I am having an issue with this question. Usually, I am good at these, but as a class, we were set this as a challenge. It might be easy for you lot but:
2n + 1 counters are in a bag: n are red, the rest are blue. Show that the probability of a different colour is (n + 1 / 2n + 1)
Thanks.
On
I assume that you want to find the probability of choosing $2$ balls of different colour.
In that case. The probability is=$1-[\mathbb{P}(R)+\mathbb{P}(B)]$ Where $\mathbb{P}(R)$=Probability that both balls are Red and $\mathbb{P}(B)$=Probability that both balls are Blue.
Now $\mathbb{P}(R)$=$\frac{n(n-1)}{(2n+1)(2n)}$
Now $\mathbb{P}(B)$=$\frac{n+1)n}{(2n+1)(2n)}$
hence the Answer is $1-\frac{n(n-1)}{(2n+1)(2n)}-\frac{n+1)n}{(2n+1)(2n)}$=$\frac{n+1}{(2n+1)}$
And that's your answer
Perhaps using numbers might help. For example: if there are 14 counters in a bag, 4 are red and the rest are blue. Show the probability of a different color is $\frac{10}{14}$. Are you able to see it now? If choosing a counter is equally likely, then the probability of choosing a blue counter is $\frac{10}{14}$ since there are 10 counters which are equally likely to be selected.
However, instead of the numbers I showed you with, they give you variables. So, $2n+1$ total counters, $n$ of which are red, this implies that $(2n+1)-n$ are the number of blue counters. This yields $n+1$ blue counters. (notice that the total probability will be the number of red counters + the number of blue counters which is $n + (n+1) = 2n+1)$ which is what is stated in the beginning of the problem.
So the probability of choosing blue counter is this number of blue counters/total number of counters which is $\frac{n+1}{2n+1}$