Question

A large industrial firm allows a discount on any invoice that is paid within 30 days. of all invoices, 10% receive the discount. in a company audit, 15 invoices are sampled at random. the probability that fewer than 2 of the 10 sampled invoices receive the discount is:

Answer 1

This situation has two outcomes: either an invoice is discounted or not. This two outcomes satisfy what we call a binomial distribution (note "bi" in binomial).

The binomial distribution tells us the probability that a randomly selected sample will have the outcome of success. In this case, we consider receiving the discount as the "success" outcome while not receiving it means "failure". The distribution takes the form:

$P(X=x)= _{n}C_{x} p^{x} q^{n-x}$
where n is the total number of samples, x is the number of samples you'll expect to have a success outcome, p is the probability of success, q is the probability of failure, and nCx is the combination of n samples taken x at a time.

You have an inconsistency regarding the total number of samples by the way, but I will take the first mentioned sample of 15 as I answer (you can just follow through the same process for another value).

For the next step, let's digest the problem to get the needed variables.

The total number of samples, n, is equal to 15. The probability of success (receiving a discount), p, is 10% or 0.1, and the probability of failure (not receiving a discount) is 90% or 0.9.

For P(X=x), we need to find the probability that less than two of the samples have success outcomes therefore we need to find the probability that NO invoice will receive the discount plus the probability that ONE invoice will receive the discount. This is equivalent to saying
$P(X\ \textless \ 2)=P(X=0)+P(X=1)$

Calculating these probabilities we'll get:
$P(X=0)= _{15}C_{0} (0.1)^{0} (0.9)^{15}=0.206$
$P(X=1)= _{15}C_{1} (0.1)^{1} (0.9)^{14}=0.343$
$P(X\ \textless \ 2)=P(X=0)+P(X=1)=0.206+0.343=0.549$

ANSWER: The probability that fewer than 2 sampled invoices will receive the discount is 0.549 or 54.9%.