Probability color ball problem

In summary, this question involves finding the probability of picking a white ball from a bag containing (n+7) tennis balls, where n of the balls are yellow and the remaining 7 balls are white. The first part involves proving that a given statement about the probability of picking a white ball is incorrect. The second part involves using the probability of picking different colored balls to solve a quadratic equation.
  • #1
Trail_Builder
149
0
this question is contained on a gcse past paper i am doing and i think I am approaching it from the wrong angle but have tried a few times and don't know where i am going wrong :( hope you can help

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

Homework Equations





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 realize i shouldn't 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
 
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  • #2
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
 
  • #3
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.
 
  • #4
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.
 
  • #5
Sorry buddy, never realized, will take note for the future.
 
  • #6
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
 
  • #7
Pagey said:
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 realized where i went wrong, that's utter rubbish.

The outcome is going to be

<< 2nd complete solution deleted by berkaman >>
 
Last edited by a moderator:
  • #8
Pagey said:
i realized where i went wrong, that's 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.
 
  • #9
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 . . .
 
  • #10
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...
 
  • #11
Pagey said:
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.
 
  • #12
Pagey said:
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.
 
  • #13
Ok, thanks, i do respect the rules and i think i now understand how to answer properly, thanks for the advice . . .
 

1. What is the probability of drawing a specific colored ball from a bag?

The probability of drawing a specific colored ball from a bag depends on the total number of balls in the bag and the number of balls of that specific color. It can be calculated by dividing the number of balls of that color by the total number of balls in the bag.

2. Can the probability of drawing a specific colored ball change?

Yes, the probability of drawing a specific colored ball can change if the number of balls in the bag or the number of balls of that specific color changes. For example, if more balls of that color are added to the bag, the probability of drawing that color increases.

3. What is the probability of drawing multiple colored balls in a row?

The probability of drawing multiple colored balls in a row can be calculated by multiplying the probabilities of drawing each ball separately. For example, if the probability of drawing a red ball is 1/4 and the probability of drawing a blue ball is 1/2, the probability of drawing a red ball followed by a blue ball would be (1/4) * (1/2) = 1/8.

4. How does the number of balls in the bag affect the probability of drawing a specific color?

The number of balls in the bag affects the probability of drawing a specific color because it changes the total number of balls and the number of balls of that specific color. As the number of balls in the bag increases, the probability of drawing a specific color decreases.

5. What is the difference between theoretical probability and experimental probability?

Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely. Experimental probability is based on actual results from conducting an experiment. It may differ from theoretical probability due to chance or other factors.

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