# 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

• PS: The count of red marbles follows a Binomial Distribution. Have you studied those yet? – Graham Kemp Mar 22 '17 at 1:30
• Just got to it today. Still learning the material – Lady T Mar 22 '17 at 1:33

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 :

$$\Pr(X>7) ~=~ \Pr(X=8)+\Pr(X=9)+\Pr(X=10)$$

• would $k$=$8$ based on this formula? – Lady T Mar 22 '17 at 1:42
• Yes, just substitute the appropriate values. – Graham Kemp Mar 22 '17 at 2:02
• $1$ more question. Is binomial distribution the same as binomial experiment? – Lady T Mar 22 '17 at 2:22
• @LadyT A binomial distribution is that of the count of successes among a sequence of independent Bernoulli trials, each with identical success rate. We write that " $X$ is binomially distributed with parameters of amount $n$, and rate $p$ " as : $X\sim \mathcal{Bin}(n,p)$ . – Graham Kemp Mar 22 '17 at 2:30

You are correct, that the probabilities do not change from trial to trial.

There are three scenarios : 0,1 or 2 white balls.

Picking $0$ white balls happens with probability $\left(\frac{8}{10}\right)^{10}$.

Picking $1$ white ball happens with probability $10 \times \frac{2}{10} \left(\frac{8}{10}\right)^{9}$, since the trial on which the white ball is picked can be chosen in $10$ ways.

Picking $2$ white ball happens with probability $\binom{10}2 \times \left(\frac{2}{10}\right)^2 \left(\frac{8}{10}\right)^{8}$, since the trials on which the white balls can be picked , can now be chosen in $\binom{10}{2}$ ways.

Hence, the answer is the sum of these i.e. $\binom{10}2 \times \left(\frac{2}{10}\right)^2 \left(\frac{8}{10}\right)^{8} + 10 \times \frac{2}{10} \left(\frac{8}{10}\right)^{9} + \left(\frac{8}{10}\right)^{10}$.

• why when we choose $0$ the probability is $\frac{8}{10}$ – Lady T Mar 22 '17 at 1:32
• Is it because we must have more than $7$ – Lady T Mar 22 '17 at 1:33
• @LadyT If there are no white balls, then all the picked balls must have been red. The probability that a ball is red is $\frac 8{10}$, and since there are ten trials, we have to take the tenth power. – астон вілла олоф мэллбэрг Mar 22 '17 at 1:35