Question

About 1111​% of the population of a large country is math phobicmath phobic. If two people are randomly​ selected, what is the probability both are math phobicmath phobic​? What is the probability at least one is math phobicmath phobic​?

Answer 1

a)
`P(X=2) = (2C2) (0.11)^2 (1-0.11)^(2-2)= 0.0121`

b)
`P(X=0) = (2C0) (0.11)^0 (1-0.11)^(2-0)= 0.7921`

`P(X \geq 1)= 1-P(X<1) = 1- P(X=0)=1-0.7921= 0.2079`

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

Part a

Let X the random variable "number of people math phobicmath phobic" , on this case we now that the distribution of the random variable is:

`X \sim Binom(n=2, p=0.11)`

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!)`

And we want this probability:

`P(X=2)`

If we use the probability mass function we got:

`P(X=2) = (2C2) (0.11)^2 (1-0.11)^(2-2)= 0.0121`

Part b

For this case we want this probability:

`P(X \geq 1)= 1-P(X<1) = 1- P(X=0)`

We can find first this probability

`P(X=0) = (2C0) (0.11)^0 (1-0.11)^(2-0)= 0.7921`

`P(X \geq 1)= 1-P(X<1) = 1- P(X=0)=1-0.7921= 0.2079`