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binjip
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Throwing a die 7 times
Throw a die 7 times.
i) What is the probability that you get number 6 twice and all other outcomes once. (e.g. one possible set of outcomes would be 6, 6, 5, 4, 3, 2, 1)
ii) What is the probability that you get all the numbers of a die? (e.g. 6, 5, 4, 3, 2, 1, x where x is {1, ..., 6})
P() = # relevant outcomes / # of all possible outcomes
i)
There are 67 possible ways of arranging the set of 7 outcomes (# of permutations).
There 6*5*4*3*2*1*1 = 6! ways of arranging the numbers under the given conditions.
P(two 6, and all other outcomes) = 6! / 67 = 5! / 66
ii) # of permutations is 67. No change here.
The # of ways to arrange the numbers under give conditions changes:
6*6*5*4*3*2*1 = 6*6!
P(all 6 numbers of a die) = 6*6! / 67 = 5! / 65
Is this correct? Would appreciate any comments. Thanks
Homework Statement
Throw a die 7 times.
i) What is the probability that you get number 6 twice and all other outcomes once. (e.g. one possible set of outcomes would be 6, 6, 5, 4, 3, 2, 1)
ii) What is the probability that you get all the numbers of a die? (e.g. 6, 5, 4, 3, 2, 1, x where x is {1, ..., 6})
Homework Equations
P() = # relevant outcomes / # of all possible outcomes
The Attempt at a Solution
i)
There are 67 possible ways of arranging the set of 7 outcomes (# of permutations).
There 6*5*4*3*2*1*1 = 6! ways of arranging the numbers under the given conditions.
P(two 6, and all other outcomes) = 6! / 67 = 5! / 66
ii) # of permutations is 67. No change here.
The # of ways to arrange the numbers under give conditions changes:
6*6*5*4*3*2*1 = 6*6!
P(all 6 numbers of a die) = 6*6! / 67 = 5! / 65
Is this correct? Would appreciate any comments. Thanks
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