Q.Three coins are tossed simultaneously. Let $X$ denote the number of heads. Construct the probability distribution of $X$ and verify that it is a valid distribution.

$P(X<1.5)=\int_0^{1.5}\frac{x}{8}\,dx = \left[\frac{x^2}{16}\right]_0^{1.5} = \frac{2.25}{16} = \frac{9}{64} \approx 0.1406$.

The probability density function is

$f(x) = \begin{cases} \dfrac{x}{8}, & 0 < x < 4 \\[4pt] 0, & \text{otherwise.} \end{cases}$

For a continuous random variable, the probability over an interval is the integral of the p.d.f.:

$P(X < 1.5) = \int_{0}^{1.5} f(x)\,dx = \int_{0}^{1.5} \frac{x}{8}\,dx.$

Integrating:

$\int_{0}^{1.5} \frac{x}{8}\,dx = \frac{1}{8}\left[\frac{x^2}{2}\right]_{0}^{1.5} = \left[\frac{x^2}{16}\right]_{0}^{1.5}.$

Substituting the limits:

$= \frac{(1.5)^2}{16} - 0 = \frac{2.25}{16} = \frac{9}{64} = 0.140625.$

For a continuous r.v., $P(a<x<b)=\int_a^b f(x)\,dx$. Here $f(x)=\frac{x}{8}$ on $(0,4)$, so integrating from $1$ to $2$ gives $\frac{3}{16}$.

Since $X$ is a continuous random variable with probability density function $f(x)=\dfrac{x}{8}$ for $0<x<4$, the required probability is the area under the curve between $x=1$ and $x=2$:

$P(1<x<2)=\displaystyle\int_1^2 f(x)\,dx=\int_1^2 \frac{x}{8}\,dx$

Integrating term by term,

$=\dfrac{1}{8}\left[\dfrac{x^2}{2}\right]_1^2=\dfrac{1}{16}\left[x^2\right]_1^2$

$=\dfrac{1}{16}\left(2^2-1^2\right)=\dfrac{1}{16}(4-1)=\dfrac{3}{16}$

Because $f(x)=0$ for $x\ge 4$, $P(x>2)=\int_2^4 \frac{x}{8}\,dx=\frac{3}{4}$.

The density is $f(x)=\dfrac{x}{8}$ only on $0<x<4$ and is $0$ elsewhere, so the event $x>2$ contributes probability only over $2<x<4$:

$P(x>2)=\displaystyle\int_2^{4} f(x)\,dx=\int_2^{4}\frac{x}{8}\,dx$

Integrating,

$=\dfrac{1}{8}\left[\dfrac{x^2}{2}\right]_2^{4}=\dfrac{1}{16}\left[x^2\right]_2^{4}$

$=\dfrac{1}{16}\left(4^2-2^2\right)=\dfrac{1}{16}(16-4)=\dfrac{12}{16}=\dfrac{3}{4}$

Apply $\int_{-\infty}^{\infty} f(x)\,dx = 1$: $\int_0^2 kx\,dx = 2k = 1$, so $k = \dfrac{1}{2}$.

For $f(x)$ to be a probability density function, it must be non-negative and satisfy

$\displaystyle \int_{-\infty}^{\infty} f(x)\,dx = 1$.

Since $f(x) = 0$ outside $(0,2)$,

$\displaystyle \int_0^2 kx\,dx = 1$.

Evaluating the integral,

$k\left[\dfrac{x^2}{2}\right]_0^2 = k\cdot\dfrac{4}{2} = 2k$.

So $2k = 1 \Rightarrow k = \dfrac{1}{2}$ (positive, keeping $f(x)\ge 0$).