A machine can throw n balls of different color up in the air. The probability of throwing up r balls is directly proportional to r. If it is given that a particular ball is the first to be thrown up in the air, then what is the probability that all the balls have been thrown up by the machine.
Answer:$\frac{2}{n+1}$
I have no idea on how to even begin solving this problem and any suggestions or solutions would be highly appreciated.
Let one of the colors be red and let $R$ be the event that the first ball is red. Let $n$ be the event that $n$ balls are thrown.
Then the probability the the red ball is first given that all $n$ are thrown up must be $$P(R|n)= \frac{1}{n}$$ by symmetry. Similarly, for any number $1\le j \le n$ of balls thrown, by symmetry the probability the red ball is first must be $1/n$ so $$P(R|i) = \frac{1}{n}$$ for all $i.$
By Bayes, we have $$ P(n|R) = \frac{P(R|n)P(n)}{\sum_{i}P(R|i)P(i)} = \frac{P(n)}{\sum_{i=1}^n P(i)}= P(n).$$
So the probability that there are $n$ balls thrown given that the first is red is the same as the probabilty there are $n$ balls thrown. (Which you can compute by normalizing $P(n)\propto n$ is $2/(n+1)$. To do this set $P(i) = \alpha i$ and fix $\alpha$ so that $\sum_{i=1}^n P(i) = 1$.)
When you think about it, this makes sense. The fact that the red ball is first gives you no additional information about $n$.