IanStewart wrote:

GODSPEED wrote:

OA is 1/3....

I took little diff approach....please explain wht's wrong with it???

Probab of 1 letter to correct add ENV = 4C1 / 4! = 1/6, while answer is just double of this WHY???

The number of ways to put exactly one letter in the right envelope is not equal to 4C1. There are 4C1 ways to choose which letter goes in the right envelope, but you then need to work out how many ways the remaining letters can be placed in the wrong envelope. For the first of these three letters, there are 2 wrong envelopes you could choose. Now, you still have one letter left which has its matching envelope unused; you must put this letter in the wrong envelope, so you have only 1 choice for this letter, and finally for the last letter you have only 1 choice for where to put it. So you have 2*1*1 = 2 ways to assign the remaining letters incorrectly, which is why you need to multiply your answer by 2.

I posted a slightly different solution to BTG a while ago, which I'll paste here:

One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left. 2/3 chance the next letter goes in the wrong envelope, 1/2 the next one does, and 100% the last one does- its envelope must have been used already.

(2/3)(1/2) = 1/3.

Slightly old post, but very interesting explanation...

Just would like to add 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

No#2. W - C - W - W

No#3. W - W - C - W

No#4. W - W - W - C

There are 4 ways, it can be done

So, probability of 1 letter going in correct envelope: 4* (1/4) * (2/3) * (1/2) * 1 =

1/3