Suppose we have $4$ red $7$ green and $4$ yellow balls. Draw repeatedly at random and with replacement. What is the probability of drawing a green ball before seeing any red ball?
I did it as follows:
Let $A_n$ be the event that you draw at least one green ball and no red balls for $n-1$ draws, before finally seeing a red ball on draw $n$. Then to calculate $P(A_n)$ it is equal to $1 – $ the probability that you see no green balls and no red balls, times $(\frac 4 {15})$ the probability of seeing a red ball.
I calculated the probability that you see no green balls and no red balls as $\frac {4} {15}$ – that is the probability of drawing only yellow balls.
Hence, the probability $A_n$ is given as $(\frac {11} {15})^{n-1} \cdot \frac 4 {15}$.
Then we sum from $n = 1$ to $\infty$ and find that the solution is $\frac 4 {15} \cdot \frac {15} 4$…which is $1$, clearly wrong. Which part in the solution went wrong?
Best Answer
Your probability for $A_n$ is wrong. $$\left(\dfrac{11}{15}\right)^{n-1}$$ could be the probability that you receive no red balls, but it is not the probability that you also receive at least one green ball.
Let $B_n$ be the event that you draw no red balls in the first $n-1$ draws. Let $C_n$ be the event that you draw at least one green ball in the first $n-1$ draws.
These are not independent events. So, $$P(B_n\cap C_n) \neq P(B_n)P(C_n)$$
This probability would be difficult to calculate. It is certainly possible, but not really necessary. An easier way to calculate your original probability:
You draw a green ball on the first draw. Or you draw a yellow ball followed by a green. Or you draw a yellow, then a yellow, followed by a green. Etc.
Let $P$ be the probability you are looking for.
$$P = \dfrac{7}{15}\sum_{n=0}^\infty \left(\dfrac{4}{15}\right)^n = \dfrac{7}{11}$$
It does not matter what comes after you draw the first green. You know that you drew it before you ever draw a red. By the law of large numbers, if you kept drawing, you would eventually draw a red.