Question

A nationwide survey of seniors by the University of Michigan reveals that almost 18.0% disapprove of daily pot smoking. If 8 seniors are selected at random, what is the probability that at least 2 disapprove of daily pot smoking.

Answer 1

`P(X\geq 2)=1- P(X<2)= 1-[P(X=0) +P(X=1)]`

And using the probability mass function we can find the individual probabilities:

`P(X=0)=(8C0)(0.18)^0 (1-0.18)^(8-0)=0.2044`

`P(X=1)=(8C1)(0.18)^1 (1-0.18)^(0-1)=0.3590`

And replacing we got:

`P(X\geq 2)=1 -[0.2044 +0.3590]= 0.4366`

Then the probability that at least 2 disapprove of daily pot smoking is 0.4366

Explanation:

Let X the random variable of interest "number of seniors who disapprove of daily smoking ", on this case we now that:

`X \sim Binom(n=8, p=0.18)`

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 to find this probability:

`P(X\geq 2)=1- P(X<2)= 1-[P(X=0) +P(X=1)]`

And using the probability mass function we can find the individual probabilities:

`P(X=0)=(8C0)(0.18)^0 (1-0.18)^(8-0)=0.2044`

`P(X=1)=(8C1)(0.18)^1 (1-0.18)^(0-1)=0.3590`

And replacing we got:

`P(X\geq 2)=1 -[0.2044 +0.3590]= 0.4366`

Then the probability that at least 2 disapprove of daily pot smoking is 0.4366