probability

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jainrahul1985: Master | Next Rank: 500 Posts; Posts: 228; Joined: Sun Aug 17, 2008 8:08 am; Thanked: 4 times

by jainrahul1985 » Sun Feb 27, 2011 9:37 pm

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>a+1 ?

a) 1/4 b) 1/3 c) 1/2 d) 3/5 e) 2/3

OA A

Can someone please explain how to approach this question in the easiest way .

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Source: Beat The GMAT — Problem Solving | Sun Feb 27, 2011 9:37 pm

manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

by manpsingh87 » Sun Feb 27, 2011 10:09 pm

jainrahul1985 wrote: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>a+1 ?

a) 1/4 b) 1/3 c) 1/2 d) 3/5 e) 2/3

OA A

Can someone please explain how to approach this question in the easiest way .

x^2+y^2=1; represents equation of circle whose center lies at the origin (0,0) and having radius =1;

if we draw the circle considering origin as the center we will find out that it touches the XY axis at the point (1,0),(-1,0),(0,1),(0,-1) of these values;

Now looking at the conditions b>a+1; only portion of area where this thing can be achieved lies between the points (-1,0) and (0,1), for all other portions points will lie outside the circle,

Therefore of all the 4 portions (i.e. we are dividing circle into 4 parts (I,II,III,IV) quadrant) only second quadrant points yields the desired result hence porbability is 1/4..!!

I hope it helps..!!!

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Sun Feb 27, 2011 10:14 pm

It can be more clear if we draw the figure.

The circle has its center at the origin and radius 1 (as given in the question)
a^2 + b^2 = 1 and b = a + 1 intersect at the points (0, 1) and (-1, 0)
{As a^2 + (a + 1)^2 = 1 or 2(a^2 + a) = 0 or a = 0, -1 and b = 1, 0}
When b > a + 1, b is the y-coordinate and it will be greater than a + 1 at all points above the line y = x + 1 (see the figure), which includes points in Quadrant 2. Hence, required probability = 1/4

[spoiler]The correct answer is A.
[/spoiler]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Feb 27, 2011 10:16 pm

jainrahul1985 wrote: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>a+1 ?

a) 1/4 b) 1/3 c) 1/2 d) 3/5 e) 2/3

OA A

Can someone please explain how to approach question in the easiest way .

The easiest approach is to draw the circle and the line y=x+1:

Every point (a,b) above the line y=x+1 satisfies the given condition that b > a+1. As shown above, a quarter of the circle (the portion shaded in blue) will lie above y=x+1. Thus, the probability is of picking a point (a,b) on the circle such that b>a+1 is 1/4.

The correct answer is A.

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jainrahul1985: Master | Next Rank: 500 Posts; Posts: 228; Joined: Sun Aug 17, 2008 8:08 am; Thanked: 4 times

by jainrahul1985 » Sun Feb 27, 2011 10:45 pm

awesome approach by both experts . you made this question so simple for me . thanks a ton

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HSPA: Legendary Member; Posts: 1101; Joined: Fri Jan 28, 2011 7:26 amFollowed by: 13 members; Thanked: 47 times; GMAT Score:640

by HSPA » Mon Feb 28, 2011 10:01 pm

Are a,b integers here??

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