Apple's Binomially Distributed? Here is the question:
"Apples are packaged automatically in 3-pound bags. Suppose that 4% of the time the bag of apples weighs less than 3 pounds. If you select bags randomly and weigh them in order to discover one underweight bag of apples, find the probability that the number off bags that must be selected is at least 20."
I have tried to use the binomial distribution to approach this problem, but then I would find myself trying to do 20 different calculation for x. The reason being because I am thinking of this: $\binom{x}{1}(.04)^{1}(.96)^{x-1} $ what is the easier way of doing this, or am I wrong altogether?
 A: To get at least 20 bags selected, this means all of the first 19 must be of weight more than three pounds, which has probability $0.96$ per bag. So ...
A: It is not binomial, because you don't have a fixed number of weighings. You weigh the bags till you find the underweight bag. This is a geometric distribution. Let $X$ be the number of weighings required. What is the probability that $X = k$?
Let's see... $X = 1$ if you find the underweight bag in the first weighing itself (and at least one weighing is required, so this is the minimum possible value of $X$). In the first weighing, the probability of getting the underweight bag is $0.04$. So $P(X = 1) = 0.04$.
What is the probability of $X = 2$? Well, $X = 2$ only if you did not find the underweight bag in the first weighing (which has probability $0.96$), and then in the second one, you did (which has probability $0.04$). So $P(X = 2) = 0.96 \times 0.04$.
Similarly, $P(X = 3) = (0.96)^2 0.04$ - two failures and (finally) one success.
In general, $P(X = k) = (0.96)^{k - 1}0.04$.
Now, you need the probability that the number of weighings required to find the underweight bag is at least $20$. That is
$
\begin{align}
P(X \ge 20) & = 1 - P(X < 20)\\
& = 1 - [P(X = 0) + \cdots + P(X = 19)]\\
& = 1 - [0.04 + (0.96)0.04 + (0.96)^2 0.04 + \cdots + (0.96)^{18} 0.04]\\
& = 1 - 0.04\dfrac{1 - (0.96)^{19}}{1 - 0.96}\\
& = 1 - [1 - (0.96)^{19}]
\end{align}\\
\boxed{P(X \ge 20) = (0.96)^{19}}
$
Or
$\begin{align}
P(X \ge 20) & = P(X = 20) + P(X = 21) + \cdots\\
& = (0.96)^{19} 0.04 + (0.96)^{20} 0.04 + \cdots\\
& = \dfrac{(0.96)^{19} 0.04}{1 - 0.04}
\end{align}\\
\boxed{P(X \ge 20) = (0.96)^{19}}
$
Or
At least $20$ weighings are required if and only if the first $19$ weighings failed to find the underweight bag. This has probability $\boxed{(0.96)^{19}}$.
Note: This is the shortest solution, but also the least general. It is more important to know the geometric distribution, and understand when to apply it instead of the binomial distribution.
A: The distribution is not binomial.  A binomial distribution requires the following model assumptions:


*

*Each trial is a dichotomous outcome ("yes"/"no," "success"/"failure," 0/1, etc).

*The outcome of each trial is independent of any other trial.

*Each trial has the same probability distribution for the two possible outcomes.

*The number of trials is some fixed parameter $n$.

*We are interested in counting the number of "successes" out of $n$ trials.


Your question does not satisfy requirements 4 and 5, because you are interested in the number of trials needed to observe some fixed number of successes, not the number of successes in a fixed number of trials.
In such a case, you are looking at a negative binomial distribution.  A special case of this is when you want the number of trials you need to observe until the first success, in which case the distribution is geometric.
If $N$ counts the random number of total trials needed to observe the first underweight bag, then $N$ is geometric with probability mass function $$\Pr[N = k] = p(1-p)^{k-1},$$ where $p$ is the probability of observing an underweight bag for any single trial.  This makes sense, because in order to require exactly $k$ trials, you must observe $k-1$ bags that are not underweight, and then the final bag you observe is underweight.  Since each trial is independent and identically distributed, and the probability of being underweight is $p$, and the probability of not being underweight is $1-p$, we easily get the above result.  Then the question asks for $$\Pr[N \ge 20] = \sum_{k=20}^\infty (0.04)(0.96)^{k-1},$$ which I leave to you to compute.
