Question

g When a customer places an online order with Amazon, a computerized accounting information system (AIS) automatically checks to see if the customer has exceeded his or her credit limit. Past records indicate that the probability of customers exceeding their credit limit is 0.05. On a given day, 20 customers place orders. Assume that the number of customers that the AIS detects as having exceeded their credit limit is distributed as a binomial random variable. What is the probability that at least two of the customers exceed their limit

Answer 1

Answer:

The probability that at least two of the customers exceed their limit is 0.2642.

Step-by-step explanation:

We are given that Past records indicate that the probability of customers exceeding their credit limit is 0.05.

On a given day, 20 customers place orders.

Let X = the number of customers who exceed their credit limit

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r}; x = 0,1,2,......$

where, n = number of trials (samples) taken = 20 customers

            r = number of success = at least two

            p = probability of success which in our question is the probability 

                  of customers exceeding their credit limit, i.e; 0.05.

So, X ~ Binom(n = 20, p = 0.05)

Now, the probability that at least two of the customers exceed their limit is given by = P(X $\geq$ 2)

        P(X $\geq$ 2) = 1 - P(X < 2) = 1 - P(X = 0) - P(X = 1)

=  $1- \binom{20}{0}\times 0.05^{0} \times (1-0.05)^{20-0}- \binom{20}{1}\times 0.05^{1} \times (1-0.05)^{20-1}$

=  $1- (1 \times 1 \times 0.95^{20})- (20 \times 0.05^{1} \times 0.95^{19})$

=  0.2642