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Re: If a real number x is randomly chosen between 0 and 2, inclusive, and [#permalink]
Kinshook wrote:
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


Can you elaborate on the calculation. Why are you taking 4/6 and 2/6? And why the probability is 1/2 for interval [0;2]?

I would be really grateful if you could answer.
Probability is my weakest area((
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Re: If a real number x is randomly chosen between 0 and 2, inclusive, and [#permalink]
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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


sherxon wrote:
Kinshook wrote:
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


Can you elaborate on the calculation. Why are you taking 4/6 and 2/6? And why the probability is 1/2 for interval [0;2]?

I would be really grateful if you could answer.
Probability is my weakest area((
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If a real number x is randomly chosen between 0 and 2, inclusive, and [#permalink]
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.


Kinshook wrote:
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


sherxon wrote:
Kinshook wrote:
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


Can you elaborate on the calculation. Why are you taking 4/6 and 2/6? And why the probability is 1/2 for interval [0;2]?

I would be really grateful if you could answer.
Probability is my weakest area((
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Re: If a real number x is randomly chosen between 0 and 2, inclusive, and [#permalink]
1
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ChinmayBohra01 wrote:
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.




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

Posted from my mobile device
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Re: If a real number x is randomly chosen between 0 and 2, inclusive, and [#permalink]
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Expert Reply
sadievj wrote:
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


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.
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If a real number x is randomly chosen between 0 and 2, inclusive, and [#permalink]
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.
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Re: If a real number x is randomly chosen between 0 and 2, inclusive, and [#permalink]
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
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Re: If a real number x is randomly chosen between 0 and 2, inclusive, and [#permalink]
Expert Reply
sadievj wrote:
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

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
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Re: If a real number x is randomly chosen between 0 and 2, inclusive, and [#permalink]
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Re: If a real number x is randomly chosen between 0 and 2, inclusive, and [#permalink]
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