SOLUTION: I APPRECIATE ANY HELP AT ALL!!! I just don't know what to do to start with... Each week you buy a cereal box of a certain company that costs 5�. The company advertises that

Question 868984: I APPRECIATE ANY HELP AT ALL!!! I just don't know what to do to start with...
Each week you buy a cereal box of a certain company that costs 5�.

The company advertises that inside some "lucky" boxes, a "gift's coupon" of value of 10� is included.

It is given that 1% of the boxes of the company includes a gift's coupon.

a) How much is your average profit in the first 100 weeks of purchase?

b) How many boxes on average you must buy until you find one box with a gift's coupon? How much money will you have spent on average?

c) Find the mininmum number n, of weeks for which you must keep on buying boxes so that you find a coupon in that time with a probability >= 50%.

d) Let's say that n-1 weeks have passed and you have not found yet a coupon, what is the probability that you will find one at the next week (n is the number of weeks that you counted on previous question).

e) Count the probability that you have found exactly 2 coupons in 100 weeks, using the distribution of Poisson.

f) Count the probability that you have NOT found any coupon in 100 weeks.
Found 2 solutions by ewatrrr, ScientificArt:
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

Re TY, Yes, statistically (and in life), expected value(profit) can be a negative number.
Had .1% in my head somehow
My thoughts...using .01
p(success) = .01 0r 1%
a) E = = -485.05� More politically correct (on the win: net 5�)
b) m = np 0r p = m/n , .01 = 1/100, n = 100, m = 1
On an average 100 boxes to win one (Cost for the Boxes 500� ): Net cost: 490�(with a win)
c)P(X ≥ 1) = 1 - binomcdf(69, .01, 0) = .5002, min 69
d) P = 1C1(.01)^1(.999)^0 = .01
e)average 1/100 weeks P(x = 2) = .1839 Poisson (average: 1)
f) P(x = 0) = .3679 Poisson (average:1)
Answer by ScientificArt(6) (Show Source): You can put this solution on YOUR website!

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