In a previous thread, IanStewart posted a solution to this question:

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tanya-prepared-4-different-letters-to-4-different-addresses-83683.html)

"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. "

I would like to point out that, although the final result is correct (1/3), the explanation is not correct.

"One letter goes in the right envelope. It doesn't matter which envelope this is- there are three envelopes left." This assumption is nowhere reflected in the chain of computations.

In the above computations, 2/3 is what is called a conditional probability, meaning that under the assumption that a certain letter is placed correctly, then the probability of placing one of the remaining 3 letters incorrectly is 2/3, then the probability of placing one of the remaining two letters incorrectly after two were already placed, one correctly and one incorrectly, etc.

The correct chain should be: 1/4 for the probability of a certain letter to be placed correctly, 2/3 for one of the remaining three placed incorrectly, 1/2 for one of the remaining two to be misplaced, and 1/1 for the last one which will be surely misplaced. This gives (1/4)*(2/3)*(1/2)*(1/1)=1/12 for the probability of exactly one particular letter to be placed correctly.

Since there are four possibilities to chose the one letter placed correctly, the required probability is 4*(1/12) = 1/3.

There is a collection of questions with all the possible scenarios for placing the 4 letters:

letter-arrangements-understanding-probability-and-combinats-84912.html?hilit=letters%20envelopesCan we use a similar reasoning used by IanStewart (for exactly 1 correctly placed letter) to find the probability of placing exactly 2 letters correctly?

The reasoning would go like this: consider two letters placed correctly, doesn't matter which ones, then the probability of placing one of the remaining letters incorrectly is 1/2, and then the fourth letter incorrectly is 1/1. Can we conclude that 1/2 is the probability of placing exactly two letters correctly?

The correct answer is 1/4. So, what's wrong with this reasoning?

I claim, we have to take into account the probability of placing a given pair of letters correctly. This probability is 1/12, since the probability of one of the letters to be placed correctly is 1/4, then the other one to be placed correctly is 1/3, which gives the (1/4)*(1/3) = 1/12 (or, either only the two letters are placed correctly or all four are, which means a probability of 2/24 = 1/12).

Then, the probability of one of the remaining two letters to be misplaced is 1/2, and the last one is for sure going to be misplaced (probability 1/1).

Since there are 4C2=(4*3)/2 = 6 possibilities to chose a pair of letters from the given 4 letters, the probability of placing exactly any two letters correctly is given by (1/12) * (1/2) * 6 = 1/4, which is the correct answer.

My point is: all the assumptions should be reflected in the computations, and first of all, we should make all the necessary and correct assumptions, regardless whether we are dealing directly with probabilities or counting number of possibilities. Easier said than done...but we have to try :O)

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PhD in Applied Mathematics

Love GMAT Quant questions and running.