Question

In a certain city district, the need for money to buy drugs is stated as the reason for 38% of all thefts.

Find the probability that among the next 7 theft cases reported in this district, (a) exactly 3 resulted from the need for money to buy drugs; (b) at most 4 resulted from the need for money to buy drugs.

Answer 1

a)
`P(X=3)=(7C3)(0.38)^3 (1-0.38)^(7-3)=0.28378`

b)
`P(X\leq 4) =P(X=0)+P(X=1)+P(X=2) +P(X=3)+P(X=4)`

`P(X=0)=(7C0)(0.38)^0 (1-0.38)^(7-0)=0.0352`

`P(X=1)=(7C1)(0.38)^1 (1-0.38)^(7-1)=0.1511`

`P(X=2)=(7C2)(0.38)^2 (1-0.38)^(7-2)=0.2778`

`P(X=3)=(7C3)(0.38)^3 (1-0.38)^(7-3)=0.28378`

`P(X=4)=(7C4)(0.38)^4 (1-0.38)^(7-4)=0.17393`

And adding we got:

`P(X \leq 4) = 0.92181`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=7, p=0.38)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

Part a

For this case we want this probability

`P(X=3)`

And using the pmf we got:

`P(X=3)=(7C3)(0.38)^3 (1-0.38)^(7-3)=0.28378`

Part b

`P(X\leq 4) =P(X=0)+P(X=1)+P(X=2) +P(X=3)+P(X=4)`

`P(X=0)=(7C0)(0.38)^0 (1-0.38)^(7-0)=0.0352`

`P(X=1)=(7C1)(0.38)^1 (1-0.38)^(7-1)=0.1511`

`P(X=2)=(7C2)(0.38)^2 (1-0.38)^(7-2)=0.2778`

`P(X=3)=(7C3)(0.38)^3 (1-0.38)^(7-3)=0.28378`

`P(X=4)=(7C4)(0.38)^4 (1-0.38)^(7-4)=0.17393`

And adding we got:

`P(X \leq 4) = 0.92181`