The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________
0.25
Probability of Sum of Uniform Variables
This problem requires us to calculate the probability of the sum of two independent random variables, x and y, being greater than 20. Both variables follow a uniform probability distribution. We will use a geometric approach to solve this, which involves finding the area of the sample space and the area of the favorable region.
Understanding Uniform Distributions
First, let's understand the properties of the given uniform probability distributions for x and y:
- The variable x is uniformly distributed between 0 and 10. This means its probability density function (PDF) is constant within this range: $$ \text{f}_{x}(\text{x}) = \begin{cases} \frac{1}{10} & \text{for } 0 \le \text{x} \le 10 \\ 0 & \text{otherwise} \end{cases} $$
- The variable y is uniformly distributed between 0 and 20. Its probability density function (PDF) is also constant within its range: $$ \text{f}_{y}(\text{y}) = \begin{cases} \frac{1}{20} & \text{for } 0 \le \text{y} \le 20 \\ 0 & \text{otherwise} \end{cases} $$
Defining the Sample Space
Since x and y are independent random variables, their joint probability density function is the product of their individual PDFs, i.e., $\text{f}(\text{x}, \text{y}) = \text{f}_{x}(\text{x}) \times \text{f}_{y}(\text{y})$.
The sample space for the pair (x, y) is a rectangular region defined by the ranges of x and y:
- $0 \le \text{x} \le 10$
- $0 \le \text{y} \le 20$
The total area of this sample space represents the total possible outcomes. We calculate it as:
$$ \text{Area}_{\text{total}} = (\text{Maximum x} - \text{Minimum x}) \times (\text{Maximum y} - \text{Minimum y}) $$ $$ \text{Area}_{\text{total}} = (10 - 0) \times (20 - 0) = 10 \times 20 = 200 $$The joint PDF within this rectangular region is $\text{f}(\text{x}, \text{y}) = \frac{1}{10} \times \frac{1}{20} = \frac{1}{200}$.
Identifying the Favorable Region (x + y > 20)
We are interested in the probability that the sum of the variables (x + y) is greater than 20. This condition, $\text{x} + \text{y} > 20$, defines the favorable region within our sample space.
To visualize this region, consider the line $\text{x} + \text{y} = 20$. We need to find the area within the sample space where points satisfy $\text{x} + \text{y} > 20$.
- Let's find where the line $\text{x} + \text{y} = 20$ intersects the boundaries of our sample space:
- If $\text{x} = 0$ (left boundary), then $0 + \text{y} = 20 \Rightarrow \text{y} = 20$. So, the point is $(0, 20)$.
- If $\text{x} = 10$ (right boundary), then $10 + \text{y} = 20 \Rightarrow \text{y} = 10$. So, the point is $(10, 10)$.
The region $\text{x} + \text{y} > 20$ consists of all points above the line segment connecting $(0, 20)$ and $(10, 10)$, and within the bounds of the sample space. The vertices of this favorable region are:
- $(0, 20)$
- $(10, 20)$ (the top-right corner of the sample space rectangle)
- $(10, 10)$
This forms a right-angled triangle. Let's calculate its area:
- The base of this triangle can be considered along the line $\text{y} = 20$, spanning from $\text{x} = 0$ to $\text{x} = 10$. $$ \text{Base} = 10 - 0 = 10 $$
- The height of the triangle is the perpendicular distance from the point $(10, 10)$ to the line $\text{y} = 20$. $$ \text{Height} = 20 - 10 = 10 $$
The area of this favorable region is calculated using the formula for the area of a triangle:
$$ \text{Area}_{\text{favorable}} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 10 \times 10 = \frac{1}{2} \times 100 = 50 $$Calculating the Probability
For a uniform probability distribution over a continuous region, the probability of an event is the ratio of the area of the favorable region to the total area of the sample space:
$$ \text{P}(\text{x} + \text{y} > 20) = \frac{\text{Area}_{\text{favorable}}}{\text{Area}_{\text{total}}} $$Substituting the calculated areas:
$$ \text{P}(\text{x} + \text{y} > 20) = \frac{50}{200} $$ $$ \text{P}(\text{x} + \text{y} > 20) = \frac{1}{4} $$ $$ \text{P}(\text{x} + \text{y} > 20) = 0.25 $$Thus, the probability of the sum of variables (x + y) being greater than 20 is 0.25.
Important Questions from Continuous Distributions
Suppose X is a continuous random variable with probability density function
\(f(x)=\frac{1}{\pi} \frac{1}{1+(x+1)^2}\), -∞ < x < ∞.
Define
\(Y=\left\{\begin{array}{cc} \frac{X}{|X|}, & \text { if } X \neq 0 \\ 0, & \text { if } X=0 \end{array}\right.\)
Then which of the following statements are true?
Let X1, X2, ..., Xn be a random sample from an absolutely continuous distribution with the probability density function
\(f(x \mid \theta)=\left\{\begin{array}{cl} e^{\theta-x}, & \text { if } x \geq \theta \\ 0, & \text { if } x<\theta \end{array},\right.\)
where θ ∈ ℝ is unknown. Define \(\bar{X}=\frac{1}{n} \sum_{i=1}^n X_i\) and X(1) = min{X1, ..., Xn}. Then
which of the following statements are true?
Suppose that X is a continuous random variable with probability density function given by:
f(x) = \(\left\{ {\begin{array}{c} {\frac{x}{8},}&{x \in \left[ {0,2} \right)}\\ {\frac{1}{4},}&{x \in \left[ {2,4} \right)}\\ { - \frac{x}{8} + \frac{3}{4},}&{x \in \left[ {4,6} \right)} \end{array}}\right.\)
Find the mean of X.
Probability density function of a random variable X is given below
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{if\;1 \le x \le 5}\\ 0&{otherwise} \end{array}} \right.\)
P (X ≤ 4) is
- The annual precipitation data of a city is normally distributed with mean and standard deviation as 1000 mm and 200 mm, respectively. The probability that the annual precipitation will be more than 1200 mm is
