GCSE Maths Help - probability

[JMatiman]

Hi all, not sure where else to turn at this point 😆

My step lad has his mocks coming up and its been a while since I sat my GCSE's obviously.

Between myself and him we have worked our way through many many books and test papers etc but forever getting stuck on this kind of question. Not seen a solid example in any book or online. The question reads as follows.......

There are 3 green balls and some other balls in a bag. The probability of picking a green ball is 3/N (3 over N), where N is the total number of balls in the bag. One green ball is removed. The probability of picking a green ball decreases by 0.1. Work out the two possible values of N.

Through my many hours of book reading / googling and trying my best to blow off the old cobwebs in my mind. I can only see a reasonable solution where we know the total amount of balls already in the bag which we would a tree diagram to work it out. Really finding it difficult to work this out without knowing that info, currently 1am and lost the will to live whilst he is happily sleeping 😆

any help or pointers and massively welcomed.

cheers

I'm not much good at probabilities, but I make the answer 5 or 6 (but it's late so my maths may be utterly wrong!)

Probability of picking first green ball = 3/N
Probability of picking second green ball = 2/(N-1)
Decrease in probability 3/N - 2/(N-1) = 1/10
therefore 30/N - 20/(N-1) =1
then (30(N-1) - 20N)/(N(N-1)) =1

which is a quadratic equation ie. N2 -11N-30 =0 ( by N2 I mean NxN - sorry, don't know how to do N squared when typing!)

Solving the quadratic gives you values of 5 or 6 for N

I agree with @pinkgown .

Expanding her missing steps:
30N-30-20N=N^2 - N
10N-30=N^2-N
N^2 -11N +30=0
(N-5)(N-6)=0
N = 5 or 6.

There was a similar question in Edexcel 2015 called 'Hannah's Sweets'.
Search for that to check he can also do that one.

If you get stuck again. I made this to help parents out