# Solving for N in a binomial distribution

Mid-term study...

Two dice are rolled. How many times must the dice be rolled so that the probability of getting a sum of 10 or greater on at least one roll is larger than 0.9?

So am I correct in thinking that it would be: $$\sum_{j=1}^{n}\sum_{k=1}^{j}{n \choose k}P^k(1-P)^{j-k} \geq .9$$

Given that the probability of getting a sum of 10+ is $\frac{6}{36}$

I get:

$$\sum_{j=1}^{n}\sum_{k=1}^{j}{n \choose k} \left( \frac{6}{36}\right)^k \left( \frac{30}{36}\right)^{j-k} \geq .9$$

Am I right in my formula? If not, what should it be? And then how do I solve for N?

Thanks.

EDIT / RESPONSE:

What I wrote above is the same as saying

$${n \choose 0}P^0(1-P)^n < .1$$

The first parts are both just 1, which plugging in numbers leaves me with

$$\left(\frac{30}{36} \right)^n < .1$$

To solve, I just need to do the following

$$\log_{\frac{30}{36}}{\left(\frac{30}{36} \right)^n} < log_{\frac{30}{36}}{(.1)}$$

Which according to my calculations should be 12.6, which means 13.

Hint:

Translate into: how many times the dice must roll so that the probability of getting a sum of $10$ or greater none of the times is less than $0.1$?

edit:

Let $p$ denote the probability that by rolling $2$ dice we will not get a sum of $\geq10$.

This $p$ is easy to find. You can do it.

Then $p^n$ is the probability that by $n$ rollings it does not happen that we meet a sum of $\geq 10$.

Then the question is: what is the minimal $n\in\mathbb N$ with $p^n<0.1$?

You can use logarithms to find this $n$.

• Any chance you could show me the process? My brain is getting a little fried, and if I don't learn from your work, I'll just hurt myself when I take the test. Or at least get me started? Commented Oct 31, 2014 at 10:15
• Do you agree that the question in my answer is exactly the same as the question in your question? The events: 'it happens at least once' and 'it happens none of the times' are complements of each other. Commented Oct 31, 2014 at 10:18
• Yes, I'm just not even sure how to reverse it. And then there's still the solving for n when there's a factorial involved that I don't know how to do. Commented Oct 31, 2014 at 10:19
• I have edited something to explain. Commented Oct 31, 2014 at 10:27
• Counting $\left(5,5\right)$ twice is wrong! The probability that both dice give a $5$ is $\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$. Just like the probability that the first dice gives a $4$ and the second a $6$. Commented Oct 31, 2014 at 11:12

So am I correct in thinking that it would be

$$\sum_{j=1}^{n}\sum_{k=1}^{j}{n \choose k}P^k(1-P)^{j-k} \geq .9$$

No, not quite.

Rolling two die such that their sum is less than 10 is a trial with probability of success $p$.

The number of successes, $N_n$, in a series of $n$ trials has a binomial probability distribution. $N_n\sim \mathcal{Bin}(n, p)$.

You simply want to find the number of trials, $n$, such that $\mathsf P(N_n\geq 1)\geq 0.9$. That is:

$$\sum_{k=1}^n {n\choose k}p^k(1-p)^{n-k}\geq 0.9$$

However the probability of the complement event is easier to much calculate, so find $n$ such that: $\mathsf P(N_n=0) \leq 0.1$. That is:

$$(1-p)^n \leq 0.1$$