A company takes out an insurance policy to cover accidents that occur at its manufacturing plant. The probability that one or more accidents will occur during any given month is 3/5. The number of accidents that occur in any given month is independent of the number of accidents that occur in all other months. Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs.
Can anyone please solve this using the N-Binomial formula with X trials needed until the rth success and not with Y failures before rth success since the former is the way i learnt the formula. or both methods if that makes the explanation even clearer. Much thanks in advance.
If we let $X$ represent the $NegBin(r,p)$, where $X$ is the number of trails until the $r$th success, where each success occurs with a probability $p$ , then if I understand your problem correctly, you want the probability $P(X > 4)$, or the probability it takes more than 4 months, for the first ($r = 1$) "success" (which is having a month where an accident occurs), which happens with a probability $p = \frac{3}{5}$.
In this case, we can solve for $P(X > 4)$ in two ways:
You can calculate out the first method by hand and you get $\boxed{\frac{16}{625}}$
By the second method you get the following geometric sum: $(\frac{2}{5})^{4} * (\frac{3}{5}) + (\frac{2}{5})^{5} * (\frac{3}{5}) + (\frac{2}{5})^{6} * (\frac{3}{5}) + ... = \frac{(\frac{2}{5})^{4} * (\frac{3}{5})}{ 1 - \frac{2}{5}} = (\frac{2}{5})^{4} = \boxed{\frac{16}{625}}$
As you can see both methods yield the same answer.