M20-12

Question

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

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

Bunuel · Accepted Answer

Official Solution:

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

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

Consider the diagram below:
 
 https://gmatclub.com/gmat-practice-tests/resources/import/img/m20-12.png 
 
The circle represented by the equation  x^2+y^2 = 1  is centered at the origin and has a radius of  r=\sqrt{1}=1 .
 
So, set  T  is the circle itself (red curve).
 
The 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)  that satisfy this condition (belong to  T  and have a y-coordinate greater than x-coordinate + 1) lie above the line  y = x + 1  (blue line). You can see that the portion of the circle which is above the line is  \frac{1}{4}  of the whole circumference, hence the probability  P = \frac{1}{4} .
 
Note that while Geometry is not tested on GMAT Focus, Coordinate Geometry is tested under the Functions and Graphing sections found in the Official Guide for GMAT Focus Edition.

Answer: A

awal_786 · Answer

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

Bunuel · Answer

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.

vinnik · Answer

Hi Bunuel,

Shouldn't it be 1/4 of the area rather than circumference ?

Please confirm.

Regards

Bunuel · Answer

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} .

Hope it's clear.

sathishm07 · Answer

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

Bunuel · Answer

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.

ShravyaAlladi · Answer

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.

Senthil7 · Answer

I think this is a  high-quality  question and I agree with explanation.

niks18 · Answer

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

Subham10 · Answer

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

How did u conclude this? Please show calculation

Bunuel · Answer

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.

asthagupta · Answer

Hi Bunuel,

A small doubt. Though i agree with solution and got the same answer but in case if we are selecting 1/4 of the circumference, we are including either the point (-1,0) or the point (0,1) and none of them satisfies the condition.
So shouldn't exact answer be a little less than 1/4.

Bunuel · Answer

A point has no dimension, so it does not matter. The same way a line has only one dimension, length, but no area.

Miracles86 · Answer

Hi Bunnel,

I've made the exact graphical thought process as you and got the the answer Pi-2/4Pi, which is not in the answer options. It's not clear to me why "Original question does not ask about the area, it asks about the portion of the circumference" as the question asks "Set T consists of all points (x,y) such that x2+y2=1. If point (a,b) is selected from set T at random, what is the probability that b>a+1b>a+1?"

An area can be defined as an infinite number of point. Thus, for me the favorable cases should indeed be an area, since the Set T consists in ALL points that obey to a certain equation.

Could you please clarify?

Thanks a lot!

Bunuel · Answer

Set T consists of all the points on the circumference of the circle only (x^2 + y^2  =  1), NOT the all the points within the circle (x^2 + y^2  <  1), so we should find the probability that the point is ON the circumference but above y = x + 1, NOT the probability that the point is IN the circle but above y = x + 1.

Hope it's clear.

RK007 · Answer

I think this is a high-quality question and I agree with explanation. Such a bomb of a question! So simple, yet deceiving! Thanks for the question!

coreyander · Answer

I think this is a  high-quality  question and I agree with explanation.

ZZhang · Answer

x and y lie between -1 and 1, that is: -1 --- 0 --- 1

Probability than b > a is 50% since they both have an equal chance to be larger than one another

If b > a + 1, then b cannot be negative and must be in the range of 0 --- 1 for the statement to hold true

The number of values that b can hold is reduced by half, and thus the probability is reduced by half, to 25%, or 1/4.

M20-12

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