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Hi all. I can't think that I would get a question like this if i'm understanding this correctly
Question: There are 5 50c coins and 3 20c coins in a tin label the 50c coins f_1 to f_5 and the 20c coins t_1 to t_3. suppose you put your hand in the tin and take out 3 coins without looking. Write down the set that represents the event "I have at least one 50c coin amongst the 3 coins that I drew"
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I don't think I was totally clear. So my problem is. It seems like that would be a very big set? And not something I would be expected to do in a exam?
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hi Deon588,
If the 50c were just labelled F (all five of them) and the 20c similarly labelled T, then it wouldn't be so bad. I agree with you that writing all the possibilities using f_1 etc would make a long list.
If all eight coins are distinct and order is not relevant, then there are (8 x 7 x 6) /(3 x 2 x 1) = 56 ways of listing them all. So if you reduce this list to exclude any which have no 50c coins ... how many is that? Must be 56 - 1 = 55.
Did this come out of a book or from a teacher? If the latter, go and check.
I don't think it would come up in an exam.
Bob
Last edited by Bob (2011-10-02 20:01:20)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob good to know i'm understanding this correctly... This came from revision exercises my lecturer created so i'll have to check with her
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