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?
Best Answer
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.