You're saying that the custom object has lookup to Lead but then you say Subjects are generic and somehow should be displayed on every Lead page? I don't think it'll work.

Stuff appears on related list only when field Subject_c.Lead_c will be populated with "this" Lead's Id. (please note I've made best guess at the field name). So you'd need to insert separate data for each Lead which can quickly blow your storage usage and will be a pain in the a$$ to maintain later. Is it only for displaying? Or do you plan to later capture some kind of survey results for each Lead?

If it's just for display I think you'll need to embed a Visualforce page in the Lead page layout to achieve that in a saner way. The subjects are specific to current viewing user? Or it's more like a general list, just 3 subjects for whole organisation?

P.S. "object" is like a table in normal database. I think you mixed a bit the difference between table and records / rows of data stored in it.

Created Custom objects on salesforce not showing up under Leads
I am unsure with question c/d/e/f but I will give my answers for all questions I have attempted.

The random variables $X$ and $Y$ have joint probability density function

$f_{X,Y}(x,y) = ke^{-(x+y)},\ x&gt;0,\ y&gt;0,\ 0&lt;x&lt;y.$

(a) Sketch the region over which the joint probability density function is non-zero.

(b) Show that $k = 2.$

(c) Find the marginal Distribution of $X$; name the distribution and state its mean and variance.

(d) Find the conditional distribution of $Y|X = x$ for $x&gt;0$. Pay particular attention to the range of $Y$ conditional on $x$.

(e) Calculate $Pr(X + Y&lt;1)$.

(f) Given that $E(Y) = 1.5$ and $Var(Y) = 1.25$, calculate the correlation between $X$ and $Y$.

ANSWERS:

(a) Don't know how to draw on here but it's the area above $y=x$ and $y&gt;0$.

(b) Here I used double integration

$\int_0^\infty \int_0^y ke^{-(x+y)}dxdy = 1$

$\int_0^\infty [-ke^{-(x+y)}]_0^y dy = \int_0^\infty[-ke^{-y}+ke^{-2y}]dy$

$[ke^{-y}-\frac{k}{2}e^{-2y}]_0^\infty = [ke^{-0} - \frac{k}{2}e^{-2\cdot0}]-[ke^{-\infty}-\frac{k}{2}e^{-2\cdot \infty}] = [k - \frac{k}{2}]-[0] = \frac{k}2$

Hence $\frac{k}{2}=1$ and $k=2$

(c) $P_x(x)=\int_yP_{X,Y}(x,y)dy=\int_0^\infty2e^{-(x+y)}dy$

$[-2e^{-(x+y)}]_0^\infty= -2e^{-x}$

Presuming that the integration I have done is correct the marginal distribution of $X$ is $-2e^{-x}$ and hence this is a gamma distribution as the limits are $(0,\infty)$ and it just doesn't match the binomial distribution either.

Gamma distribution: $\frac{1}{\Gamma(\alpha)\cdot \beta^\alpha}\cdot x^{\alpha-1}\cdot e^{-\frac{x}{\beta}}$

Hence $\frac{1}{\Gamma(\alpha)\cdot \beta^\alpha}\cdot x^{\alpha-1}\cdot e^{-\frac{x}{\beta}} = -2e^{-x}$

This is where I have gone wrong as using $\beta = 1$ to leave $e^{-x}$ as that i then need to find:

$\frac{1}{\Gamma(\alpha)\cdot 1^\alpha}\cdot x^{\alpha-1} = -2$ and using $\alpha =1$ to make $x^0$ and getting rid of $x$ but this makes $1=-2$.

(d) unsure how to do this question also.

(e) Adding the line $y=1-x$ to the graph from *(a)*we get a new region to find

$\int_0^1 \int_x^{1-x}2e^{-(x+y)}dydx$

$\int_0^1[-2e^{-(x+y)}]_x^{1-x}dx$

$\int_0^1[(-2e^{-2x})-(-2e^{-1})]dx$

$[e^{-2x}+2xe^{-1}]_0^1$

$e^0 + 0 - e^{-2} + 2e^{-1} = 1-\frac{1}{e^2}+\frac{2}{e}$

Hence $Pr(x+y&lt;1) = 1-\frac{1}{e^2}+\frac{2}{e}$

(f) Unsure how to do.
