Find total number of balls with conditional probability

Question

So, this is the problem

You have an urn with balls of two different colors and you take two without replacement. The probability of getting two of the same color is the same as getting one from each color. How many balls from each color if we know that in total we have 31?

What I do is (X,Y denote the different colors, index denotes different drawings)

$P(X_2|X_1) = P(Y_2|X_1) = P(X_2\cap X_1)/P(X_1) = P(Y_2\cap X_1)/P(X_1) \\ P(X_2\cap X_1) = P(Y_2\cap X_1) \rightarrow\\ \frac{X-1}{29}\frac{X}{30} = \frac{Y}{29}\frac{X}{30} \rightarrow\\ X-1=Y \rightarrow X= 16, Y=15 $

Is this correct ? I am worried, that one could choose instead of $P(Y_2\cap X_1)/P(X_1)$ to have $P(X_2\cap Y_1)/P(Y_1)$ and then maybe my solution would not hold...

Answer 1

If there is $x$ of one color, there is $(31-x)$ of other.

Number of ways of picking balls of the same color,

$ \displaystyle x(x-1) + (31-x) (30-x)$

Number of ways of picking one ball of each color,

$= 2 x (31-x)$

If you equate the two, you get $ \displaystyle x = \frac{31 \pm \sqrt {31}}{2}$.

So there is in fact no way to have $31$ balls in total of two different colors such that the probability of picking two balls of same color is same as probability of picking one ball of each color (without replacement).

Answer 2

Two of the same colour is the event: $(X_1\cap X_2)\cup(Y_1\cap Y_2)$.

One from each colour is the event: $(X_1\cap Y_2)\cup(Y_1\cap X_2)$

Remember that there are $x$ balls of one colour and $31-x$ of the other.