Here's the question you clicked on:
Kaederfds
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.
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 \[x \approx 13\]
oh, multiply 15*28 in each side
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... \[\frac{x}{x+15} + \frac{1}{15} = \frac{x+13}{x+13+15}\]
do you get this @Kaederfds ??
i got 9 what about you?
no! the x represent the green balls but green balls is in 15 balls!
thank you for pointing that out @AndrewNJ
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
@ParthKohli your expertise are needed ;)
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.