In a bag there are a certain number of black balls and white balls. Th

Hi, I did it with another method, it took a lot of trial and error to find a way to solve this one. At first I took the pure algebraic way to solve it, but no joy. Then I kind of tweaked my method a bit. I have taken a method similar to chetan2u. As we have been asked the ratio of Black and White balls, let's call the ratio k.

Let the number of Black balls = kx
Let the number of White balls = x

Now let's write the probability expression

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

on simplifying we get

\(x^2(k^2 + 1 - 2k) = x(k + 1)\)

Here, we know that x(number of white balls) is positive and an integer. So we can simply cut x here

\(x(k^2 + 1 - 2k) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1 - 2k)}\)

Now, we know that x has to be an integer. From here, we can try values for k from options such that x turns out to be an integer.

Only (B) satisfies.

Hi,

Your solution using variable is awesome! This algebraic way always excites me

However, I've just found out an error in a half-way step, thus making the final answer not as you expected.

It runs smoothly until this step:

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

In fact, after simplifying the equation, we will get the following:

\(x^2(k^2 + 1) = x(k + 1)\)
(In your post, you just made an extra \(-2kx^2\))

Moving on, because x(number of white balls) is a positive integer, we can simply cut x:

\(x(k^2 + 1) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1)}\)

x is a positive integer only when k=1. Therefore, the ratio of the number of black balls to white balls is 1:1

(C) must be the correct answer if the above probability equation (in red color) is correct!

Still not understand why OA is (B). Any thought?

In the highlighted part above LHS should be multiplied by 2 because we can have BW as well as WB cases. Having said that the question is not good. chetan is right, the ratio could be 6:3 (2:1) or 3:6 (1:2) among many other cases. So, you can ignore this question and move on.

TOPIC IS LOCKED.