Question

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a given day, 20 online retail orders are placed. Assume that the number of online retail orders that turn out to be fraudulent is distributed as a binomial random variable, what is the probability that two or more online retail orders will turn out to be fraudulent

Answer 1

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

`P(x=k)=\dbinom{n}{k}p^k(1-p)^(n-k)=\dbinom{20}{k}\cdot0.08^k\cdot0.92^(20-k)`

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

`P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^(0)\cdot0.92^(20)=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^(1)\cdot0.92^(19)=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483`

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.