# What is the probability of drawing more than $15$ red marbles?

Question: An urn contains $2$ white marbles and $8$ red marbles. A marble is drawn $20$ times in succession with replacement. What is the probability of drawing more than $15$ red marbles?

I had a problem just like this a while back (posted on here) but the marbles was drawn randomly in succession though. So I am stuck on this because it is not random. Also I was told to use a calculator on this so my answer can be accurate without mistakes. Thing is, I need help using the binomial distribution on my calculator (TI-$84$ plus silver edition). Need help.

Actually, there is no need to be confused. In succession just means one by one or one after another, and we are told it's with replacement. They are still drawn randomly. Does that clarify your confusion?

To find the binomial coefficient ${_nC_r} = \binom{n}{r} = \frac{n!}{r!(n-r)!}$ on the TIs, you usually go to an option called "PRB" and it should be something like ${_nC_r}$ or $C(n,r)$.

Here is one guide I found on the TI-84 PLUS.

• my $n$=$20$ and $r$=$16$? – Lady T May 4 '17 at 5:13
• Yes, except that will just give you the coefficient for $P(X = 16)$. You still need to add up 17 through 20 – Em. May 4 '17 at 5:15
• too add it up, it should be $\frac{16}{20}$ for $X$=$16$? and so forth – Lady T May 4 '17 at 5:21
• Not completely. Maybe you remember from you "previous question" but $P(X = 16) = \binom{20}{16} (8/10)^{16}(2/10)^{20-16}$, from the binomial distribution. – Em. May 4 '17 at 5:25
• Sure thing. Good luck. – Em. May 4 '17 at 5:35

The number of red marbles, call it $X$, follows a $\text{Binomial}(20, 8/10)$ distribution. (Why?)

The probability you seek can be computed as $$P(X > 15) = P(X=16) + P(X=17) + P(X=18) + P(X=19)+P(X=20).$$ Use the definition of the binomial distribution to compute each term on the right-hand side.