Question

It is known that diskettes produced by a cer- tain company will be defective with probability .01, independently of each other. 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 moreIf someone buys 3 packages, what is the probability that he or she will return exactly 1 of them?

Answer 1

Answer:

1.27%

Step-by-step explanation:

To solve this problem, we may consider a binomial distribution where a customer can either accept or reject (and return) the diskette package.

Lets consider  some aspects:

1. From the formulation of the exercise we know that a package is accepted if it has at most 1 defective diskette. So our event A is defined as:

A = 0 or 1 defective diskette

2. The probability of a diskette being defective is 0.01

3. Each package contains 10 diskettes.

If X is defined as number of defective diskettes in the package, the probability of X is given by a binomial distribution with probability 0.01 and n=10

X ~ Bin(p=0.01, n=10)

Let us remember the calculation of probability for the binomial distribution:

$P(X=x)=nCx*p^{x}*(1-p)^{(n-x)}$ with x = 0, 1, 2, 3,…, n

Where

n: number of independent trials

p: success probability  

x: number of successes in n trials

In our case success means finding a defective diskette, therefore

n=10

p=0.01

And for x we just need 0 or 1 defective diskette to reject the package

Hence,

$P(X=x)=10Cx*0.01^{x}*(1-0.01)^{(10-x)}$ with x = 0, 1

So,

$P(A)=P(X=0)+P(X=1)$

$P(A)=10C0*0.01^{0}*(1-0.01)^{(10-0)} + 10C1*0.01^{1}*(1-0.01)^{(9)}$

$P(A)=0.99^{10}+10*0.01*0.99^{9}$

$P(A)=0.9957$

Now, because we have 3 packages and we might reject just 1 of them, we can find this probability like this:

$3*(1-P(A))*P(A)*P(A) = (1-0.9957)*0.9957*0.9957=0.0127$

Finally, we have that the probability of returning exactly one of the three packages is 1.27%