## Kaederfds 2 years ago A bag contains 15 balls of which x balls are green. If 13 more green balls are added to the bag, the probability of drawing a green ball at random from the bag is increased by 1/15. Find the value of x.

1. AndrewNJ

initial p for pick a green p=x/15 after probability p' is increased by 1/15 so (x/15)+(1/15)=(13+x)/(15+13) (x+1)/15=(13+x)/28 multiply 15+28 in each side so we can have 28(x+1)=15(13+x) 13x=167 $x \approx 13$

2. AndrewNJ

oh, multiply 15*28 in each side

3. lgbasallote

there were originally x green and 15 not green therefore probability is $\frac{x}{x+15}$ when 13 is added then green becomes x+ 13 and 15 still not green therefore the probability now is $\frac{x+13}{x+ 13 + 15}$ it says that the new probability is the old probability plus 1/15 therefore... $\frac{x}{x+15} + \frac{1}{15} = \frac{x+13}{x+13+15}$

4. lgbasallote

do you get this @Kaederfds ??

5. Kaederfds

i got 9 what about you?

6. AndrewNJ

no! the x represent the green balls but green balls is in 15 balls!

7. AndrewNJ

@Kaederfds

8. lgbasallote

oh yes my mistake

9. lgbasallote

thank you for pointing that out @AndrewNJ

10. AndrewNJ

:)

11. lgbasallote

then it shall be first probability is $\frac{x}{15}$ then when 13 is added $\frac{x+13}{28}$ then it says that new probability is old probability plus 1/15 $\frac{x}{15} + \frac{1}{15} = \frac{x+13}{28}$ since they are common denoms you can add x/15 and 1/15 $\frac{x+ 1}{15} = \frac{x+ 13}{28}$ so now you cross multiply $28(x+1) = 15(x+13)$ distribute... $28x + 28 = 15x + 195$ thisdoesnt look like it results to a whole number though

12. lgbasallote

@ParthKohli your expertise are needed ;)

13. phi

Things work out better if instead of If 13 more green balls are added we say 3 more green balls are added. So maybe a typo in the problem, or whoever posed the problem neglected the constraint that the answer have an integer solution.