Find the probability that the $4$th ball removed from the box is white

Question

A box has $10$ balls, $6$ of which are black and $4$ of which are white. $3$ balls are removed from the box, their color unnoted. Find the probability that a fourth ball removed from the box is white, Assume that the $10$ balls are equally likely to be drawn from the box.

I got the correct result, but I used binomial coefficients, which were introduced in the next chapter, so I wanted to ask if there's a simpler way to solve the exercise, I did;

$\displaystyle\Pr(4^{th}\text{Ball is white})=\frac{\binom{6}{3}\frac47+\binom{6}{2}\binom{4}{1}\frac37+\binom{6}{1}\binom{4}{2}\frac27+\binom{4}{3}\frac17}{\binom{10}{3}}=\frac25$

where the $1.$ summand, $\displaystyle\frac{\binom{6}{3}\frac47}{\binom{10}{3}}$ is the probability, that $4^{th}\text{Ball is white}$, if the first $3$ were black. etc.

Is it pure chance that $\frac25=\frac{4}{10}=$(number of white balls divided by total number of balls) is the correct result ?

Answer 1

Any ball is just as likely to be the fourth ball removed as any other. So the probability is $\frac{4}{10}$.

Remark: Good question! It is useful to have done the problem the "hard" way. The instinct for pain avoidance helps one to remember the easy way.

Answer 2

You can make a tree diagram:

$$\frac{2}{21}+\frac{1}{14}+\frac{1}{14}+\frac{1}{14}+\frac{1}{35}+\frac{1}{14}+\frac{1}{35}+\frac{1}{35}+\frac{1}{210}=\frac{2}{5}$$

Answer 3

I agree with André, since the colours of the previous balls are not noted, the events are independent. This means it doesn't matter whether we are drawing 1st ball or the last, the probability of drawing $4$th ball as "white" will be $\frac{4}{10}$. Also, the tree given by Asdruba is well explaining the cases..