Question

In a certain city district, the need for money to buy drugs is stated as the reason for 75% of all thefts. Find the probability that among the next 5 theft cases reported in this district, (a) exactly 2 resulted from the need for money to buy drugs; (b) at most 3 resulted from the need for money to buy drugs.

Answer 1

Answer:

a) 0.0879 = 8.79% probability that among the next 5 theft cases reported in this district, exactly 2 resulted from the need for money to buy drugs.

b) 0.3672 = 36.72% probability that among the next 5 theft cases reported in this district, at most 3 resulted from the need for money to buy drugs.

Step-by-step explanation:

For each theft, there are only two possible outcomes. Either it was motivated by a need to buy drugs, or it wasnt. Thefts are independent, which 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.

In a certain city district, the need for money to buy drugs is stated as the reason for 75% of all thefts.

This means that $p = 0.75$

Next 5 theft cases

This means that $n = 5$

(a) exactly 2 resulted from the need for money to buy drugs;

This is P(X = 2).

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

$P(X = 2) = C_{5,2}.(0.75)^{2}.(0.25)^{3} = 0.0879$

0.0879 = 8.79% probability that among the next 5 theft cases reported in this district, exactly 2 resulted from the need for money to buy drugs.

(b) at most 3 resulted from the need for money to buy drugs.

This is

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

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

$P(X = 0) = C_{5,0}.(0.75)^{0}.(0.25)^{5} = 0.0010$

$P(X = 1) = C_{5,1}.(0.75)^{1}.(0.25)^{4} = 0.0146$

$P(X = 2) = C_{5,2}.(0.75)^{2}.(0.25)^{3} = 0.0879$

$P(X = 3) = C_{5,3}.(0.75)^{3}.(0.25)^{2} = 0.2637$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0010 + 0.0146 + 0.0879 + 0.2637 = 0.3672$

0.3672 = 36.72% probability that among the next 5 theft cases reported in this district, at most 3 resulted from the need for money to buy drugs.