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# 1) if a monkey is given 11 letters that can form the word

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1) if a monkey is given 11 letters that can form the word [#permalink]

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25 Apr 2005, 12:12
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1) if a monkey is given 11 letters that can form the word "probability", what is the probability that the monkey will form the word "probability"?

2)A player selects six numbers from 1 to 49. What is the probability of picking the six winning numbers?

3) A typewriter has 48 keys (including space, numbers, etc). What is the probability that a baby, who strikes keys randomly, will correctly type the word "hamlet"?
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25 Apr 2005, 12:41
1) 1/11!
2) 6!/(49*48*47*46*45*44)
3) 1/48^6

will explain if correct...
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25 Apr 2005, 13:20
1) 1/(11!/2!*2!)

2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.
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25 Apr 2005, 14:02
1) 1/(11!/2!*2!) = 4/11!
Correct.

2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.
Correct.

1/49C6 = 6! x 43! / 49! = 6! / (49x48x47x46x45x44)

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25 Apr 2005, 14:08
mckenna wrote:
1) 1/(11!/2!*2!) = 4/11!
Correct.

2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.
Correct.

1/49C6 = 6! x 43! / 49! = 6! / (49x48x47x46x45x44)

oh right ! too late here in germany you can approach this problem with combination- as well as with permutation-reasoning...
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25 Apr 2005, 14:53
Can you guys explain 1?
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25 Apr 2005, 15:19
rc1979 wrote:
Can you guys explain 1?

divide by 2! and 2! because there are 2 bs and 2 s. for example probabilty with the first b in 4th position and the second b in 6th position is 1 way and when you change the b`s you get a second way, but it is the same word. so you divide 2.
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25 Apr 2005, 19:31
the number 2) was not so easy, but I got 4!/11! and 1/48^6

rc1979, christoph explained it well, there are 2 letters B (le'ts call it B1 and B2 for example)
Usually you have to work with permutation concerning words problems because order normally matters, in that case there will be no difference if you choose B1 or B2, it will be the same word whereas each B is considered as a specific letter. So you'll have twice more chances with B and it's the same idea for I.

finally, normally for a word with 11 different litters the prob will be 1/11! but since you have 2 repetitions of 2 letters, it will be 1*2*2/11!

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25 Apr 2005, 20:04
mckenna wrote:
2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.
Correct.

it is a permutation not a combiation problem. so it is
=1/(49P6)
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