Question

According to national data, 5.1% of burglaries are cleared with arrests. A new detective is assigned to six different burglaries. What is the probability that at least one of them is cleared with an arrest

Answer 1

Answer:

26.95% probability that at least one of them is cleared with an arrest

Step-by-step explanation:

For each burglary, there are only two possible outcomes. Either it is cleared, or it is not. The probability of a burglary being cleared is independent of other burglaries. So 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.

5.1% of burglaries are cleared with arrests.

This means that $p = 0.051$

A new detective is assigned to six different burglaries.

This means that $n = 6$

What is the probability that at least one of them is cleared with an arrest?

Either none are cleared, or at least one is. The sum of the probabilities of these events is 100% = 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$

Then

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

In which

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

$P(X = 0) = C_{6,0}.(0.051)^{0}.(0.949)^{6} = 0.7305$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7305 = 0.2695$

26.95% probability that at least one of them is cleared with an arrest