# math

posted by
**ctmorris**
.

A new chewing gum has been developed that is helpful to those who want to stop smoking. If 60 percent of those people chewing the gum are successful in stopping smoking, what is the probability that in a group of four smokers using the gum at least one quits smoking?

The only combination that would not include at least one person being successful is if all are not successful. If they have a .6 success rate, there is a .4 failure rate. Because all probabilities add up to 1.00.

P(person 1 failure) and P(person 2 failure) and P(person 3 failure) and P(person 4 failure)

And in probability means to multiply. What do we do with this?"

The probability of at least one quitting smoking is equal to 1-prob(all smoking)

Pr(all smoking)=.4^4

Pr(at least one quitting smoking)=1-.4^4

= greater than 96 percent (you do the calculation)

0.1041666