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by [email protected] » Tue Sep 25, 2007 11:39 am
As there are total 4 letters and 4 envelopes
1 letter can be put in 4 diff envelopes in which only 1 way of putting is correct
So total no of ways letters can be put = 4*4= 16
As 1 letter can be put in the correct envelope in 1 way and there are four envelopes
the prob = 4/16 = 1/4
whats the OA
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by raulverde » Wed Sep 26, 2007 9:49 am
There are four Letters {L1,L2,L3,L4}
There are four Envelopes{E1,E2,E3,E4}

Ways to put 1 Letter in the correct Envelope ?

(L1 GOES INTO E1) OR (L2 GOES INTO E2) OR (L3 GOES INTO E3) OR (L4 GOES INTO E4)

(Event of picking the right combination ) = 1/4
(Event of picking the wrong combination) = 2/3 * 1/2 *1/1 = 1/3

Probability of all these events is the same 1/12 .

1/12 + 1/12 + 1/12 + 1/12 = 4*(1/12) = 1/3
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by wongee » Sat Sep 29, 2007 12:54 pm
hey I lost the method in the portion:

(Event of picking the right combination ) = 1/4
(Event of picking the wrong combination) = 2/3 * 1/2 *1/1 = 1/3 (how does this come about?)

Probability of all these events is the same 1/12 . (and this)

1/12 + 1/12 + 1/12 + 1/12 = 4*(1/12) = 1/3 (this i get)

THANKS IN ADVANCE!
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by raulverde » Mon Oct 01, 2007 11:51 pm
wongee,

Say you have 3 Envelopes left (e1,e2,e3) the corresponding letters for them are (l1,l2,l3)

now if we were to pick a wrong combinations . For the first wrong combination. we would have 3 choices of envelopes and 2 choices of letters ( We dont include the 1 right letter )

Hence 2/3

The next wrong combination would be 1/2 and the last one 1/1

HTH
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by ash g » Fri Mar 07, 2008 5:59 pm
hi, I would like to slightly twist the question in an effort to better understand probability.

[ Refering to raulverde explanation]
If the QUESTION WAS: What is the probablity that none of the letters match the correct envelope, would the solution be as simple as below or would it be more complicated.

(E1 doesnt match) & (E2 doesnt match) & (E3 doesnt match) & (E4 doesnt match)

(3/4) (2/3) (1/2) (1) = 1/4 <<==Answer
!E1 !E2 !E3 !E4

The reason for doubt is simply that it may so happen that L2 matched with E1, so the second term could be (3/3) rather than (2/3). Do we cater for such a event ?
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by ash g » Fri Mar 07, 2008 6:09 pm
hi, I would like to slightly twist the question in an effort to better understand probability.

[ Refering to raulverde explanation]
If the QUESTION WAS: What is the probablity that none of the letters match the correct envelope, would the solution be as simple as below or would it be more complicated.

(E1 doesnt match) & (E2 doesnt match) & (E3 doesnt match) & (E4 doesnt match)

(3/4) (2/3) (1/2) (1) = 1/4 <<==Answer
!E1 !E2 !E3 !E4

The reason for doubt is simply that it may so happen that L2 matched with E1, so the second term could be (3/3) rather than (2/3). Do we cater for such a event ?
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by GMATters1001 » Sat Dec 20, 2008 10:04 pm
anyone have a better explanation for this?
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by ronniecoleman » Sun Dec 21, 2008 7:59 am
Nice question!!!

Ok

Choose one letter which should be have a right address... 4C1

left three letters
Total ways: 3!
but it contains ways where one address is correct, so total 3 ways
+ where all letters are correct

so 3!-4

Now

( 4C1 *(3!-4) ) / 4!

= 1/ 3
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by gmatapril » Fri Mar 25, 2011 3:17 pm
why we have to do the last step 1/12* 4. since we are asked about probability of only one letter to be put in correct envelope address
raulverde wrote:There are four Letters {L1,L2,L3,L4}
There are four Envelopes{E1,E2,E3,E4}

Ways to put 1 Letter in the correct Envelope ?

(L1 GOES INTO E1) OR (L2 GOES INTO E2) OR (L3 GOES INTO E3) OR (L4 GOES INTO E4)

(Event of picking the right combination ) = 1/4
(Event of picking the wrong combination) = 2/3 * 1/2 *1/1 = 1/3

Probability of all these events is the same 1/12 .

1/12 + 1/12 + 1/12 + 1/12 = 4*(1/12) = 1/3
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