Sweets in a bag probability problem

[thebosonbreaker]

Andrei has a bag of x sweets.
He removes two sweets from the bag simultaneously (without replacement).
He now removes a third sweet.
The probability that the third sweet is red is (x/2) - 1.
How many red sweets were in Andrei's bag to begin with?

Could somebody please tell me if (and how) it is possible to solve this problem?
Thank you very much for your help in advance!

 

[scottdave]

I am not sure if probability, is the correct term to use here (a probability is a number less than 1, like 1/2 or 1/3). Could the value (x/2) -1 be a number between 0 and 1? If probability is the word that they intend, then Well yes if x/2 is between 1 and 2. What does this tell you about possible values of x? Would it help you to determine how many start out as red? Do they give you any information about the original sweets, which were pulled? You will need to know if they were red, or not.

 

[WWGD]

I agree with @scottdave, the problem does not seem to make sense; what if x=10 so that x/2-1 =4 , or x >10? Are you sure you there isn't a mistake somewhere?

 

[ssd]

I think, we have to deduce that 0≤x/2-1≤1. ⇒2≤x≤4. Again three sweets are already drawn⇒3≤x≤4. Now consider Case1: x=3. Therefore, prob of third being sweet is red is either 0 or 1 (according as last sweet left in box non red or red after withdrawal of 1st two). This does not satisfy the given prob condition as the given prob=1/2. Alternatively, there are 3 possible cases of the 1st drawn 2 sweets ³C₂. That can not produce prob for the last one left=1/2. Case2: x=4. ⇒x/2-1=1.Last two in the box must be red. Therefore the 1st two must also be red. ⇒ all the 4 sweets must be red. Since, if there is one non red, and two reds are drawn in first draw, the prob of red in third draw is 1/2 which does not satisfy the given prob condition. Similarly, there can not be 2 or 3 non reds.
One trivial case remains for x=2. After the 1st two drawn there will none for the third draw. Therefore the prob of red in 3rd draw is 0, which satisfies the prob condition. Then there can be 0,1,2 possible reds.

 

[scottdave]

...One trivial case remains for x=2. After the 1st two drawn there will none for the third draw. Therefore the prob of red in 3rd draw is 0, which satisfies the prob condition. Then there can be 0,1,2 possible reds.

Nice working through. After reading this, and re-reading the problem, I say that x=2 is not a possibility, because the problem tells you that a 3rd piece is removed. This makes it only one possible answer.

 

[ssd]

Actually, I added the case of x=2 on the basis of the given prob condition (third item is red with prob x/2-1=0). A non existent item (here, the third sweet) can as well be drawn with prob=0 (which, in turn implies, it is impossible to draw it).