More statistics.
Binomial Probability
$$P(r)=\frac{n!}{r!(n-r)!}p^rq^{n-r}=C_{n,r}p^rq^{n-r}$$
The problem states 10% of adults deliberately do a one time fling (purchase clothing, wear 'em to an event, and return 'em).

In a group of 7 adults what is the probability that anyone has done a one time fling?

I figured that:
n=7
r=0

I plugged those into the equation:
$$P(0)=\frac{7!}{0!(7-0)!}p^0q^{7-0}=C_{7,0}p^0q^{7-0}$$

What about p and q?
The probability of success (p) is 0.1 (1% of people do one time flings)?
and q would be 0.9? Would that be right?

Question

More statistics. 
Binomial Probability
$$P(r)=\frac{n!}{r!(n-r)!}p^rq^{n-r}=C_{n,r}p^rq^{n-r}$$
The problem states 10% of adults deliberately do a one time fling (purchase clothing, wear 'em to an event, and return 'em). 

In a group of 7 adults what is the probability that anyone has done a one time fling?

I figured that:
n=7
r=0

I plugged those into the equation: 
$$P(0)=\frac{7!}{0!(7-0)!}p^0q^{7-0}=C_{7,0}p^0q^{7-0}$$

What about p and q? 
The probability of success (p) is 0.1 (1% of people do one time flings)?
and q would be 0.9? Would that be right?

kropot72 · Answer

$$P(0\ out\ of\ 7)=7C0\times 0.1^{0}\times 0.9^{7}=0.9^{7}$$
$$P(at\ least\ 1\ out\ of\ 7)=1.0000-0.9^{7}=\ ?$$

kropot72 · Answer

$$q=(1-p)$$