That method gives the right answer when there are just 2 or 3 envelopes, but it falls down when there are 4 or more. To see this in action, let's try it with 4 envelopes. Your approach gives
P(at least one letter is in the right envelope) = 1 - (3/4)(2/3)(1/2) = 3/4
Now, if you grab a pen and paper and write down all the ways in which you can allocate letters to envelopes so that there's at least one correct match, you find there are 15 possible arrangements. There are 4! ways of arranging the letters in total, so this gives
P(at least one letter is in the right envelope) = 15/4! = 15/24 = 5/8
This is different to the answer above.
The reason for the difference is that it's trickier to calculate the probability that none of the letters ends up in the right envelope than you might expect.
Let's start with the first envelope, envelope A. There are 4 letters that can go into envelope A and 3 of them will be incorrect, so P(A is incorrect) = 3/4.
Now let's move onto envelope B. There are 3 remaining letters that can go into envelope B, and it's tempting to say that 2 of them will be incorrect. The trouble is, the real answer all depends on what went into envelope A. As an example, if we put letter B into envelope A leaving behind letters A, C and D, then the probability that envelope B will be filled incorrectly is 3/3; it's only 2/3 if envelope A was incorrectly filled with letter C or D. This means that overall, the probability of incorrectly filling envelopes A and B is not (4/3)(2/3), which is why you get a different final answer.
I have a similar but difficult problem that's got me baffled for a while now. Please help me if you can. Thanx
There are 6 sealed envelopes, whose sizes are listed here:
4.125" x 9.5" Business envelope
3.5" x 9.25" Chou 30 envelope (from Japan)
4" x 5.5" Greeting card envelope
9" x 12" Large manila envelope
7.5" x 10.5" Small manila envelope
9.125" x 9.125" Square envelope
Each envelope has a single digit written on the outside of it. Some of the envelopes are inside of others; there's even one envelope that's inside an envelope that's inside another one. All of the envelopes are lying flat; They haven't been folded, cut, rolled, wrinkled, or otherwise modified them to make them fit.
You'll need to figure out what number is written on each envelope, from the clues below:
Exactly four of the envelopes are inside of other envelopes. One of the two "outside" envelopes has a transparent window, through which the number 7 is visible. Except for that window, the envelopes are opaque.
The number on the greeting card envelope equals the number on one of the other envelopes; there are no other duplicate numbers.
The product of the 6 numbers is 20160.
The number on the square envelope equals the number of envelopes inside it.
The number on the Chou 30 envelope is the sum of two of the other numbers.
The number on the large manila envelope equals the average of the numbers on the envelopes inside it.