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# anyone want to take a shot at this Tanya prepared 4

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anyone want to take a shot at this Tanya prepared 4 [#permalink]

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17 Jun 2007, 19:01
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

anyone want to take a shot at this

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelop with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

a) 1/24
b) 1/8
c) ¼
d) 1/3
e) 3/8

Dan
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17 Jun 2007, 19:12
I got 1/12 which is not given in the choices!
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17 Jun 2007, 21:03
I got it

P = 4(1/4*2/3*1/2) = 1/3
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18 Jun 2007, 08:19
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18 Jun 2007, 09:28
i got 1/4. what is the OA? how do you solve it?
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18 Jun 2007, 15:13
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This is how I looke at it:

Chance that 1st letter goes into correct address is 1/4
Chance that 2nd letter goes into incorrect address is 2/3 ( becos out of remaining 3 covers 2 carry incorrect addr for 2nd letter)
Chance that 3rd letter goes into incorrect address is 1/2 ( becos out of remaining 2 covers 1 carries incorrect addr for 3rd letter)
If all this happens, 4th letter should go into the wrong cover for sure. Chance = 1

For prob that one letter gets into correct address and others into wrong = 1/4*2/3*1/2*1 = 1/12 ---> This is where I stopped when I initially solved the problem.

But since there are 4 such permutations prob = 4*1/12 = 1/3

I don't know if this is the OA or not
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18 Jun 2007, 15:33
Thanks

OA: d) 1/3

Dan
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18 Jun 2007, 20:08
L1, L2, L3 and L4 are four letters
A1, A2, A3 and A4 are four addresses

with L1 in A1, each of the other three letters can go to two of the remaining envelopes. So, the total number of favourable outcomes are

1*2*2*2 = 8

total number of ways = 4!=24

P=8/24=1/3
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18 Jun 2007, 20:08
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