Suppose a bag has $3$ red balls, $3$ green balls and $3$ blue balls. If $2$ balls are picked at random without replacement, what is the probability that they are the same colour?
My solution is in $2$ distinct ways:
- For the first ball I have $9$ choices, while for second ball I have $8 $ choices. So $2$ balls can be picked in $9 \times 8 = 72$ ways, but since here the order of the result doesn't matter, the sample space only has $\frac{72}{2} = 36$ outcomes. Now let $A$ be the event "I pick $2$ red balls". This can be done in $3 \times 2$ ways but because order doesn't matter, in fact only $3\choose 2$ outcomes correspond to $A$. Similarly, I have $3\choose 2$ outcomes for picking $2$ blue balls, and $3\choose 2$ outcomes for picking $2$ green balls.
Therefore the required probability $P$ is $P = 3 \times \frac{3\choose 2}{36} = \frac{1}{4}$.
- My second argument is simply that: suppose the first ball is drawn, of any colour. Then the probability of picking a second ball the same colour as first must be $\frac{2}{8}$, since no matter what colour was picked first, there are only $2$ balls remaining of that same colour, out of a total of $8$. Hence we again get the required probability $P = \frac{1}{4}$.
Are both of my solutions entirely correct?