Question 1202976
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The response from the other tutor arrives at the correct answer only by chance; the method used is faulty.  If we use that method on the problem with only 3 envelopes, the answer we get is<br>
3C1/3!+3C2/3!+3C3/3! = 3/6 + 3/6 + 1/6 = 7/6<br>
which of course is not a valid number for a probability.<br>
The probability that at least one letter gets into the right envelope is 1, minus the probability that none of them get in the right envelope.<br>
If no letter gets in the right envelope, we have a derangement -- an arrangement in which none of the letters are in the right place.  The number of derangements of 4 items is<br>
{{{4!(1-1/1!+1/2!-1/3!+1/4!)}}}
{{{24(1-1+1/2-1/6+1/24)}}}
{{{12-4+1 = 9}}}<br>
(N.B. -- Do an internet search on "derangement" if you want to try to understand that formula....)<br>
Since there are 4! = 24 possible arrangements of the 4 letters, the probability that we have a derangement (none of the letters is in the right envelope) is 9/24 = 3/8<br>
So the probability that at least one letter gets in the right envelope is 1-3/8 = 5/8.<br>
ANSWER: 5/8<br>