• Support PF! Buy your school textbooks, materials and every day products Here!

Probability color ball problem

this question is contained on a gcse past paper i am doing and i think im approaching it from the wrong angle but have tried a few times and dunno where i am going wrong :( hope you can help

1. Homework Statement

A bag contains (n+7) tennis balls.
n of the balls are yellow.
The other 7 balls are white.

John will take at random a ball from the bag.
He will look at its colour and then put it back in the bag.

part a) Bill states that the probability that John will take a white ball is 2/5
Prove that Bill's statement cannot be correct.

part b) After John has put the ball back into the bag, Mary will then take at random a ball from the bag.
She will note its colour.

Given the probability that John and Mary will take balls with different colours is 4/9, prove that 2n^2 - 35n + 98 = 0

2. Homework Equations



3. The Attempt at a Solution

a) (I think I got this bit right)

2/5 = 7/(n+7)
n+7 = 7/(2/5)
n+7 = 7/0.4
n+7 = 17.5
n = 10.5

Bill's statement cannot be write because there cannot be a non-interger number of balls.

b)

I did a probability tree thing and worked out there are 2 ways that they will have different colours, each with a 7n/(n+7) probability.

therefore, there is a 14n/(n+7) chance of them selecting different colours.

(this is where i might be going wrong)

so, 14n/(n+7) = 4/9

126n/(n+7) = 4
n+7 = 126n / 4
n+7 = 31.5n
7 = 30.5n

??

this is where i get stuck cause i realise i shouldnt being working out n should I?

is it i have to factorise it out somewhere? get it in quadradic form or whatever?

hope you can help

thnx
 

Answers and Replies

18
0
I think you had a typo and meant 2n^2 - 3.5n + 98 = 0

Part (a) looks correct

Part (b)

Probability of John or Mary getting yellow : n/n+7
White : 7/n+7


Probability of John(yellow) and M(b) = (7/n+7)(n/n+7)
M(y) J(b) = (n/n+7)(n/n+7)

Therefore the outcome is going to be the same.

so expand

(7/n+7)(n/n+7) = 7n / n^2+14n+49

Therefore 7n / n^2+14n+49 = 4/9

63n = 4(n^2 + 14n + 49)
4n^2 + 56n + 196 -63n = 0
4n^2 - 7n + 196 = 0 (/2)
2n^2 - 3.5n + 98 = 0
 
Dick
Science Advisor
Homework Helper
26,258
618
Your first step in b) is dubious. 7n/(n+1)>1 for n>1. Can that be a probability? Hint: it maybe clearer to compute the probability they both pick the same color.
 
cristo
Staff Emeritus
Science Advisor
8,056
72
Pagey, welcome to PF. Please note that we do not give out full solutions to homework problems, but instead attempt to guide the original poster to the solution.
 
18
0
Sorry buddy, never realised, will take note for the future.
 
o rite, yeh i see where i went wrong at begining, stupid error (grrrr)

i still confused as to how you got ...3.5n... pagey?

i check and it wasnt a typo
 
18
0
Probability of John(yellow) and M(b) = (7/n+7)(n/n+7)
M(y) J(b) = (n/n+7)(n/n+7)

Therefore the outcome is going to be the same.

so expand

(7/n+7)(n/n+7) = 7n / n^2+14n+49

Therefore 7n / n^2+14n+49 = 4/9
i realised where i went wrong, thats utter rubbish.

The outcome is going to be

<< 2nd complete solution deleted by berkaman >>
 
Last edited by a moderator:
berkeman
Mentor
56,085
6,139
i realised where i went wrong, thats utter rubbish.

The outcome is going to be

<< 2nd complete solution deleted by berkaman >>
Pagey, you were warned above not to post complete solutions. Stop that! Our task here on the PF is to offer tutorial help and guidance, hints, suggested paths to look down, etc. NOT to do the student's work for them.

I'm issuing you a warning point. Not a good way to start your tenure at the PF, but hopefully your future posts will be useful in a tutorial way.
 
18
0
Sorry dude, but i don't see what i'm supposed to say, i didn't answer, i tried to take note of the fact that i wasn't supposed to answer the question and tryed to provide information that would help him. Sorry for the incovience nevertheless . . .
 
pagey i think you're supposed to say stuff like, "this line should read like this.." and then leave the rest, or say like, "you forgot to do this to that", etc...
 
cristo
Staff Emeritus
Science Advisor
8,056
72
Sorry dude, but i don't see what i'm supposed to say, i didn't answer, i tried to take note of the fact that i wasn't supposed to answer the question and tryed to provide information that would help him. Sorry for the incovience nevertheless . . .
If you're unsure as to what constitutes "help," then just take a look at some of the threads in the homework forum and see what help is given to the original poster.
 
berkeman
Mentor
56,085
6,139
Sorry dude, but i don't see what i'm supposed to say, i didn't answer, i tried to take note of the fact that i wasn't supposed to answer the question and tryed to provide information that would help him. Sorry for the incovience nevertheless . . .
If you are ending each post with a question, that's a good indicator that you are providing the right kind of help. Having the final answer at the end of your post is a bad indicator.
 
18
0
Ok, thanks, i do respect the rules and i think i now understand how to answer properly, thanks for the advice . . .
 

Related Threads for: Probability color ball problem

Replies
25
Views
2K
Replies
2
Views
7K
Replies
11
Views
622
Replies
7
Views
727
Replies
11
Views
2K
Top