You type four letters to four different people and address the envelopes. If you then insert the letters into the envelopes randomly, what is the probability that exactly three letters will go into the correct envelopes?
I think this is trick question.
If 3 are correct, how can the fourth be wrong?
So all 4 will have to be right.
prob of that = 1/4! = 1/24
Exactly 3 is impossible. If 3 were right, 4 would have to be right. The chances of exactly 3 would be 0.
To solve this question, we can use the principle of derangements, which is a combinatorial technique that calculates the number of permutations with no elements in their original position.
Let's start by finding the total number of possible ways to insert the four letters into the envelopes randomly. Since there are four letters, we have four choices for the first letter, three choices for the second letter, two choices for the third letter, and one choice for the last letter. Thus, the total number of permutations is 4! = 24.
Now, let's find the number of permutations where exactly three letters go into the correct envelopes. To calculate this, we need to determine the number of derangements for three letters. Using a derangement formula, we have:
D(3) = 3! × (1 - 1/1! + 1/2! - 1/3!) = 3 × (1 - 1 + 1/2 - 1/6) = 3 × (1 - 1 + 1/2 - 1/6) = 3 × (1 - 2/6 + 1/2 - 1/6) = 3 × (3/6 - 2/6) = 3 × 1/6 = 1
Therefore, there is only 1 permutation where exactly three letters go into the correct envelopes.
Finally, we can find the probability by dividing the number of favorable outcomes (1) by the total number of possible outcomes (24):
Probability = 1/24 = 0.0417, or approximately 4.17%
So, the probability that exactly three letters will go into the correct envelopes is 0.0417, or approximately 4.17%.