I was looking at the solution to this problem below and I don't understand how they used a negative binomial R.V. to solve the problem.

A research study is concerned with the side effects of a new drug. The drug is given to patients, one at a time, until two patients develop side effects. If the probability of getting a side effect from the drug is $1/6$, what is the probability that eight patients are needed?

The answer that is given to the question is below.

Let $Y$ be the number of patients who do not show side effects. Then $Y$ follows a negative binomial distribution with $r=2$, $y=6$ and $p=1/6$. Thus, $$P(Y=6)={(2+6)-1\choose 6}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^6 = 0.0651.$$

Now this is where I'm getting confused. I have my notes on the negative binomial R.V. below:

Suppose that independent trials each having probability $0<p<1$ of being a success are performed until a total of $r$ successes is accumulated. If we let $X$ equal the number of trials required then $$ P(X=n)={n-1\choose r-1}p^r (1-p)^{n-r}.$$

  • Why does it seem like $6$ is used for $r$ in one part of the answer but $2$ is used for $r$ in another part of the answer?
  • If the value for $Y$ that is used in the answer is supposed to be the same as the value for $n$ in my notes why do they add $2$ to $n$?
  • Why is the value for $n-r$ equal to $6$ instead of $4=6-2$?

Any clarifying explanations would be really appreciated.

  • $\begingroup$ Hint: pay attention to the information in the problem, which involves the numbers two and eight (not two and six). $\endgroup$
    – whuber
    Commented Sep 7, 2014 at 17:32

2 Answers 2


You want to know what the probability is that you need to try 8 patients until you have 2 "successes" (patients with side effects). So you want 6 failures (y), and 2 successes (r), for a total of 8 (x) trials, where the success probability (prob. of side effect) is 1/6 (p).

To phrase in the form of your notes:

Suppose that independent trials each having probability $p=1/6$ of having a side effect are performed until a total of $r=2$ side effects is accumulated. If we let X equal the number of trials required then ... use your formula with $n=8$ for $2+6=8$ total trials

The binomial coefficient makes sense if you note that ${n-1\choose r-1}={n-1\choose n-1-(r-1)}={n-1\choose n-r}$, so ${7\choose 1}={7\choose 6}$.


The trick here is that the binomial coefficient has a symmetry property that allows it to be written in two equivalent ways. In this case, the relevant symmetry is that:

$${n-1 \choose r-1} = {n-1 \choose (n-1)-(r-1)} = {n-1 \choose n-r}.$$

The other thing to recognise here is that with $r=2$ "successes" (side effects from drug) and $y=6$ "failures" (no side effects from drug) you have $n=r+y=8$ total patients. If you substitute all these values into the probability formula (but using the alternative expression of the binomial coefficient) then you should get the result stated in your solution.


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