Re: HSC 2015 3U Marathon
I'm used to doing P(Event) = Favourable outcomes/Total outcomes
The word is [ENTERTAINMENT]
For all cases, total outcomes = 13!/(3!3!3!) = 28828800. This was established in the above post in the image.
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For a) This is done by treating all the E's as one letter. Note that the E's are not say E_1, E_2, E_3 so there is only 1! way of arranging the E's.
Our favourable outcomes then becomes: 11!/(3!3!) = 1108800
So the probability of this is 1108800/28828800 = 1/26
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For b) In a similar way, two of the E's are treated as one letter. The words can be seen as "12 letter words" in this regard. As pointed out again in the above post, we now have a deviation of cases.
In 2 ways, we can fix one of the E's at the ends. Then, the other E has 10 possible placements. We then arrange every other letter, which can be done in 10!/(3!3!) ways. 2*10*10!/(3!3!) = 2016000
In 10 ways, we can fix one of the E's specifically not at the ends. Then, the other E now has 9 possible placements. We then arrange every other letter, which can stil be done in 10!/(3!3!) ways. 10*9*10!/(3!3!) = 9072000
Hence, the required probability is (2016000+9072000)/28828800 = 5/13
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For c) Inspection shows that the probability of 1 E apart from 2 E's together is the exact same thing as part (b). Hence, this just requires complementary probabilities.
The required probability is 1 - 1/26 - 5/13 = 15/26
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For d) Two E's have already been fixed. Note however, that the final E is no longer a repeated letter. Hence our favourable outcomes becomes 11!/(3!3!) again.
Because the result is identical to part (a), the answer is 1/26
$ for b) and c)? Why not $n$?)
