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:
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,
with x = 0, 1
So,
Now, because we have 3 packages and we might reject just 1 of them, we can find this probability like this:
Finally, we have that the probability of returning exactly one of the three packages is 1.27%
