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Tanya prepared 4 different letters to 4 different addresses

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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink] New post 30 Aug 2013, 23:33
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Hi Guys,
Can anybody shot an explanation to my query raised above ? :(

I'd much appreciate if any Quant expert kindly explain this issue to me!
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink] New post 31 Aug 2013, 05:35
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bagdbmba wrote:
Hi Guys,
Can anybody shot an explanation to my query raised above ? :(

I'd much appreciate if any Quant expert kindly explain this issue to me!


This case is solved in the following way:
Counting all incorrect, or 0 correct:
P(all incorrect)=1-(1 correct)-(2 correct)-(3 correct)-(4 correct)=1-8/24-6/24-0-1/24=9/24.

Don;t know what I can add to that...
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink] New post 31 Aug 2013, 09:51
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink] New post 15 Sep 2013, 11:23
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We are trying to find the probability of 1R3W.

Probability = number of ways to get 1R3W/number of ways total

number of ways total is 4! = 24. Imagine stuffing envelopes randomly. Stacy can put any of 4 letters into the first envelope, any of the remaining 3 into the next, either of the remaining 2 into the next, and has no choice to make on the last, or 4*3*2*1.

number of ways to get 1R3W : She could fill the first envelope with the right letter (1 way), then put either of the 2 wrong remaining letters in the next (2 ways), then put a wrong letter in the next (1 way). That's 1*2*1*1 = 2.

But since it doesn't have to be the first envelope that has the Right letter, it could be any of the 4 envelopes (i.e. we could have RWWW, WRWW, WWRW, WWWR), the total ways to get 1R3W is 4*2 = 8.

Probability is 8/24 = 1/3.
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink] New post 21 May 2014, 04:35
Aximili85 wrote:
GODSPEED wrote:
My 2 cents:

For 1 letter to go in right envelope - probability = 1/4
For 2nd letter to go in wrong envelope - probability = 2/3
For 3rd letter to go in wrong envelope - probability = 1/2
For 4th letter to go in wrong envelope - probability = 1

Now, in the order this arrangement can be done:
C - Letter going into correct envelope
W - Letter going into wrong envelope

No#1. C - W - W - W
............in correct envelope: 4* (1/4) * (2/3) * (1/2) * 1 = 1/3


Thats the only explanation i was able to understand =(
I'm worried because if I don't learn the combinatrics formula way of doing it I might make a logical inconsistency while solving under a time limit.


Just wondering, is the above method correct? What if the correct is not the first one? Does it change the procedure?

Please advice
Cheers!
J :)
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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink] New post 21 May 2014, 04:35
Aximili85 wrote:
GODSPEED wrote:
My 2 cents:

For 1 letter to go in right envelope - probability = 1/4
For 2nd letter to go in wrong envelope - probability = 2/3
For 3rd letter to go in wrong envelope - probability = 1/2
For 4th letter to go in wrong envelope - probability = 1

Now, in the order this arrangement can be done:
C - Letter going into correct envelope
W - Letter going into wrong envelope

No#1. C - W - W - W
............in correct envelope: 4* (1/4) * (2/3) * (1/2) * 1 = 1/3


Thats the only explanation i was able to understand =(
I'm worried because if I don't learn the combinatrics formula way of doing it I might make a logical inconsistency while solving under a time limit.


Just wondering, is the above method correct? What if the correct is not the first one? Does it change the procedure?

Please advice
Cheers!
J :)
Re: Tanya prepared 4 different letters to 4 different addresses   [#permalink] 21 May 2014, 04:35
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