I have two bowls A and B.
Bowl A has 6 green balls and 2 black balls.
Bowl B has 4 green balls and 3 black balls.
We roll a dice, If the result is 1. We pull a ball from Bowl A. Else we pull from B.
We repeat this experiment 100 times.
What is the probability of getting at least 1 Black ball.
Is this correct?
$ p = 1-(1 - (\frac{1}{6}*\frac{2}{8} )+1-(\frac{5}{6}*\frac{3}{7}))^{100}$
No.
The probability of "no black ball" for $1$ experiment is: $$P(E)=P(D=1)P(E\mid D=1)+P(D\neq1)P(E\mid D\neq1)=\frac16\times\frac68+\frac56\times\frac47$$(here $E$ denotes the event of "no black ball")
So the probability of "no black ball" for $100$ experiment is: $$\left(\frac16\times\frac68+\frac56\times\frac47\right)^{100}$$ Then the probability of "at least on black ball" for $100$ experiment is: $$1-\left(\frac16\times\frac68+\frac56\times\frac47\right)^{100}$$