Question

Suppose apples are rotten with probability 0.01, independently of one another. Raider Joes sells the apples in bags of 10 and offers a money-back guarantee that at most 1 of the 10 apples in the bag will be rotten. The guarantee is that a customer can return an entire bag of apples if he or she finds more than 1 rotten apple in it. If someone buys 3 bags, what is the probability that he or she will return exactly 1 of the bags?

Answer 1

Answer:

There is a 1.25% probability that he or she will return exactly 1 of the bags.

Step-by-step explanation:

The first step to solve this problem, is finding the probability that a single bag is returned.

For each apple, there are only two possible outcomes. Either it is rotten, or it is not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Raider Joes sells the apples in bags of 10 and offers a money-back guarantee that at most 1 of the 10 apples in the bag will be rotten. The guarantee is that a customer can return an entire bag of apples if he or she finds more than 1 rotten apple in it.

There are 10 apples, so $n = 10$

Suppose apples are rotten with probability 0.01, so $p = 0.01$

The probability that somene will return a bag is $P(X > 1)$

Either the number of rotten apples is one or less, or it is larger than 1. The sum of the probabilities of each case is decimal 1. So:

$P(X \leq 1) + P(X > 1) = 1$

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.01)^{0}.(0.99)^{10} = 0.9044$

$P(X = 1) = C_{10,1}.(0.01)^{1}.(0.99)^{9} = 0.0914$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.9044 + 0.0914 = 0.9958$

So, the probability that one bag has to be returned is

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.9958 = 0.0042$

If someone buys 3 bags, what is the probability that he or she will return exactly 1 of the bags?

Now we have $n = 3$ and $p = 0.0042$.

We have to find $P(X = 1)$.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{3,1}.(0.0042)^{1}.(0.9958)^{2} = 0.0125$

There is a 1.25% probability that he or she will return exactly 1 of the bags.