1
$\begingroup$

Assume balls of the same color are all identical. condition: There are at least $7$ blue balls.

Can anyone help me to understand this problem intuitively?

The way I look at it is that since $7$ blue balls must be chosen that leaves us $5$ balls to choose out of $12$. Those $5$ balls can be either red, blue, or green. The textbook seems to suggest this can be viewed as the distribution of the same objects into different boxes, which gives us the answer : $$\frac {(5+3-1)!}{(5!\cdot2!)}$$
I have problem understanding it intuitively because there are unlimited numbers of balls that's gonna give us something like :

$$\frac{C(\infty, 3) \cdot C(\infty, 1) \cdot C(\infty, 1)}{\infty \cdot \infty \cdot \infty}$$

$\endgroup$
  • $\begingroup$ Is each color equally probable? Assuming so, then Hint: all that matters is Blue vs non-Blue and a ball is Blue with probability $\frac 13$. Use the binomial distribution. $\endgroup$ – lulu Oct 2 '17 at 16:02
  • 2
    $\begingroup$ Don't be confused by the infinities. If it makes it clearer, just imagine that there are three balls (one Blue, two non-Blue) and that you replace the balls after you draw them. $\endgroup$ – lulu Oct 2 '17 at 16:03
  • 1
    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. The fact that the number of balls of each type is unlimited means that you can use as many balls of each color as needed. $\endgroup$ – N. F. Taussig Oct 2 '17 at 16:09
  • $\begingroup$ @N.F.Taussig Thank you. I will take a look at it . $\endgroup$ – Snailwalker Oct 2 '17 at 16:12
3
$\begingroup$

Let $x_1$, $x_2$, $x_3$ be the number of red, blue and green balls in the selection. Then we are looking for the number of integral solutions to $$ x_1+x_2+x_3=12;\quad x_1\geq 0,\, x_2\geq 7, \,x_3\geq 0 . $$ Letting $y_1=x_1$, $y_2=x_2-7$ and $y_3=x_3$, we seek equivalently the number of integral solutions to $$ y_1+y_2+y_3=5;\quad y_i\geq 0.$$

Now use stars and bars.

$\endgroup$
  • $\begingroup$ wow, That's a great way to look at it ! Thanks a lot ! $\endgroup$ – Snailwalker Oct 2 '17 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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