Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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