Probability of drawing spheres

[ghostfirefox]

We have 27 balls in the container, some of which are white and some black. How many white balls in the container must be at least, so that the probability that two balls were drawn at random without a draw was less than 23/30?

[Guest]

I missed few words. Now its correct - We have 27 balls in the container, some of which are white and some black. How many white balls in the container must be at least, so that the probability that two black balls were drawn at random without a return was less than 23/30?

[Guest]

Denote by [tex]x[/tex] the number of white balls in the container. So the probability that the two balls drawn are black is:

[tex]\frac{27-x}{27} \cdot \frac{26-x}{26}[/tex]

and this probability must be less than [tex]\frac{23}{30}[/tex].

As a result we can get the following quadratic inequality:

[tex]x^{2} - 53x + 163.8 < 0[/tex].

Second-order polynomial equation roots from the left hand side of the above inequality can be found, for example, with the help of the online quadratic formula solver https://ezcalc.me/quadratic-formula-calculator/. As a result we have:

[tex]x_{1 } = 3.3[/tex], [tex]x_{2 } = 47.7[/tex] .

The second root is irrelevant, so, using the first one, we come to the answer: [tex]x = 4[/tex]. The minimal number of white balls in the container is 4!

[Guest]

Ooops! The second root is 49.7 ! But it doesn't matter

[ghostfirefox]

Thanks for yours replies.