(adsbygoogle = window.adsbygoogle || []).push({}); Throwing a die 7 times

1. The problem statement, all variables and given/known data

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})

2. Relevant equations

P() = # relevant outcomes / # of all possible outcomes

3. The attempt at a solution

i)

There are 6^{7}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! / 6^{7}= 5! / 6^{6}

ii) # of permutations is 6^{7}. 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! / 6^{7}= 5! / 6^{5}

Is this correct? Would appreciate any comments. Thanks

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# Homework Help: Throwing dice 7 times

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