Question

In a certain city district the need for money to buy drugs is stated as the reason for 70% of all thefts. Find the probability that among the next 7 theft cases reported in this district, exactly 3 of them resulted from the need to buy drugs.

Answer 1

Answer:

9.72% probability that among the next 7 theft cases reported in this district, exactly 3 of them resulted from the need to buy drugs.

Step-by-step explanation:

For each theft, there are only two possible outcomes. Either the need to buy drugs is the reason of the theft, or it is not. Each theft is independent of each other. 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.

In a certain city district the need for money to buy drugs is stated as the reason for 70% of all thefts.

This means that $p = 0.7$

Find the probability that among the next 7 theft cases reported in this district, exactly 3 of them resulted from the need to buy drugs.

This is P(X = 3) when n = 7. So

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

$P(X = 3) = C_{7,3}.(0.7)^{3}.(0.3)^{4} = 0.0972$

9.72% probability that among the next 7 theft cases reported in this district, exactly 3 of them resulted from the need to buy drugs.