Question 1054322: Although 49 states prohibit the sale of cigarettes to minors, one study found that 63% of underage children could buy them. Purchasers by children account for $221 million in annual profit for the tobacco industry. If seven underage children are attempting to buy cigarettes, what is the probability that all seven are successful in buying cigarettes
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 63% of underage children can buy cigarettes, so the probability that any one underage child, chosen at random, will be able to purchase cigarettes is equal to .63.
the probability that the child will not be able to purchase cigarettes is equal to 1 - .63 = .37
you have:
p = .63
q = 1 - p which makes q = .37
you have 7 children.
this makes n = 7
this is a binomial probability question.
the formula for binomial probability is p(x) = c(n,x) * p^x * q^(n-x).
you have:
n = 7
p = .63
q = .37
the probability that all 7 will be able to buy cigarettes makes:
x = 7
the formula becomes p(7) = c(7,7) * .63^7 * .37^0 which is equal to .0393898064.
that'a approximately 3.94%.
the sum of all possible probabilities is equal to 1.
the following table shows you that.
c(n,x) is the combination formula of n! / (x! * (n-x)!).
this tells you the number of ways you can get x things out of a total of n things when order is not important.
when n = 7 and x = 7, the formula becomes c(7,7) = 7! / (7! * 0!) which becomes c(7,7) = 7! / 7! which makes c(7,7) = 1.
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