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Pocatello

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If all the letters making up the word Wollongong are placed in to a hat and drawn out, what is the probability of:
a) The word beginning with an 'o'
b) The word having no two identical letters in a row
c) At least one 'l' placed between the 'g''s


Thanks
 

damo676767

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Pocatello said:
If all the letters making up the word Wollongong are placed in to a hat and drawn out, what is the probability of:
a) The word beginning with an 'o'
b) The word having no two identical letters in a row
c) At least one 'l' placed between the 'g''s


Thanks
interesting

part a)
3/10


part b)
prob of l's in a row
9/<sup>10</sup>C<sub>2</sub>
1/5

prob of g's in a row
9/<sup>10</sup>C<sub>2</sub>
1/5

prob of n's in a row
9/<sup>10</sup>C<sub>2</sub>
1/5

prob of o's in a row
71/<sup>10</sup>C<sub>3</sub>
71/120


prob of l's and g's
1/5 * 1/5
1/25

prob of l's and n's
1/5 * 1/5
1/25

prob of l's and 0's
1/5 * 7/120
7/600

prob of g's and n's
1/5 * 1/5
1/25

prob of g's and o's
1/5 * 7/120
1/600

prob of n's and o's
1/5 * 1/120
1/600

prob of l's and g's and n's
1/5 * 1/5 * 1/5
1/125

prob of l's and g's and o's
1/5 * 1/5 * 7/120
7/3000

prob of l's and o's and n's
1/5 * 1/5 * 7/120
7/3000

prob of o's and g's and n's
1/5 * 1/5 * 7/120
7/3000

prob of all
1/5 * 1/5 * 1/5 * 7/120
7/15000


ther fore prob of none
p = 1 - (1/5 + 1/5 + 1/5 + 7/120 - 1/25 - 1/25 - 7/600 - 1/25 - 1/600 - 1/600 - 1/125 - 7/3000 - 7/3000 - 7/3000 - 7/15000)
= 3691/7500

ther must be an eaier way to do it

part c)
i'l sleep on it
 

Pocatello

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What do you think?

a) (9!/3!2!2!2!)/(10!/3!2!2!2!)
 
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hereSIR

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This can't be correct, since you are ignoring the cases where only two of the three o's occur together
 

Affinity

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You dig up a stale thread just for this! :p
 

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