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Two real numbers are chosen at random between $0$ and $2$. What is the probability that the sum of their squares is no more than $4$? Express your answer as a common fraction in terms of $\pi$.

 Apr 3, 2021
 #1
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The probability is 2*pi/3.

 Apr 3, 2021
 #2
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Sorry, it says that was incorrect, but I have one more try. Can you please show your work so I can look for errors?

RiemannIntegralzzz  Apr 3, 2021
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You should not be plugging in answers without knowing how they were come by !!

Melody  Apr 3, 2021
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Sorry @Melody and @Guest, they where both wrong. The solution it gave was:

 Apr 3, 2021
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Sorry, my image isn't displaying.. I'll try to type it out by hand

RiemannIntegralzzz  Apr 3, 2021
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We let the two numbers be x and y. The set of all possible pairs (x,y) satisfies the inequalities 0

 

We want to find the area of the set of points which also satisfies x^2+y^2<=4. The set of ordered pairs that satisfy this inequality is a circle with radius 2 centered at the origin. The overlap between the circle and the square is the quarter-circle in the first quadrant with radius 2. This region has area 1/4(2)^2$\pi$=$\pi$

 

Thus the desired probability is $\boxed{\frac{\pi}{4}}$

RiemannIntegralzzz  Apr 3, 2021
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I had thought of that but I had a brain freeze.  Thanks.

 Apr 3, 2021

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