Category: Science > Math
Asked by: gyrocopter-ga
List Price: $10.00
26 Apr 2004 20:03 PDT
Expires: 26 May 2004 20:03 PDT
Question ID: 336752
A scratchcard has ten windows. Nine are prize icons which include two that match. One is a card-invalidating message. The object is to uncover the two matching icons without uncovering the invalidating message. What are the chances of winning? How did you figure it?
Answered By: mathtalk-ga on 03 May 2004 14:21 PDT
Hi, gyrocopter-ga: For the sake of determining the chances of winning, we can ignore the unmatched prize icons. Scratching off any of these doesn't affect winning or losing; it's irrelevant to the outcome one way or the other. So we consider only the order in which the two matching prizes and the card-invalidating message would be chosen, if we devise our "strategy" in advance. Of course we will always have won or lost before the last window is scratched off, but it still makes sense to consider a plan for that "last" unscratched window as the last choice. To win the knockout message must be third among the three windows that matter, ie. the two matching prize icons and the card-invalidating message. Therefore the chance of winning is one in three. To see this in a slightly different manner, there are 3! orders in which three windows that matter can be chosen. Two of these are winners (picking the two matching prize icons first, but in either order). 2/3! = 1/3. regards, mathtalk-ga
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From: rdf-ga on 03 May 2004 07:05 PDT
Probability of picking 1st matching icon from all 10 uncovered = 1/10 Probability of then picking 2nd matching icon from the remaining 9 uncovered = 1/9 Therefore, probability of picking both matching icons = 1/10 x 1/9 = 0.01111 ://www.google.co.uk/search?hl=en&ie=UTF-8&oe=UTF-8&q=%281%2F10%29*%281%2F9%29&meta=
From: richard-ga on 03 May 2004 08:54 PDT
This problem is a lot harder than the above comment would indicate. For an example of a similar but easier problem, see page 9 of http://www.ex.ac.uk/cimt/mepres/allgcse/as5act1.pdf
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