Question

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.

Answer 1

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
\[x \approx 13\]

Answer 2

oh, multiply 15*28 in each side

Answer 3

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...

\[\frac{x}{x+15} + \frac{1}{15} = \frac{x+13}{x+13+15}\]

Answer 4

do you get this @Kaederfds ??

Answer 5

i got 9
what about you?

Answer 6

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

Answer 7

@Kaederfds

Answer 8

oh yes my mistake

Answer 9

thank you for pointing that out @AndrewNJ

Answer 10

:)

Answer 11

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

Answer 12

@ParthKohli your expertise are needed ;)

Answer 13

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.