Question

An individual repeatedly attempts to pass a driving test. Suppose that the probability of passing the test with each attempt is 0.25, and that the results of successive tests are independent. Let X be the number of tests taken until the individual passes (a) Find the probability mass function of X (b) Evaluate the probability of passing the test with three or less attempts. (c) Evaluate the probability of passing the test with five or more attempts.

Answer 1

a) Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

`P(X= k) = (1-p)^(k-1) p , k = 1,2,3,...`

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

`P(X= 1) = (1-0.25)^(1-1) *0.25 = 0.25`

`P(X= 2) = (1-0.25)^(2-1) *0.25 = 0.1875`

`P(X= 3) = (1-0.25)^(3-1) *0.25 = 0.1406`

And adding the values we got:

`P(X \leq 3) =0.25+0.1875+0.1406=0.578`

c)
`P(X \geq 5) = 1-P(X<5) = 1- P(X\leq 4)= 1-[P(X=1) +P(X=2) +P(X=3) +P(X=4)]`

And we can find the individual probabilities:

`P(X= 1) = (1-0.25)^(1-1) *0.25 = 0.25`

`P(X= 2) = (1-0.25)^(2-1) *0.25 = 0.1875`

`P(X= 3) = (1-0.25)^(3-1) *0.25 = 0.1406`

`P(X= 4) = (1-0.25)^(4-1) *0.25 = 0.1055`

`P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316`

Explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

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

Let X the random variable that measures the number of trials until the first success, we know that X follows this distribution:

`X\sim Geo (1-p)`

Part a

Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

`P(X= k) = (1-p)^(k-1) p , k = 1,2,3,...`

Part b

We want this probability:

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

We find the individual probabilities like this:

`P(X= 1) = (1-0.25)^(1-1) *0.25 = 0.25`

`P(X= 2) = (1-0.25)^(2-1) *0.25 = 0.1875`

`P(X= 3) = (1-0.25)^(3-1) *0.25 = 0.1406`

And adding the values we got:

`P(X \leq 3) =0.25+0.1875+0.1406=0.578`

Part c

For this case we want this probability:

`P(X \geq 5)`

And we can use the complement rule like this:

`P(X \geq 5) = 1-P(X<5) = 1- P(X\leq 4)= 1-[P(X=1) +P(X=2) +P(X=3) +P(X=4)]`

And we can find the individual probabilities:

`P(X= 1) = (1-0.25)^(1-1) *0.25 = 0.25`

`P(X= 2) = (1-0.25)^(2-1) *0.25 = 0.1875`

`P(X= 3) = (1-0.25)^(3-1) *0.25 = 0.1406`

`P(X= 4) = (1-0.25)^(4-1) *0.25 = 0.1055`

`P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316`