 2^{t}},pars{x,partiald{}{x}}^{2}sum_{h = 0}^{t}
{t choose h}x^{h},rightvert_{ x = 1}
\[5mm] = &
left.{1 over 2^{t}},pars{x,partiald{}{x}}^{2}pars{1 + x}^{t},rightvert_{ x = 1} =
left.{1 over 2^{t}},x,partiald{}{x}txpars{1 + x}^{t - 1}
,rightvert_{ x = 1}
\[5mm] = &
{t over 2^{t}},xbracks{%
pars{1 + x}^{t - 1} + xpars{t - 1}pars{1 + x}^{t - 2}}
_{ x = 1}
\[5mm] = &
{t over 2^{t}}bracks{2^{t - 1} + pars{t - 1}2^{t - 2}}
=
{1 over 2},t + {1 over 4},tpars{t - 1} =
bbx{tpars{t + 1} over 4}
end{align}
answered Dec 4 at 20:21
Felix Marin
66.8k7107139
add a comment |
up vote
1
down vote
up vote
1
down vote
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{h: HEAD. t: TOSSES}$.
begin{align}
&bbox[10px,#ffd]{sum_{h = 0}^{t}{t choose h}
pars{1 over 2}^{t - h}pars{1 over 2}^{h}h^{2}} =
left.{1 over 2^{t}},pars{x,partiald{}{x}}^{2}sum_{h = 0}^{t}
{t choose h}x^{h},rightvert_{ x = 1}
\[5mm] = &
left.{1 over 2^{t}},pars{x,partiald{}{x}}^{2}pars{1 + x}^{t},rightvert_{ x = 1} =
left.{1 over 2^{t}},x,partiald{}{x}txpars{1 + x}^{t - 1}
,rightvert_{ x = 1}
\[5mm] = &
{t over 2^{t}},xbracks{%
pars{1 + x}^{t - 1} + xpars{t - 1}pars{1 + x}^{t - 2}}
_{ x = 1}
\[5mm] = &
{t over 2^{t}}bracks{2^{t - 1} + pars{t - 1}2^{t - 2}}
=
{1 over 2},t + {1 over 4},tpars{t - 1} =
bbx{tpars{t + 1} over 4}
end{align}
answered Dec 4 at 20:21
Felix Marin
66.8k7107139
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{h: HEAD. t: TOSSES}$.
begin{align}
&bbox[10px,#ffd]{sum_{h = 0}^{t}{t choose h}
pars{1 over 2}^{t - h}pars{1 over 2}^{h}h^{2}} =
left.{1 over 2^{t}},pars{x,partiald{}{x}}^{2}sum_{h = 0}^{t}
{t choose h}x^{h},rightvert_{ x = 1}
\[5mm] = &
left.{1 over 2^{t}},pars{x,partiald{}{x}}^{2}pars{1 + x}^{t},rightvert_{ x = 1} =
left.{1 over 2^{t}},x,partiald{}{x}txpars{1 + x}^{t - 1}
,rightvert_{ x = 1}
\[5mm] = &
{t over 2^{t}},xbracks{%
pars{1 + x}^{t - 1} + xpars{t - 1}pars{1 + x}^{t - 2}}
_{ x = 1}
\[5mm] = &
{t over 2^{t}}bracks{2^{t - 1} + pars{t - 1}2^{t - 2}}
=
{1 over 2},t + {1 over 4},tpars{t - 1} =
bbx{tpars{t + 1} over 4}
end{align}
answered Dec 4 at 20:21
Felix Marin
66.8k7107139
answered Dec 4 at 20:21
Felix Marin
66.8k7107139
answered Dec 4 at 20:21
Felix Marin
66.8k7107139
answered Dec 4 at 20:21
Felix Marin
66.8k7107139
66.8k7107139
add a comment |
add a comment |
up vote
0
down vote
In general for $n$ tosses
$$
mathbf{E} = sum_{i=0}^n{n choose i}left(frac1{2}right)^{!!n} i^2
$$
answered Dec 4 at 20:09
karakfa
1,943811
I think simplifies to $frac{n(n+1)}{4}$
– karakfa
Dec 4 at 20:19
add a comment |
up vote
0
down vote
In general for $n$ tosses
$$
mathbf{E} = sum_{i=0}^n{n choose i}left(frac1{2}right)^{!!n} i^2
$$
answered Dec 4 at 20:09
karakfa
1,943811
I think simplifies to $frac{n(n+1)}{4}$
– karakfa
Dec 4 at 20:19
add a comment |
up vote
0
down vote
up vote
0
down vote
In general for $n$ tosses
$$
mathbf{E} = sum_{i=0}^n{n choose i}left(frac1{2}right)^{!!n} i^2
$$
answered Dec 4 at 20:09
karakfa
1,943811
In general for $n$ tosses
$$
mathbf{E} = sum_{i=0}^n{n choose i}left(frac1{2}right)^{!!n} i^2
$$
answered Dec 4 at 20:09
karakfa
1,943811
answered Dec 4 at 20:09
karakfa
1,943811
answered Dec 4 at 20:09
karakfa
1,943811
answered Dec 4 at 20:09
karakfa
1,943811
1,943811
I think simplifies to $frac{n(n+1)}{4}$
– karakfa
Dec 4 at 20:19
add a comment |
I think simplifies to $frac{n(n+1)}{4}$
– karakfa
Dec 4 at 20:19
I think simplifies to $frac{n(n+1)}{4}$
– karakfa
Dec 4 at 20:19
I think simplifies to $frac{n(n+1)}{4}$
– karakfa
Dec 4 at 20:19
add a comment |
up vote
0
down vote
In this particular case the probabilities of 0, 1, 2, and 3 heads are 1/8, 3/8, 3/8, 1/8 and so
$$ E(Y) = (1/8) times 0^2 + (3/8) times 1^2 + (3/8) times 2^2 + (1/8) times 3^2 = 3. $$
In general for $n$ tosses, you have $E(Y) = E(X^2)$ where $X$ is a binomial(n, 1/2) random variable. Furthermore $E(X^2) = E(X)^2 + Var(X)$. It's well-known that $E(X) = n/2, Var(X) = n/4$, and so you have $E(X)^2 = (n/2)^2 + n/4 = n(n+1)/4$. Felix Marin already proved this by a different route.
answered Dec 4 at 20:47
Michael Lugo
17.9k33576
add a comment |
up vote
0
down vote
In this particular case the probabilities of 0, 1, 2, and 3 heads are 1/8, 3/8, 3/8, 1/8 and so
$$ E(Y) = (1/8) times 0^2 + (3/8) times 1^2 + (3/8) times 2^2 + (1/8) times 3^2 = 3. $$
In general for $n$ tosses, you have $E(Y) = E(X^2)$ where $X$ is a binomial(n, 1/2) random variable. Furthermore $E(X^2) = E(X)^2 + Var(X)$. It's well-known that $E(X) = n/2, Var(X) = n/4$, and so you have $E(X)^2 = (n/2)^2 + n/4 = n(n+1)/4$. Felix Marin already proved this by a different route.
answered Dec 4 at 20:47
Michael Lugo
17.9k33576
add a comment |
up vote
0
down vote
up vote
0
down vote
In this particular case the probabilities of 0, 1, 2, and 3 heads are 1/8, 3/8, 3/8, 1/8 and so
$$ E(Y) = (1/8) times 0^2 + (3/8) times 1^2 + (3/8) times 2^2 + (1/8) times 3^2 = 3. $$
In general for $n$ tosses, you have $E(Y) = E(X^2)$ where $X$ is a binomial(n, 1/2) random variable. Furthermore $E(X^2) = E(X)^2 + Var(X)$. It's well-known that $E(X) = n/2, Var(X) = n/4$, and so you have $E(X)^2 = (n/2)^2 + n/4 = n(n+1)/4$. Felix Marin already proved this by a different route.
answered Dec 4 at 20:47
Michael Lugo
17.9k33576
In this particular case the probabilities of 0, 1, 2, and 3 heads are 1/8, 3/8, 3/8, 1/8 and so
$$ E(Y) = (1/8) times 0^2 + (3/8) times 1^2 + (3/8) times 2^2 + (1/8) times 3^2 = 3. $$
In general for $n$ tosses, you have $E(Y) = E(X^2)$ where $X$ is a binomial(n, 1/2) random variable. Furthermore $E(X^2) = E(X)^2 + Var(X)$. It's well-known that $E(X) = n/2, Var(X) = n/4$, and so you have $E(X)^2 = (n/2)^2 + n/4 = n(n+1)/4$. Felix Marin already proved this by a different route.
answered Dec 4 at 20:47
Michael Lugo
17.9k33576
answered Dec 4 at 20:47
Michael Lugo
17.9k33576
answered Dec 4 at 20:47
Michael Lugo
17.9k33576
answered Dec 4 at 20:47
Michael Lugo
17.9k33576
17.9k33576
add a comment |
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026066%2fexpected-values-of-squares%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
Where do the factors of $frac 16$ come from? For example, there is only one way to get $Y=3$ so the coefficient of $9$ should be $frac 18$.
– lulu
Dec 4 at 19:57