Determine the probability of a conditional p.d.f
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So the joint p.d.f is f(x,y) = $c(x^2 + y^2)$ for $0leq xleq 1$ and $0leq y leq 1$
Now the steps I worked out is applying the formula $f(x|y) = frac{f(x,y)}{f_{2}(y)}$
Which gave me
f(x|y) = $frac{3x^2+ 3y^2}{1+3y^2}$.
Assuming that is correct, the answer Pr(X$<frac{1}{2}$|$Y=frac{1}{2}$)should be $frac{2}{7}$.
I solved this to get that answer: $int_{0}^{1/2} frac{3x^2+ 3(frac{1}{2})^2}{1+3(frac{1}{2})^2}$
But the correct answer is $frac{1}{3}$. I don't understand what I'm missing.
probability-theory probability-distributions conditional-probability
edited Nov 18 at 0:40
mathnoob
1,133115
asked Nov 18 at 0:24
Sir lethian
163
add a comment |
up vote
2
down vote
favorite
So the joint p.d.f is f(x,y) = $c(x^2 + y^2)$ for $0leq xleq 1$ and $0leq y leq 1$
Now the steps I worked out is applying the formula $f(x|y) = frac{f(x,y)}{f_{2}(y)}$
Which gave me
f(x|y) = $frac{3x^2+ 3y^2}{1+3y^2}$.
Assuming that is correct, the answer Pr(X$<frac{1}{2}$|$Y=frac{1}{2}$)should be $frac{2}{7}$.
I solved this to get that answer: $int_{0}^{1/2} frac{3x^2+ 3(frac{1}{2})^2}{1+3(frac{1}{2})^2}$
But the correct answer is $frac{1}{3}$. I don't understand what I'm missing.
probability-theory probability-distributions conditional-probability
edited Nov 18 at 0:40
mathnoob
1,133115
asked Nov 18 at 0:24
Sir lethian
163
2
Seems good to me. Maybe you were calculating the wrong thing or the answer you were provided is wrong.
– Stan Tendijck
Nov 18 at 0:50
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
So the joint p.d.f is f(x,y) = $c(x^2 + y^2)$ for $0leq xleq 1$ and $0leq y leq 1$
Now the steps I worked out is applying the formula $f(x|y) = frac{f(x,y)}{f_{2}(y)}$
Which gave me
f(x|y) = $frac{3x^2+ 3y^2}{1+3y^2}$.
Assuming that is correct, the answer Pr(X$<frac{1}{2}$|$Y=frac{1}{2}$)should be $frac{2}{7}$.
I solved this to get that answer: $int_{0}^{1/2} frac{3x^2+ 3(frac{1}{2})^2}{1+3(frac{1}{2})^2}$
But the correct answer is $frac{1}{3}$. I don't understand what I'm missing.
probability-theory probability-distributions conditional-probability
edited Nov 18 at 0:40
mathnoob
1,133115
asked Nov 18 at 0:24
Sir lethian
163
So the joint p.d.f is f(x,y) = $c(x^2 + y^2)$ for $0leq xleq 1$ and $0leq y leq 1$
Now the steps I worked out is applying the formula $f(x|y) = frac{f(x,y)}{f_{2}(y)}$
Which gave me
f(x|y) = $frac{3x^2+ 3y^2}{1+3y^2}$.
Assuming that is correct, the answer Pr(X$<frac{1}{2}$|$Y=frac{1}{2}$)should be $frac{2}{7}$.
I solved this to get that answer: $int_{0}^{1/2} frac{3x^2+ 3(frac{1}{2})^2}{1+3(frac{1}{2})^2}$
But the correct answer is $frac{1}{3}$. I don't understand what I'm missing.
probability-theory probability-distributions conditional-probability
probability-theory probability-distributions conditional-probability
edited Nov 18 at 0:40
mathnoob
1,133115
asked Nov 18 at 0:24
Sir lethian
163
edited Nov 18 at 0:40
mathnoob
1,133115
asked Nov 18 at 0:24
Sir lethian
163
edited Nov 18 at 0:40
mathnoob
1,133115
edited Nov 18 at 0:40
mathnoob
1,133115
edited Nov 18 at 0:40
mathnoob
1,133115
1,133115
asked Nov 18 at 0:24
Sir lethian
163
asked Nov 18 at 0:24
Sir lethian
163
asked Nov 18 at 0:24
Sir lethian
163
163
2
Seems good to me. Maybe you were calculating the wrong thing or the answer you were provided is wrong.
– Stan Tendijck
Nov 18 at 0:50
add a comment |
2
Seems good to me. Maybe you were calculating the wrong thing or the answer you were provided is wrong.
– Stan Tendijck
Nov 18 at 0:50
2
2
Seems good to me. Maybe you were calculating the wrong thing or the answer you were provided is wrong.
– Stan Tendijck
Nov 18 at 0:50
Seems good to me. Maybe you were calculating the wrong thing or the answer you were provided is wrong.
– Stan Tendijck
Nov 18 at 0:50
add a comment |
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Seems good to me. Maybe you were calculating the wrong thing or the answer you were provided is wrong.
– Stan Tendijck
Nov 18 at 0:50