# Number of trials before 3 successes

I have a couple of similar questions, which I think I know how to solve (but wish to see if I am right), and suspect can be solved using a Negative Binomial distribution, but not sure if this is the case and how to define the distribution and it's parameters.

1. We toss a dies until it falls on "6" three times. What is the probability we will need to toss it: a) 4 times b) 5 times

2. Somebody shoots a target until he hits it 4 times. The probability of hitting the target in a single shot is 0.7. What is the probability that he will need: a) 5 shots b) 6 shots

What I though to do, was (if to take the first question as an example), to say that needing 4 tosses means that I didn't get "6" either in my first toss, second, or third. So the probability is: 3*(5/6)(1/6)(1/6)*(1/6). Am I correct ?

• Yes, that is correct. – Empy2 Dec 25 '15 at 14:24
• thank you. How can I get the same number using the negative binomial? – user3275222 Dec 25 '15 at 14:28

Yes, your calculation of the first question is correct as Micheal points out.

In order to apply the negative binomial, you need to first specify the probability model. In the usual scenario of question (1), we use $r$ to represent the total number of failures before the experiment stopped (for this example $r=3$), and $k$ is for the number of success with $p$ as success probability. Thus the model is given by

$P(K=k)=\binom{k+r-1}{k}p^{k}(1-p)^{r}$, for $k=0,1,2,3...$.

For this question, given total number of trials $n=k+r=4$, $r=3$, $p=\frac{5}{6}$, thus $P(K=1)=\binom{3}{1}(\frac{5}{6})^1(\frac{1}{6})^3=3(\frac{5}{6})^1(\frac{1}{6})^3$ which is the answer you have. The beauty of specifying the model lies in simple and automatic calculation. The rest you need to do is just changing the parameter values to get an answer automatically for corresponding problem. All things are programmed in software like $R$.

The usual difficult lies in determining the binomial coefficients but if you learn some combination in prob theory you will be fine. Hope this help a bit.

if the $r_{th}$ success occurs on the $x_{th}$ trial, and the probability of success on an individual trial is $p,$
$Pr = \binom{x-1}{r-1}p^r(1 - p)^{x - r}$
so for 3rd success on 4th trial, $Pr = \binom32(1/6) ^3\cdot(5/6)$