Question

A manager randomly selected 25 large cash transactions at a bank that were made in January. The manager then carefully tracked down the details of these transactions to see that the correct procedures for reporting these large transactions were followed. If the chance for a procedural error is 10%, what is the probability that the manager finds more than two such transactions

Answer 1

Answer:

0.4629 = 46.29% probability that the manager finds more than two such transactions

Step-by-step explanation:

For each transaction, there are only two possible outcomes. Either there is a procedural error, or there is not. The probability of a procedural error on a transaction is independent of any other transaction. This means that we use the binomial probability distribution to solve this question.

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 combinations 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.

A manager randomly selected 25 large cash transactions at a bank that were made in January.

This means that $n = 25$

The chance for a procedural error is 10%

This means that $p = 0.1$

What is the probability that the manager finds more than two such transactions?

This is:

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

In which

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{25,0}.(0.1)^{0}.(0.9)^{25} = 0.0718$

$P(X = 1) = C_{25,1}.(0.1)^{1}.(0.9)^{24} = 0.1994$

$P(X = 2) = C_{25,2}.(0.1)^{2}.(0.9)^{23} = 0.2659$

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0718 + 0.1994 + 0.2659 = 0.5371$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5371 = 0.4629$

0.4629 = 46.29% probability that the manager finds more than two such transactions