2
$\begingroup$

Suppose there are $3$ red balls and $1$ blue ball. What's the probability that if I draw a ball $10$ times, I don't draw a blue ball?

Is it $1-\Pr(\text{no red ball}) = 1-\big(\frac{3}{4}\big)^3$?

Also, what is the chance that I draw at least $1$ blue ball within the $10$ draws?

  • 1
    Do you mean 1 - (3/4)^$\color{red}{10}$?2012-08-13

1 Answers 1