Q.Two balls are drawn in succession without replacement from an urn containing four red balls and three black balls. Let $X$ be the number of red balls drawn. Find the probability mass function and the mean for $X$.

Computes mean and variance of the given uniform-type density on $(0,l)$ using $E[X]$ and $E[X^2]-\left(E[X]\right)^2$.

1. $\text{Mean}=E[X]=\displaystyle\int_0^lx\,f(x)\,dx=\int_0^lx\cdot\dfrac1l\,dx=\dfrac1l\left[\dfrac{x^2}2\right]_0^l=\dfrac1l\cdot\dfrac{l^2}2=\dfrac l2$.
2. $E[X^2]=\displaystyle\int_0^lx^2\,f(x)\,dx=\int_0^lx^2\cdot\dfrac1l\,dx=\dfrac1l\left[\dfrac{x^3}3\right]_0^l=\dfrac1l\cdot\dfrac{l^3}3=\dfrac{l^2}3$.
3. $\text{Variance}=E[X^2]-\left(E[X]\right)^2=\dfrac{l^2}3-\left(\dfrac l2\right)^2=\dfrac{l^2}3-\dfrac{l^2}4$.
4. Common denominator $12$: $\dfrac{4l^2}{12}-\dfrac{3l^2}{12}=\dfrac{l^2}{12}$.
5. So Mean $=\dfrac l2$, Variance $=\dfrac{l^2}{12}$.

A constant has no spread around its mean, so its variance is identically zero, regardless of the constant's value.

1. Variance measures the expected squared deviation from the mean: $\text{Var}(X)=E[(X-E[X])^2]$.
2. For a constant $c$ (viewed as a "random variable" that always takes the value $3$), $E[c]=c=3$, and $(c-E[c])^2=0$ always.
3. So $\text{Var}(3)=E[0]=0$.

Since $\text{Var}(X)=E(X^2)-[E(X)]^2$, with $\text{Var}(X)=4$ and $E(X)=2$, we get $E(X^2)=8$.

1. The variance formula is $\text{Var}(X)=E(X^2)-[E(X)]^2$.
2. Given $\text{Var}(X)=4$ and $E(X)=2$.
3. Substitute: $4=E(X^2)-(2)^2=E(X^2)-4$.
4. Solve for $E(X^2)$: $E(X^2)=4+4=8$.

Solving the two expectation equations gives $C=-2$.

1. By linearity of expectation, $E(X+C)=E(X)+C=8$ and $E(X-C)=E(X)-C=12$.
2. Subtracting the second equation from the first: $(E(X)+C)-(E(X)-C)=8-12\Rightarrow 2C=-4$.
3. So $C=-2$.
4. Check: adding the two equations gives $2E(X)=20\Rightarrow E(X)=10$; then $E(X)+C=10-2=8$ (matches) and $E(X)-C=10+2=12$ (matches).
5. Options (b) $4$, (c) $-4$, (d) $2$ fail this check.