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Set \(T\) consists of all points \((x, y)\) such that \(x^2 + y^2 = 1\). If point \((a, b)\) is selected from set \(T\) at random, what is the probability that \(b \gt a + 1\)?

A. \(\frac{1}{4}\) B. \(\frac{1}{3}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)

Set \(T\) consists of all points \((x, y)\) such that \(x^2 + y^2 = 1\). If point \((a, b)\) is selected from set \(T\) at random, what is the probability that \(b \gt a + 1\)?

A. \(\frac{1}{4}\) B. \(\frac{1}{3}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)

Look at the diagram below:

The circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r=\sqrt{1}=1\).

So, set \(T\) is the circle itself (red curve).

Question is: if point \((a,b)\) is selected from set \(T\) at random, what is the probability that \(b \gt a+1\)? All points \((a,b)\) which satisfy this condition (belong to \(T\) and have y-coordinate > x-coordinate + 1) lie above the line \(y=x+1\) (blue line). You can see that portion of the circle which is above the line is \(\frac{1}{4}\) of the whole circumference, hence \(P=\frac{1}{4}\).

Set \(T\) consists of all points \((x, y)\) such that \(x^2 + y^2 = 1\). If point \((a, b)\) is selected from set \(T\) at random, what is the probability that \(b \gt a + 1\)?

A. \(\frac{1}{4}\) B. \(\frac{1}{3}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)

Look at the diagram below:

The circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r=\sqrt{1}=1\).

So, set \(T\) is the circle itself (red curve).

Question is: if point \((a,b)\) is selected from set \(T\) at random, what is the probability that \(b \gt a+1\)? All points \((a,b)\) which satisfy this condition (belong to \(T\) and have y-coordinate > x-coordinate + 1) lie above the line \(y=x+1\) (blue line). You can see that portion of the circle which is above the line is \(\frac{1}{4}\) of the whole circumference, hence \(P=\frac{1}{4}\).

Set \(T\) consists of all points \((x, y)\) such that \(x^2 + y^2 = 1\). If point \((a, b)\) is selected from set \(T\) at random, what is the probability that \(b \gt a + 1\)?

A. \(\frac{1}{4}\) B. \(\frac{1}{3}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)

Look at the diagram below:

The circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r=\sqrt{1}=1\).

So, set \(T\) is the circle itself (red curve).

Question is: if point \((a,b)\) is selected from set \(T\) at random, what is the probability that \(b \gt a+1\)? All points \((a,b)\) which satisfy this condition (belong to \(T\) and have y-coordinate > x-coordinate + 1) lie above the line \(y=x+1\) (blue line). You can see that portion of the circle which is above the line is \(\frac{1}{4}\) of the whole circumference, hence \(P=\frac{1}{4}\).

Answer: A

Is there any other simple solution??

I, personally, don't know any simpler one. The one presented should be simple enough after some practice.

I understood how you made this circle, but I can't understand how you made this blue line and computed probability from this

The blue line is y = x + 1. The area above it is y > x + 1. So, all the points for which y-coordinate is greater than x-coordinate + 1.
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Shouldn't it be 1/4 of the area rather than circumference ?

Please confirm.

Regards

Original question does not ask about the area, it asks about the portion of the circumference

If it were: set T consists of all points (x,y) such that \(x^2+y^2<1\) (so set T consists of all points inside the circle). If point (a,b) is selected from set T at random, what is the probability that b>a+1?

Then as the area of the segment of the circle which is above the line is \(\frac{\pi{r^2}}{4}-\frac{r^2}{2}=\frac{\pi-2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}\).

Bunuel Hi, I have a question. The line b>a+1 in the graph, contains all the values on which side of the circumference of the circle, right(Quadrant 1,3,4) or left(Quadrant 2)?

Bunuel Hi, I have a question. The line b>a+1 in the graph, contains all the values on which side of the circumference of the circle, right(Quadrant 1,3,4) or left(Quadrant 2)?

Thanks

The blue line is y = x + 1. The area above it is y > x + 1. So, all the points for which y-coordinate is greater than x-coordinate + 1.
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for choosing left and right you can substitute origin(0,0) in the equation y>x+1 => 0>1 which is not correct . Hence origin doesn't satisfy the equation thus it is on the side not containing the origin.

Last edited by ShravyaAlladi on 19 Jul 2016, 00:57, edited 1 time in total.

I have a question. The line b>a+1 in the graph, contains all the values on which side of the circumference of the circle, right(Quadrant 1,3,4) or left(Quadrant 2)?

Thanks

for choosing left and right you can substitute origin(0,0) in the equation y>x+1 => 0>1 which is not correct . Hence origin doesn't satisfy the equation thus it is on the side not containing the origin

Set \(T\) consists of all points \((x, y)\) such that \(x^2 + y^2 = 1\). If point \((a, b)\) is selected from set \(T\) at random, what is the probability that \(b \gt a + 1\)?

A. \(\frac{1}{4}\) B. \(\frac{1}{3}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)

Sir,

Do you have few more quetsions list where Geomtry is Clubbed with Probability.
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Set \(T\) consists of all points \((x, y)\) such that \(x^2 + y^2 = 1\). If point \((a, b)\) is selected from set \(T\) at random, what is the probability that \(b \gt a + 1\)?

A. \(\frac{1}{4}\) B. \(\frac{1}{3}\) C. \(\frac{1}{2}\) D. \(\frac{3}{5}\) E. \(\frac{2}{3}\)

Sir,

Do you have few more quetsions list where Geomtry is Clubbed with Probability.

Thank you for the graph and explanation. However, could u explain why this approach is incorrect?

r = 1 Circum = 2pi.

Any point that satisfies b> a+1 can be translated into if y=x+1 and we see this covers 1/4 of whole Circum.

However, can I use Prob=n/N approach where N is whole circumference. n - is represented by the arc-length - L

Inscribed angle = (90*L)/pi*radius ------> looking at the graph, ins.angle =90 & r = 1,. Then L -arc-length is pi. Thus I have Prob = n/N = pi/2pi = 1/2

x^2+y^2=1. therefore it is a circle with radius 1. (put x=0 , y=+1/-1 similarly, y=0, x=+1/-1. gives the points where the line will cross the axes). Draw the circle.

since we are concerned about points inside this circle so (x,y) should not exceed +1/-1. b>a+1 is only possible when a is negative greater than -1. this gives positive b; for eg. if a=-0.5 then b>0.5. so all points (a,b) will be in IInd quadrant. X & Y axis divides the circle in 4 parts. out of the 4 (a,b) can lie on only 1 quadrant. therefor probability will be 1/4

I think this is a high-quality question and the explanation isn't clear enough, please elaborate. "You can see that portion of the circle which is above the line is 1/4 of the circumference of the circle"--

I think this is a high-quality question and the explanation isn't clear enough, please elaborate. "You can see that portion of the circle which is above the line is 1/4 of the circumference of the circle"--

How did u conclude this? Please show calculation

This should be easy. I'll give you the hints and let you figure out the rest.

1. The circle is centred at the origin, so each quadrant has 1/4 of the circumference. 2. y = x + 1 cuts the x and y-axis at (-1, 0) and (0, 1) respectively at the same exact points the circle cuts the axis.
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