+38 There are 4 mailboxes and 4 letters, they are randomly placed in mailboxes, what is the probability that one of the mailboxes has exactly 3 letters.
There are four ways in which a mailbox could have exactly 3 letters: the first mailbox could have 3 letters, the second mailbox could have 3 letters, the third mailbox could have 3 letters, or the fourth mailbox could have 3 letters.
Let's consider each case separately:
Case 1: The first mailbox has 3 letters.
In this case, there are 3 letters left to be placed in the remaining 3 mailboxes. Each of the remaining 3 letters can be placed in any of the 3 remaining mailboxes, so there are 3 choices for each of the 3 remaining letters. Therefore, there are 3^3 = 27 ways in which the remaining letters could be placed in the remaining mailboxes.
Case 2: The second mailbox has 3 letters.
This case is similar to Case 1, but we need to account for the fact that the 3 letters that go in the second mailbox could be any 3 of the 4 letters. There are 4 ways to choose which 3 letters go in the second mailbox, and then there are 3 choices for the remaining letter, and 2 choices for where to put it. Therefore, there are 4 x 3 x 2 = 24 ways in which the letters could be placed in this case.
Case 3: The third mailbox has 3 letters.
This case is also similar to Case 1, but we need to account for the fact that the 3 letters that go in the third mailbox could be any 3 of the 4 letters. There are 4 ways to choose which 3 letters go in the third mailbox, and then there are 3 choices for the remaining letter, and 2 choices for where to put it. Therefore, there are 4 x 3 x 2 = 24 ways in which the letters could be placed in this case.
Case 4: The fourth mailbox has 3 letters.
This case is similar to Case 1, but we need to account for the fact that the 3 letters that go in the fourth mailbox could be any 3 of the 4 letters. There are 4 ways to choose which 3 letters go in the fourth mailbox, and then there are 3 choices for the remaining letter, and only 1 choice for where to put it. Therefore, there are 4 x 3 x 1 = 12 ways in which the letters could be placed in this case.
Therefore, the total number of ways in which one of the mailboxes could have exactly 3 letters is 27 + 24 + 24 + 12 = 87.
The total number of ways in which the 4 letters could be placed in the 4 mailboxes is 4^4 = 256.
Therefore, the probability that one of the mailboxes has exactly 3 letters is 87/256.
+118673 Hi guest answerer.
Looking at in as different cases was a good idea
But
One of your early statements is
"
Case 1: The first mailbox has 3 letters.
In this case, there are 3 letters left to be placed in the remaining 3 mailboxes."
there is only 4 letters, if three go in the first letter box then there is only 1 more letter.
+118673 There are 4 mailboxes and 4 letters, they are randomly placed in mailboxes, what is the probability that one of the mailboxes has exactly 3 letters.
It sounds to me like the letter boxes are different but the letters are effectively the same
First I will look for the total number of ways
Here are the 4 letters * * * *
We have to split those into 4 piles where some piles can have no letters.
We need to seperate them using 3 bars
eg ||*|***
that would be non in the first box, none in the second, 1 in the third and 3 in the 4th.
So that would be 7C3 or 7C4 ways = 35
NOW how many of those have 3 in one box If there are 3 in the first box then there can be 1 in any of the othere 3 that is 3 ways
There are 4 letter boxes so that is 3*4 =12 ways
So I get 12/35
NOTE: This is called stars and bars method.
