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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 (Answer up to two decimal places)
Correct answer is '0.25'. Can you explain this answer?
Most Upvoted Answer
The variable x takes a value between 0 and 10 with uniform probability...
Solution:
Given that x takes a value between 0 and 10 with uniform probability distribution and y takes a value between 0 and 20 with uniform probability distribution.
We need to find the probability of the sum of variables (x + y) being greater than 20.
To solve this problem, we can use the concept of joint probability distribution.
Joint Probability Distribution:
The joint probability distribution of two random variables x and y is a function that maps each pair of outcomes (x, y) to the probability of that pair of outcomes occurring.
We can represent the joint probability distribution of x and y using a probability density function f(x, y).
For uniform probability distribution, the probability density function is given by:
f(x, y) = 1 / (10*20) = 0.005
where 10 and 20 are the ranges of x and y respectively.
Probability of Sum of Variables:
To find the probability of the sum of variables (x + y) being greater than 20, we need to integrate the joint probability density function over the region where (x + y) > 20.
P(x + y > 20) = ∫∫(x+y>20) f(x,y) dxdy
= ∫10 0 ∫20-x 0 f(x,y) dy dx
= ∫10 0 [(20-x)/20] f(x,y) dx
= ∫10 0 (20-x)/200 dx
= [x/200 - (x^2)/40]10 0
= [10/200 - (100)/40] - [0/200 - (0)/40]
= 0.05 - 2.5
= -2.45
Since probability cannot be negative, the probability of the sum of variables (x + y) being greater than 20 is 0.
Therefore, the given answer of 0.25 is incorrect.
Given that x takes a value between 0 and 10 with uniform probability distribution and y takes a value between 0 and 20 with uniform probability distribution.
We need to find the probability of the sum of variables (x + y) being greater than 20.
To solve this problem, we can use the concept of joint probability distribution.
Joint Probability Distribution:
The joint probability distribution of two random variables x and y is a function that maps each pair of outcomes (x, y) to the probability of that pair of outcomes occurring.
We can represent the joint probability distribution of x and y using a probability density function f(x, y).
For uniform probability distribution, the probability density function is given by:
f(x, y) = 1 / (10*20) = 0.005
where 10 and 20 are the ranges of x and y respectively.
Probability of Sum of Variables:
To find the probability of the sum of variables (x + y) being greater than 20, we need to integrate the joint probability density function over the region where (x + y) > 20.
P(x + y > 20) = ∫∫(x+y>20) f(x,y) dxdy
= ∫10 0 ∫20-x 0 f(x,y) dy dx
= ∫10 0 [(20-x)/20] f(x,y) dx
= ∫10 0 (20-x)/200 dx
= [x/200 - (x^2)/40]10 0
= [10/200 - (100)/40] - [0/200 - (0)/40]
= 0.05 - 2.5
= -2.45
Since probability cannot be negative, the probability of the sum of variables (x + y) being greater than 20 is 0.
Therefore, the given answer of 0.25 is incorrect.
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Community Answer
The variable x takes a value between 0 and 10 with uniform probability...
Consider x values on x-axis and y values on y-axis
0≤ x ≤ 10 and 0 ≤ y ≤ 20
Here x and y together constitute a rectangle given by OBPR.
Where, AB is the line x + y = 20
The probability of the sum of variables (x + y) being greater than 20 is given by
P[(x + y) > 20] = 
= 0.25
= 0.25
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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 (Answer up to two decimal places)Correct answer is '0.25'. Can you explain this answer?
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