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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 \(b \gt a + 1\) ?

b > a + 1 is represented by the part of circle that is in second quadrant.

Hey, the OA is A 1/4. I got that but just by pure guess work. I could not understand how the statement b>a+1 covers a quarter of the entire circle. Could you please explan how you figured that out? Thankyou

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 \(b \gt a + 1\) ?

I think it cannot be 1/4 cuz the full Qd-II doesnot fall under the area b > (a+1); it would only if b>a. For example if (a,b) is (-0.1, 0.1) doesnot fall under the area given by the above constraiant.

so the probability should be much lesser than 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 \(b \gt a + 1\) ?

I think it cannot be 1/4 cuz the full Qd-II doesnot fall under the area b > (a+1); it would only if b>a. For example if (a,b) is (-0.1, 0.1) doesnot fall under the area given by the above constraiant.

so the probability should be much lesser than 1/4.

GMATTIGER, the point from your example doesn't belong to Set T. \(a^2+b^2 \ne 1\) in your example.

I'm attaching an image to make it clearer. Line AB has equation \(y=x+1\), just like \(b = a +1\). Any points above the line AB satisfy the inequality \(b \gt a +1\). Thus we have \(\frac{1}{4}\) of all the points from Set T that satisfy the inequality.

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 \(b \gt a + 1\) ?

I think it cannot be 1/4 cuz the full Qd-II doesnot fall under the area b > (a+1); it would only if b>a. For example if (a,b) is (-0.1, 0.1) doesnot fall under the area given by the above constraiant.

so the probability should be much lesser than 1/4.

GMATTIGER, the point from your example doesn't belong to Set T. \(a^2+b^2 \ne 1\) in your example.

I'm attaching an image to make it clearer. Line AB has equation \(y=x+1\), just like \(b = a +1\). Any points above the line AB satisfy the inequality \(b \gt a +1\). Thus we have \(\frac{1}{4}\) of all the points from Set T that satisfy the inequality.

This a really good explaination...... _________________

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 \(b \gt a + 1\) ?

The circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r=\sqrt{1}=1\) (for more on this check Coordinate Geometry chapter of math book: math-coordinate-geometry-87652.html ).

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>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 1/4 of the whole circumference, hence P=1/4.

Answer: A.

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 \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}\).