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Unformatted text preview: rtainty Homework 1 Spring 2011 (Bond) Sample space, enumeration, probability The number of “three
1” die rolls is 10x25 = 250. Now, this is also true of any other number. The number of “three
2” die rolls is also 250. So the number of three
of
a
kind die rolls is n = 250 x 6 = 1500 P = 1500/7776 = 0.193 (But see note at end of problem) 3/4 CEE 202 – Engineering Risk & Uncertainty Homework 1 Spring 2011 (Bond) Sample space, enumeration, probability Problem 6 continued b) Full house (two of one kind & three of another kind) We’ve already done the hard work for this problem. There are 10 possible ways in which you could roll ‘three
1’ combinations. The other 2 dice have to provide the 2
of
one
kind. There are 5 ways in which they could do that. The number of die rolls that produce “three
1s” and “2
of
something
else” is 10 x 5 = 50. This is also true of “three
2, plus 2
of
something
else” so n = 50 x 6 = 300. P = 0.039 Note: If you know how to play Yahtzee, you would know that if you rolled 3 of one kind, and 2 of another, you wouldn’t count it as 3
of
a
kind. You would call it a...
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 Spring '08
 Clark

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