Question

It is known that diskettes produced by a certain company will be defective with probability .$01$, independently of one another. The company sells the diskettes in packages of size $10$and offers a money-back guarantee that at most$1$of the diskettes in the package will be defective. The guarantee is that the customer can return the entire package of 10 diskettes if he or she finds more than 1 defective diskette in it. If someone buys 3 packages, what is the probability that he or she will return exactly 1 of them?

Short answer

The probability that someone returns$1$ package of $3$bought is$0.01278$.

Step-by-step solution

Step 1:Given information

It is known that diskettes produced by a certain company will be defective with probability .$01$, independently of one another. The company sells the diskettes in packages of size $10$ and offers a money-back guarantee that at most $1$of the $10$ diskettes in the package will be defective. The guarantee is that the customer can return the entire package of diskettes if he or she finds more than $1$defective diskette in it

Explanation

Diskettes created by a certain company will be defective with a probability $0.01$(independently of one another). Packages selling includes $10$diskettes. The company suggests a money-back guarantee that at most$1$out of $10$diskettes is defective. If someone purchases $3$packages, we want to estimate the probability that he will return precisely $1$package. First, let us calculate the probability that $1$package is returned. We have:

$\mathrm{\mathbb{P}}(X\ge 2)=1-\mathrm{\mathbb{P}}(X\le 1)$

$=1-\sum _{k=0}^1\left(\begin{array}{c}10\\ k\end{array}\right)·0.{01}^k·0.{99}^{10-k}$

$=1-0.{99}^{10}-10·0.01·0.{99}^9$

$=0.0043$

Step 3:Explanation

It remains to calculate the probability that exactly one package is returned. Let $A$be event that package is returned, obviously $\mathrm{\mathbb{P}}\left(A\right)=0.0043$. Thus it follows:

$\mathrm{\mathbb{P}}\left(B\right)=\left(\begin{array}{l}3\\ 1\end{array}\right)·\mathrm{\mathbb{P}}\left(A\right)·(1-\mathrm{\mathbb{P}}(A){)}^2$

$=3·0.0043·(1-0.0043{)}^2$

$=0.01278$

Therefore the probability of this event is $0.01278.$

Step 4:Final information

The probability that someone returns$1$package of $3$bought is $0.01278.$