SOLUTION: On average, 40 percent of U.S. beer drinkers order light beer. (a) What is the probability that none of the next eight customers who order beer will order light beer? (b) That

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: On average, 40 percent of U.S. beer drinkers order light beer. (a) What is the probability that none of the next eight customers who order beer will order light beer? (b) That      Log On


   

Question 121859: On average, 40 percent of U.S. beer drinkers order light beer.
(a) What is the probability that none
of the next eight customers who order beer will order light beer?
(b) That one customer will?
(c) Two customers?
(d) Fewer than three?
(e) Construct the probability distribution
Please kindly help. Thank you
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
The probability that a single customer orders light beer is 0.4, therefore the probability that the customer orders something else is 1 - 0.4 or 0.6.

a) What the previous customer orders has no relation to what the current customer orders, so each of the orders of the next eight customers is an independent event. Hence the total probability is the product of the individual probabilities:

0.6%5E8=0.017

b) The probability that one customer will order light beer is again, 0.4. This situation requires that we consider the probability that 7 customers don't order light beer times the probability that one does or:

%280.6%5E7%29%280.4%29=.011

c) Similar to part b:

%280.6%5E6%29%280.4%5E2%29=.007

d) Fewer than three. This means that either zero, 1, or 2 order light beer, and the remainder do not. So the first 6 must order other than light beer, and the other two customers can order either light beer or not. The probability that a given customer will either order light beer or something else is 1 (0.4 + 0.6), so:

%280.6%5E6%29%281%5E2%29=0.47

e) Probability distribution. Not exactly sure what you want here, but here is a stab at it: