What is the correct probability for P(3<X<4|X>1)?

[Yankel]

Hello, I have this question, which I think I solve correctly, but I am getting the wrong answer.

X represent the point that the computer chooses on a scale of 2 to 5 (continuous scale) in a non-uniform way using the density:

f(x)=C*(1+x)

what is the probability P(3<X<4|X>1) ?

I solved the integral from 2 to 5 to find that C=2/27

Then using this value I did conditional probability P(3<X<4)/P(X>1). The nominator was 1/3 and the denominator was 32/27, my final result is then 9/32. The answer I have with the question is 9/27. Which one is wrong then ? Can you assist me please? Thank you in advance !

 

[Bushy]

I have

\(\displaystyle \displaystyle c\times \int_2^5 x+1 ~dx = 1\)

\(\displaystyle \displaystyle c\times \left[ \frac{x^2}{2}+x\right]_2^5 = 1\)

\(\displaystyle \displaystyle c\times \left(\left[ \frac{5^2}{2}+5\right]-\left[ \frac{2^2}{2}+2\right]\right) = 1\)

\(\displaystyle \displaystyle c\times \left(\left[ 12.5+5\right]-\left[ 2+2\right]\right) = 1\)

\(\displaystyle \displaystyle c\times \left(\left[ 17.5\right]-\left[ 4\right]\right) = 1\)

\(\displaystyle \displaystyle c = \frac{1}{13.5} = \frac{2}{27}\)

 

[Jameson]

Hello, I have this question, which I think I solve correctly, but I am getting the wrong answer.

X represent the point that the computer chooses on a scale of 2 to 5 (continuous scale) in a non-uniform way using the density:

f(x)=C*(1+x)

what is the probability P(3<X<4|X>1) ?

I solved the integral from 2 to 5 to find that C=2/27

Then using this value I did conditional probability P(3<X<4)/P(X>1). The nominator was 1/3 and the denominator was 32/27, my final result is then 9/32. The answer I have with the question is 9/27. Which one is wrong then ? Can you assist me please? Thank you in advance !

Hi Yankel,

I got the same thing for $C$ and the same numerator since $P([3<X<4] \cap [X>1])=P(3<X<4)$. For the denominator though it seems like you used this integral:
$$ \frac{2}{27} \int_{1}^{5} (1+x) \, dx = \frac{32}{27}$$
Remember though that any probability must be between 0 and 1, so this isn't a valid answer. The small mistake comes from the fact that $P(X<2)=0$ and $P(X>5)=0$, so $P(X>1)=P(2<X<5)=1$. For some reason the book chose to write the answer as 9/27 instead of 1/3, but those are equivalent answers. :)

 

[Yankel]

Oh...how did I miss this...typing numbers without thinking (Doh)

thanks !