Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The probability of picking two white balls from a lot of black and white balls is $1\over{2}$. If the number of black balls is even, what is the minimum number of black and white balls in the lot?

share|improve this question
    
How many trials do you have? Do you have to pick the two white balls on consecutive trials? What have you tried so far? –  Doug Spoonwood Jun 11 '12 at 2:04
1  
Are you sampling with replacement or without replacement? I think this could affect the answer. –  Michael Chernick Jun 11 '12 at 2:09
    
@MichaelChernick: If the sampling were with replacement, the probability of picking two white balls would never be $\frac12$, so I assumed it was without replacement. –  robjohn Jun 11 '12 at 7:41
    
@robjohn Yes that is true since it would have to be 1/sqrt(2) in each case and of course sqrt(2) is not an integer. –  Michael Chernick Jun 11 '12 at 11:09
    
@MichaelChernick: yes, if $\sqrt{2}$ were rational, it would be an integer. That is, given $p$ white balls and $q$ black balls, the probability of drawing two white balls with replacement would be $\left(\frac{p}{p+q}\right)^2$ which cannot be $\frac12$. –  robjohn Jun 11 '12 at 11:33
add comment

3 Answers 3

With $p$ white balls and $q$ black balls, the probability of picking two white balls is $$ \frac{p(p-1)}{(p+q)(p+q-1)}\tag{1} $$ This is because there are $p(p-1)$ ways to choose two white balls and $(p+q)(p+q-1)$ ways to choose any two balls.

We want $q$ to be even and $$ \frac{p(p-1)}{(p+q)(p+q-1)}=\frac12\tag{2} $$ The smallest solution to $(2)$ with $q$ even is $15$ white and $6$ black balls.


Equation $(2)$ can be transformed to $$ 2(2p-1)^2-(2(p+q)-1)^2=1\tag{3} $$ Solving the diophantine equation $(3)$ using continued fractions yields $$ \begin{align} p_n&=6p_{n-1}-p_{n-2}-2\\ q_n&=6q_{n-1}-q_{n-2} \end{align}\tag{4} $$ with $(p_1,q_1)=(1,0)$ and $(p_2,q_2)=(3,1)$.

The next solution satisfies the given conditions: $(p_3,q_3)=(15,6)$.

share|improve this answer
add comment

Assuming that you are sampling without replacement, you have $k$ white balls and $2m$ black balls. The probability of choosing two white balls is $\dfrac{k (k-1)}{(2m+k)(2m+k-1)} = \frac{1}{2}$ or $2k^2-2k= (2m+k)^2-2m-k.$ So you need to find the smallest $m$ and corresponding $k$ that satisfy this equation.

share|improve this answer
add comment

The probability of picking 1 white ball out of the box is the number of white balls over the number of total balls in the box. Which is white balls + black balls.

Now, imagine that you take out one white/black ball out of that box or whatever, you will have 1 less ball in the box. Of course, the above equation still works but you have to update it with the amount of balls you have NOW.

For example, imagine there are 8 white balls and 12 black balls. The probability of taking a white ball is: $$\frac{8}{8+12}$$

After you take away one black ball for example, you have 8 white balls on the box and 11 black balls. Therefore the probability of taking out one white ball would be: $$\frac{8}{8+11}$$

If you would've taken a white ball out instead of a black one the equation would now be: $$\frac{7}{7+12}$$

So you can clearly see what happens with the equation once you take the first ball out.

Let's now imagine that we do not know the values of the amount of white/black balls in there. Saying this, you can now see that if you have an amount of white balls W, and an amount of black balls B, your probability is: $$\frac{\omega}{\omega+\beta}$$ Now, if you take out one ball, you can update the equation accordingly as we did before, only this time we cannot directly change B and W because we never knew it's values.

This means that, after taking a white ball out of the box, the probability of taking out another white ball is now $$\frac{\omega-1}{(\omega-1)+\beta}$$

Because we just took away one white ball. This means that the amount of white balls we have is W-1. If we would've taken a black ball, the equation would be: $$\frac{\omega}{\omega+(\beta-1)}$$

Hence, we now know that the probability of taking two consecutive white balls is given by the equation

$$\frac{\omega}{\omega + \beta} * \frac{\omega-1}{(\omega-1)+\beta}$$

That is the probability of taking 2 white balls out of the box, which is 1/2 according of what you said.

$$\frac{\omega}{\omega + \beta} * \frac{\omega-1}{(\omega-1)+\beta} = \frac{1}{2}$$

I believe you can solve the rest of the problem yourself. The probability-related stuff here is done =).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.