letter and envelope

[ruplun]

If 3 letters are put in 3 envelopes with three different adresses , what is the probability that no addressee receives the correct letter?

a. 1/6
b. 1/4
c. 1/3
d. 1/2
e. 2/3

[naveen451]

IMO C

[mikebarmy]

can you explain that? My logic was that for the first letter, the probability of the address being wrong or it being the wrong letter is 2/3 and the same for the other letters. So, 2/3 x 2/3 x 2/3 but 8/27 is not an option. Can someone please explain? Thanks.

[Bek]

Answer is 1/3

2/3*1/2 = 2/6 or 1/3

[ntamhane]

No. Ways the letter can be put into envelopes = 3! = =6
Probability All are delivered to the correct address = 1/6
No. Ways one letter is delivered to the correct address = 3 ways
Probability One is delivered to the correct address = 3/6 = 1/2
Probability of a letter delivered to the correct address = 1/3+1/2 = 2/3
Probability none is delivered = [spoiler]1- 2/3 = 1/3[/spoiler]

[ruplun]

Am unable to follow the explanation ..can anyone plz explain in detail

[naveen451]

If 3 letters are put in 3 envelopes with three different adresses , what is the probability that no addressee receives the correct letter?

a. 1/6
b. 1/4
c. 1/3
d. 1/2
e. 2/3

total no of ways letters placed into their respective envelopes:3!=6 ways

3 letters into the respective envelopes:1 way
2 letters cannot be placed into the respective envelopes bcoz there r only 3 envelopes
1 letter placed into the respective envelopes:3 ways
total no ways that can be place alteast 1 letter into respective envelope=4/6
total no ways that can place letters into other than respective envelopes= 1-4/6=1/3

[ruplun]

is there any other method of solving this problem..plz suggest

[Bek]

I hope this img will help

[Bek]

1) you have three letters with two wrong envelopes = 2/3

2) you have two letters with one wrong envelope = 1/2

3) you have one letter with one wrong envelope = 1/1


2/3 * 1/2 * 1/1 = 2/6 or 1/3

[Tani]

The probability of getting the first one wrong is 2/3.
You now have two letters left and two envelopes. The probability of getting the second one wrong is 1/2.
Since you only have one envelope left, the probability on the third is 1.

Multiplying these three probabilities you get 2/3*1/2*1 = 1/3

Another way to look at it:
Three total possibilities, one favorable:
1. r,r,r
2. w,w,w
3. w,w,r
(Note: it is impossible to get r,r,w)

[tpr-becky]

I was asked to reply to this problem so here is my take:

It is probability so the basic formula is want/total and when you have mulitple events and you want a particular outcome in each event you multiply the formulas.

If you consider the envelopes A, B and C and the addressees a, b, c - since the question says there is an adressee you have to assume that the letters are individual for each addressee.

Then you do a want/total for each envelope

For envelope A there are 2/3 chances to get a letter not addressed to a.
then for envelope B there is 1/2 chances to get a letter that is not addressed to b.
then for envelope C there is a 1/1 chance to get a letter not addressed to c. (in order to get the last two numbers you have to assume that in each previous case you got something you wanted).

Then you multiply the formulas (2/3 * 1/2 * 1/1) = 2/6 which is 1/3.

[baladon99]

On a related note ,can anybody say , in how many ways one can put 3 letters in 3 envelopes?

[mayank.arora]

On a related note ,can anybody say , in how many ways one can put 3 letters in 3 envelopes?

IMO, it should be
for first envelope, 3! ways,
for second envelope 2! ways,
for third envelope, 1! ways,

same would be for letters,
for first letter, 3! ways,
for second letter 2! ways,
for third letter, 1! ways,

so finally, 2 x (3! x 2! x 1!)
2 x 12 = 24 ways....

I'm not sure if I'm right, expert advice needed though!

[Jim@Knewton]

On a related note ,can anybody say , in how many ways one can put 3 letters in 3 envelopes?

3 unique letters in 3 unique address envelopes:

(3*3)*(2*2)*(1*1) = 36 possible sequence ordered arrangements (like cards on a table)
(3*3)= 3L for each of 3E,(2*2)=2L for each of 2E and (1*1)= 1L for 1E

If order does not matter, then 36/6 = 6 ways to put these into a box!
We divide by 6 to remove the two triple counts in the ordered arrangement