Question

Probability ques (easy yet worth trying)

In an urn, there are several colored balls, with equal numbers of each color. If we were to add 14 balls of a new color (different from those in the urn) into the urn, the probability of drawing two balls, without replacement, of the same color stays constant. How many balls are there in the urn initially?

Answer 1

i'd say 14n, where n is the number of colors in the urn originally.

Answer 2

Actually there is a proper integer answer to this, and nopes,its not of the form 14n

Answer 3

let k = # distinct colors
n = # of balls per color , which is the same for each color.
so there are a total of k*n balls , then you add 14 you get
k*n + 14 balls , .
now probability of taking two balls with replacement
same color

Answer 4

is it\[\frac{2}{n+14}\] for same two balls
\[\frac{2k}{n}\] for color

Answer 5

the probability before the 14 balls is put in is
k* [n(n-1)] / ( k*n * (kn-1) )
the probability after the 14 balls for picking the same ball is
(k+1)(n+1)(n) / ( nk + 14 )

Answer 6

how about we do an example
4 red balls, 4 blue balls , 4 green balls. mix them up in an urn
the probability of picking two balls of same color (w/out replacement)
P( RR or BB or GG) = 3* 4/12 *3/11

Answer 7

now add 14 yellow balls

Answer 8

i get
k*n(n-1) / [ nk * (nk-1) ] = [ k * n (n-1) + 182] / [ (nk + 14) ( nk + 13) ]

Answer 9

By the way, there are 91 balls, methinks.

Answer 10

And there are 7 balls per colour.

Answer 11

terence, before or after the 14 are put in ?

Answer 12

ok initially

Answer 13

terence, your solution works in my equation :)

Answer 14

Yay :)
Also you bothered to spell "Terence" right XD

Answer 15

k*n(n-1)/[ nk * (nk-1) ] = [ k *n (n-1) + 182] / [ (nk + 14) ( nk + 13) ]

Answer 16

may i ask how you got your answer?

Answer 17

Nothing too fancy, too much trial and error... at least I know the answer, now to find an efficient general way to solve this :>

Answer 18

i dont see an easy way to solve my equation, i tried cross multiplying

Answer 19

i dont consider this an easy question, the guy who posted this is torturing us

Answer 20

yes i agree this is not an easy question @shubhamsrg wheres this from

Answer 21

Tentatively, I think that

Given an urn with n balls and k balls per colour, if adding r balls of a new colour would make it so that the probability of drawing two balls of the same colour remains constant, then there must have been rC2 balls initially, where C is the combination operator.
So
n = rC2

Answer 22

Well your answers are right.
And seriously its not that hard. Its from brilliant.org level 4 ques. I was able to do it there somehow. I wished to share it here.

Answer 23

I should post my solution now ?

Answer 24

please post

Answer 25

Let there be n balls, q sets where 1 set denotes 1 color, and k balls in each set, i.e.
kq = n

Now, probab of drawing 2 balls of same color initially can be seen as
for 1st ball , we have n choices -> n/n = 1
for 2nd ball , we have only k-1 choices , -> (k-1)/(n-1)

Hence probab for drawing 2 balls of same color initially = 1* (k-1)/n = (k-1)/(n-1)

After 14 balls have been put in, lets consider 2 diff cases,
1) when 2 balls are drawn from original set of n balls
2) when 2 balls are drawn from these 14 balls

probab of 1st case will be (n/(n+14)) * ((k-1)/(n+13))
probab of 2nd case will be (14/(n+14))* (13/(n+13))

Net probab = (n/(n+14)) * ((k-1)/(n+13)) + (14/(n+14))* (13/(n+13))
= (n(k-1) + 182)/((n+14)(n+13))

On equating,

(k-1)/(n-1) = (n(k-1) + 182)/((n+14)(n+13))

(k-1)(n+13)(n+14) = (n(k-1) + 182)(n-1)

(k-1)(n^2 + 27n + 182) = (n^2 -n)(k-1) + 182(n-1)

kn^2 + 27kn + 182k -n^2 -27n -182 = n^2 k -n^2 -nk +n + 182n - 182

=>28kn + 182k -210n = 0

=> 2nk + 13k - 15n = 0

=> n(2k-15) = -13k

n = (13k)/(15-2k)

Hit and trial for (n,k) gives
(1,1) <-- can not be solution since 2 balls are drawn
(13,5)
(26,6)
(91,7)

Of these only (91,7) is a solution since n/k for remaining 2 isnt integer
(recall n/k = q)

Answer 26

oh teri..... :O