If $1000 needs to be distributed between Andy and John in such a way

Question

If $1000 needs to be distributed between Andy and John in such a way that Andy receives at least as much as does John, What is the probability that Andy receives at least twice as much as does John ?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. None of these.

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ThatDudeKnows · Answer

Probability, so we need to find the fraction "win"/possibile.
How do we "win?" When Andy receives at least twice as much as John. That happens when Andy receives at least $666.67. That's 1/3 of the time. So, 1/3 is our numerator.
What are the possible ways? Some of the ways to distribute the money aren't allowed. We know that Andy receives at least as much as John, so that gets rid of half the options (all the ones where John receives more). We are left with half. So, 1/2 is our denominator.

(1/3)/(1/2) = 2/3

Answer choice D.

ChakAvi · Answer

Let Andy gets a, & John gets j So, a+j = 1000 so, a= 1000-j Possible cases: a>= j 1000-j > = j or, j <= 500 Required cases: a>=2j or 1000-j >= 2j or j<= 1000/3 Prob = 1000/(3*500) = 2/3 Option D

AlevtinaN · Answer

Sorry, maybe I didn’t understand right but why 666.67 is a 1/3 ?

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ThatDudeKnows · Answer

In order for Andy to receive at least twice as much as John, he has to receive at least $666.67. That's everything in the right hand section of the attached drawing, so one-third.

SaquibHGMATWhiz · Answer

There are a few important points that we need to understand in such questions.

1. We generally count the total number of possible cases but in this question, we should not count the scenarios.
 o One common mistake one can make here is to assume the distribution to be $1, $2, and so on. Understand there's no mention of whether we can go down to cents or not and hence this should be taken as continuous.

2. Also, in general, we take all the possible cases as denominators but in this question, we should not take it as it is mentioned

So our denominator is the probability of Andy receiving at least as much as John.
 • Now this probability can be found when we understand that the probability of Andy receiving 0 to 500 is the same as Andy receiving 500 to 1000.
 
o So we can easily say that in half of the scenarios Andy can receive equal or more than Andy

On the other hand, our numerator comes from

• The least possible scenario is when Andy receive atleast  \frac{2000}{3}  and John receives almost  \frac{1000}{3} .
 o So Andy must receive in the top  \frac{1}{3} rd portion.
 o That is from  \frac{2000}{3}  to 1000    
So the probability of that event is  \frac{1}{3}

So the answer =  \frac{\frac{1}{3}}{\frac{1}{2}  =  \frac{1}{2}

SaquibHGMATWhiz · Answer

So, I was not allowed to post any attachments or links and hence could not post this second approach here. (I did not have 5 posts by that time)

So thought to post it tonight.

This approach is more about picturizing and understanding how exactly things are happening. So this is much more intuitive and of course faster.

So do read the first part of my previous post so that you can relate to it.

Our denominator is the probability of Andy receiving at least as much as John. Thus we are looking at the space towards the right of 500 (500 is inclusive). In such a question of continuous probability think in terms of area and thus we can say that area is half so the probability directly becomes  \frac{1}{2}  
 Now for Andy to get double of John or more we are looking at the area towards the right of 2000/3 (This point included)
So we can easily understand that we are talking about 1/3rd of the area and thus the probability of this is  \frac{1}{3}

So the answer =  \frac{\frac{1}{3}}{\frac{1}{2}  =  \frac{1}{2}

https://www.dropbox.com/s/yxxbvym9lob758d/GC1.jpg?dl=1

Regor60 · Answer

Andy receiving at least as much as John means:

A>=J

and since A + J = 1000:

A>=500 and J<=500

and when A=1000, J=0, therefore:

500<=A<=1000

The range of values for A is therefore:

1000-500 = 500

The condition of Andy receiving at least twice John means:

A>=2J or J<=A/2

and since A + J = 1000:

A >= 2000/3

Therefore the range of values for which Andy receives twice as much as John is:

1000 - (2000/3) =1000/3

Therefore, the probability that he receives twice as much as John is the above range divided by the total range calculated prior:

(1000/3)/500 = 1000/1500 =

10/15 = 2/3

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If $1000 needs to be distributed between Andy and John in such a way

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If $1000 needs to be distributed between Andy and John in such a way

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615 to 715 in 15 Days (Ria's Strategies)

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