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

Answer:

$P(X\geq 2)=1- P(X$

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

Step-by-step 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:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(X\geq 2)=1- P(X$

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