If a real number x is randomly chosen between 0 and 2, inclusive, and

Question

If a real number x is randomly chosen between 0 and 2, inclusive, and a real number y is randomly chosen between 0 and 6, inclusive, what is the probability that x>y?

A) 1/12
B) 1/8
C) 1/6
D) 1/4
E) 1/3

Bunuel · Answer

Similar question to practice: https://gmatclub.com/forum/if-a-real-nu ... 10013.html

Kinshook · Answer

Asked: If a real number x is randomly chosen between 0 and 2, inclusive, and a real number y is randomly chosen between 0 and 6, inclusive, what is the probability that x>y?

For y > 2; x > y is not possible
For 0<=y<=2; It is 1/2 probability that x>y

The probability that x>y = (4/6)*0 + (2/6)*(1/2) = 1/6

IMO C

sherxon · Answer

Can you elaborate on the calculation. Why are you taking 4/6 and 2/6? And why the probability is 1/2 for interval  ?

I would be really grateful if you could answer. 
Probability is my weakest area((

Kinshook · Answer

Let the divide the time line between 0 & 6 in 2 parts

Part 1: 0 to 2 
Probability that x>y = 1/2 since x & y are equally probable, half the time x > y

Part 2: 2 to 6
Probability that x>y = 0 since x has no value in the interval

Since Part 1 (0 to 2) is 2/6th part of the complete time line of 0 to 6 and Part 2 (4 to 6) is 4/6th part of the complete timeline of 0 to 6, 
Probability that x> y = (2/6)*(1/2) + (4/6)*0 = 1/6

ChinmayBohra01 · Answer

If x and y are inclusive ranges, can someone explain why is not possible if I take total cases of (x,y) as (0,0) (0,1).......
This will form 21 cases. Out of which, x>y in 3 cases (1,0), (2,0), (2,1). So the probability comes out to 1/7.

Please help me understand.

Regor60 · Answer

Real numbers include not just integers but all numbers over the intervals

Posted from my mobile device

ThatDudeKnows · Answer

There is a 2/3 chance that y will be greater than 2. In all such cases, y will be greater than x. Eliminate those.

There is a 1/3 chance that y will be less than 2. In those cases, there is an equal probability that y>x and that x>y.
So there is a 1/3 * 1/2 chance that x>y. That's 1/6.

Answer choice C.

Suruchim12 · Answer

GMATNinja  /  Bunuel

Can you please explain this?

Also, this seems like quite a tough question, what are the chances of something like this appearing on gmat? I don't see the source specified for this.

wickedvikram · Answer

For those who are still confused, please note that the numbers are "real numbers" not integer, not natural, not whole.
Which means that these numbers can be fractions to any decimal point, so you have infinite number. 
Therefore, you can't make pairs for x>y such as (1,0), (2,0), (2,1)
you will need to consider the whole range as done here by Kinshook & TDK

SaquibHGMATWhiz · Answer

Solution:

Note:  this is a question of conditional probability where we cannot count cases

Since we are looking for condition when X is greater than Y, therefore  Y also has to be between 0 and 2  
 Probability that  Y will be between 0 and 2  = \frac{2}{6}=\frac{1}{3}   
 Now, we are  choosing both X and Y from 0 to 2  and the probability that  X will be greater than Y   =\frac{1}{2}  
 Total probability  = \frac{1}{3}*\frac{1}{2}=\frac{1}{6}

Hence the right answer is    Option C

bumpbot · Answer

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If a real number x is randomly chosen between 0 and 2, inclusive, and

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If a real number x is randomly chosen between 0 and 2, inclusive, and

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