Q:

Finding the probability of drawing more than $7$ red marbles

Question: A bucket contains $2$ white and $8$ red marbles. A marble is drawn randomly $10$ times in succession with replacement. Find the probability of drawing more than $7$ red marbles? I think since the marbles are replaced, the probability of selecting a red marble does not change from trial to trail. Am I right on this assumption? Also I think if I calculate the probability of selecting $0,1$ or $2$ white marbles, I can get an answer but I do not know how to approach this.  Need help with this

A:

Yes, you are right in that assumption.   That means the count of red marbles drawn among the ten has a Binomial Distribution.
You should know the probability mass formula for $X\sim\mathcal{Bin}(n,p)$: $$~\Pr(X=k) ~=~ \dbinom {n}k p^k(1-p)^{n-k}\qquad\Big[k\in\{0,..,n\}\Big]$$
In this case, $n=10$, $p=8/10$
The probability of drawing more than seven red marbles, is :
$$\Pr(X>7) ~=~ \Pr(X=8)+\Pr(X=9)+\Pr(X=10)$$

