document.write( "Question 868984: I APPRECIATE ANY HELP AT ALL!!! I just don't know what to do to start with...\r
\n" ); document.write( "\n" ); document.write( "Each week you buy a cereal box of a certain company that costs 5€.
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\n" ); document.write( "The company advertises that inside some \"lucky\" boxes, a \"gift's coupon\" of value of 10€ is included.
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\n" ); document.write( "It is given that 1% of the boxes of the company includes a gift's coupon.\r
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\n" ); document.write( "a) How much is your average profit in the first 100 weeks of purchase?
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\n" ); document.write( "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?
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\n" ); document.write( "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%.
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\n" ); document.write( "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).
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\n" ); document.write( "e) Count the probability that you have found exactly 2 coupons in 100 weeks, using the distribution of Poisson.
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\n" ); document.write( "f) Count the probability that you have NOT found any coupon in 100 weeks. \n" ); document.write( "

Algebra.Com's Answer #523992 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!


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