Question

If X is a continuous random variable having the Uniform probability distribution between 1 and 5, what is the probability that, of six observations taken of this random variable, three of them will be larger than or equal to 4?

Answer 1

`P(X \geq 4)= 1-P(X<4)= 1-(4-1)/(5-1) = 1-0.75=0.25`

And that would represent the probability of success p =0.25.

`P(Y=3)=(6C3)(0.25)^3 (1-0.25)^(6-3)=0.13184`

Explanation:

For this case we know that X represent a continuous random variable and we know that the distribution for X is given by:

`X \sim Unif (a= 1, b=5)`

And we can begin finsding the following probability for an individual case:

`P(X \geq 4)`

We can use the complement rule and we have this:

`P(X \geq 4)= 1-P(X<4)`

And we can use the cumulative distribution function for the continuous random variable given by:

`F(x) = (x-a)/(b-a) , a\leq X\ leq b`

And replacing we got:

`P(X \geq 4)= 1-P(X<4)= 1-(4-1)/(5-1) = 1-0.75=0.25`

And that would represent the probability of success p =0.25.

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

`Y \sim Binom(n=6, p=0.25)`

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

`P(Y)=(nCy)(p)^y (1-p)^(n-y)`

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

`nCy=(n!)/((n-y)! y!)`

And we want to find this probability:

`P(Y=3)=(6C3)(0.25)^3 (1-0.25)^(6-3)=0.13184`