# How to find number of trials in Binomial Distribution?

Suppose we have a biased coin that we are testing that has a probability of getting a tail= 0.6. Say we test the coin. Using this information how do I find the number of trials P(X>=10)<=0.05?

• To clarify, are you looking for the values of $n$ such that $P(X \ge10) \le 0.05$? Nov 4, 2020 at 4:37
• What is X here ? Nov 4, 2020 at 4:40
• @Vikash B X would represent the number of heads (successes) out of n trials Nov 4, 2020 at 4:43

I am interpreting your questions as follows.

You have tossed a bias coin, with $$p=0.6$$, and observed 10 heads. You are interested in determining the number of trials $$n$$ where $$P(X \ge 10) \le 0.05$$

In this case, you essentially want to determine which values of $$n$$ will satisfy:

$$\begin{equation} P(X \ge 10) = \sum_{x=10}^{n} {n \choose x} p^x(1-p)^{n-x} \le 0.05 \end{equation}$$

We can then use R to test out different values of $$n$$, note that $$n\ge10$$ in order for this probability to be defined.

N = 10:15
names(N) = 10:15
prb = unlist(lapply(N, function(x){sum(dbinom(10:x, x, 0.6))}))
prb[prb <= 0.05]
prb


Output:

  10          11
0.006046618 0.030233088


Setting $$n$$ as 10 or 11 will satisfy the above contraint.