Question

Past records indicate that 10% of online retail orders are fraudulent. Suppose 20 online orders are placed. What is the probability that none are fraudulent

Answer 1

Answer:

$Probability = 0.1216$

Step-by-step explanation:

Given

$n = 20$

Represent fraudulent with p

$p = 10\%$

$p = 0.10$

Required

Determine the probability that none are fraudulent

First, we need to determine the proportion of those that are not fraudulent.

Represent this with q

$p + q= 1$

$10\% + q = 1$

$q = 1 - 10\%$

$q = 90\%$

$q = 0.9$

The required probability is binomial and can be determine using the following:

$(p + q)^n = ^nC_xp^xq^{n - x}; x = 0,1,2...n$

In this case x = 0 (none);

So, the required probability is:

$Probability = ^nC_0p^0q^{n - 0}$

Substitute values for n, p and q

$Probability = ^{20}C_0 * 0.1^0 * 0.9^{20 - 0}$

$Probability = ^{20}C_0 * 0.1^0 * 0.9^{20}$

$Probability = ^{20}C_0 * 1* 0.9^{20}$

$Probability = \frac{20!}{(20 - 0)!20!} * 1* 0.9^{20}$

$Probability = 1 * 1* 0.9^{20}$

$Probability = 0.9^{20}$

$Probability = 0.12157665459$

$Probability = 0.1216$ Approximated