GO Classes DPP | GATE DA | Probability | Probability Density Function

Answer 1

To find the probability, we first determine the value of $c$ using the fact that the total probability must be $1$.

Since $f(x)=cx$ for $x\in[0,10]$, we have

$\int_0^{10} cx\,dx = 1$.

So

$c\int_0^{10} x\,dx = 1$,

$c\left[\frac{x^2}{2}\right]_0^{10} = 1$,

$c\left(\frac{100}{2}\right)=1$,

$50c=1$,

$c=\frac{1}{50}$.

Now find $P(1\le x\le 2)$:

$P(1\le x\le 2)=\int_1^2 f(x)\,dx=\int_1^2 cx\,dx$.

Substitute $c=\frac{1}{50}$:

$P(1\le x\le 2)=\int_1^2 \frac{x}{50}\,dx=\frac{1}{50}\int_1^2 x\,dx$.

So

$P(1\le x\le 2)=\frac{1}{50}\left[\frac{x^2}{2}\right]_1^2$,

$P(1\le x\le 2)=\frac{1}{50}\left(\frac{4}{2}-\frac{1}{2}\right)$,

$P(1\le x\le 2)=\frac{1}{50}\cdot\frac{3}{2}=\frac{3}{100}$.

Therefore, the probability that $x\in[1,2]$ is $\boxed{\frac{3}{100}}$.

So the correct answer is B.

Answer 2

Answer: B

Answer 3

Approach: Using Properties of Continuous Probability Density Functions (PDF)

A valid continuous probability density function $f(x)$ must satisfy the total probability axiom:

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

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Step 1: Find the value of the constant $c$

Given the piecewise definition of $f(x)$:

$$f(x) = \begin{cases} cx, & x \in [0, 10] \\ 0, & \text{otherwise} \end{cases}$$

We substitute this into our total probability integral over the non-zero domain $[0, 10]$:

$$\int_{0}^{10} cx \, dx = 1$$

Pulling out the constant $c$ and integrating $x$:

$$c \left[ \frac{x^2}{2} \right]_{0}^{10} = 1$$

$$c \left( \frac{10^2}{2} - \frac{0^2}{2} \right) = 1$$

$$c \left( \frac{100}{2} \right) = 1$$

$$50c = 1 \implies c = \frac{1}{50}$$

Thus, the fully defined PDF is:

$$f(x) = \begin{cases} \frac{1}{50}x, & x \in [0, 10] \\ 0, & \text{otherwise} \end{cases}$$

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Step 2: Calculate the Probability $P(1 \le x \le 2)$

The probability that $x$ lies in the interval $[1, 2]$ is computed by integrating the PDF over those specific bounds:

$$P(1 \le x \le 2) = \int_{1}^{2} f(x) \, dx$$

$$P(1 \le x \le 2) = \int_{1}^{2} \frac{1}{50}x \, dx$$

$$\quad = \frac{1}{50} \left[ \frac{x^2}{2} \right]_{1}^{2}$$

$$\quad = \frac{1}{100} \left[ x^2 \right]_{1}^{2}$$

$$\quad = \frac{1}{100} \left( 2^2 - 1^2 \right)$$

$$\quad = \frac{1}{100} (4 - 1)$$

$$\quad = \frac{3}{100}$$

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Correct Answer:

The probability that $x \in [1,2]$ is $\frac{3}{100}$.