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# Probability

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Hi  Max: Please take a look at this question: https://web2.0calc.com/questions/pls-halp_27

Apr 16, 2022

#1
+9369
+2

This problem is equivalent to the probability of choosing two real numbers x, y with 0 < x < y < 5 such that x < 3, y - x < 3, and 5 - y < 3.

We can view this geometrically.

Let $$D$$ be the square with vertices (0, 0), (5, 0), (0, 5), (5, 5).

Let $$\Omega$$ be the region bounded by 0 < x < 3, y - x < 3, 5 - y < 3, x < y and $$D$$.

Basically, the probability is equal to $$\dfrac{\text{area}(\Omega)}{\text{area}(D)} = \dfrac{9 - 2 - \dfrac12}{25} = \dfrac{13}{50}$$.

I believe that is the answer.

Edit: Edited. This should be correct now.

Apr 16, 2022
edited by MaxWong  Apr 16, 2022
edited by MaxWong  Apr 16, 2022

#1
+9369
+2

This problem is equivalent to the probability of choosing two real numbers x, y with 0 < x < y < 5 such that x < 3, y - x < 3, and 5 - y < 3.

We can view this geometrically.

Let $$D$$ be the square with vertices (0, 0), (5, 0), (0, 5), (5, 5).

Let $$\Omega$$ be the region bounded by 0 < x < 3, y - x < 3, 5 - y < 3, x < y and $$D$$.

Basically, the probability is equal to $$\dfrac{\text{area}(\Omega)}{\text{area}(D)} = \dfrac{9 - 2 - \dfrac12}{25} = \dfrac{13}{50}$$.

I believe that is the answer.

Edit: Edited. This should be correct now.

MaxWong Apr 16, 2022
edited by MaxWong  Apr 16, 2022
edited by MaxWong  Apr 16, 2022
#2
+9369
+3

Graph: https://www.desmos.com/calculator/t08bdptj6k

I forgot to attach the graph when I sent the response.

Edit: Updated the graph to the new one.

MaxWong  Apr 16, 2022
edited by MaxWong  Apr 16, 2022
#3
+117236
+1
May 14, 2022