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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?

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

1 Answer 1

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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.

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