GMAT Prep Question

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

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!

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

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 ?

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 ?

anyone have a better explanation for this?

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

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