Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability
1. that only 1 letter will be put into the envelope with its correct address?
[spoiler]oa 1/3[/spoiler]
P(exactly n times) = P(one way) * total possible ways.
P(one way):
One way to get a good outcome is for only the first envelope to receive the correct letter.
P(1st envelope gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(next envelope gets the wrong letter) = 2/3 (3 letters left, 2 of them wrong)
At this point, one of the remaining 2 envelopes could still get the correct letter.
P(this envelope gets the wrong letter) = 1/2 (2 letters left, 1 of them wrong)
P(last envelope gets the wrong letter) = 1/1 (1 letter left, and it must be wrong since the correct letter was placed in one of the other envelopes)
Since we need all of these events to happen, we multiply the fractions:
1/4 * 2/3 * 1/2 * 1/1 = 1/12.
Total possible ways:
The result above represents ONE WAY to get a good outcome: if the first envelope is the only one to receive the correct letter.
We need to account for ALL OF THE WAYS to get a good outcome.
Since each of the 4 envelopes could be the one to receive the correct letter, we multiply the result above by 4:
4 * 1/12 = 1/3.
2. That no letter will be put into the envelope with its correct address?
[spoiler]oa 9/24[/spoiler]
The following placements of letters are possible:
None correct
Exactly 1 correct
Exactly 2 correct
All 4 correct (If we have 3 correct, we automatically have all 4 correct.)
P(none) + P(exactly 1) + P(exactly 2) + P(all 4) = 1.
Thus, P(none) = 1 - P(exactly 1) - P(exactly 2) - P(all 4).
P(exactly 1 correct) = 1/3 (as shown above).
P(exactly 2 correct):
P(1st envelope gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(next envelope gets the correct letter) = 1/3 (3 letters left, 1 of them correct)
P(next envelope gets the wrong letter) = 1/2 (2 letters left, 1 of them wrong)
P(last envelope gets the wrong letter) = 1/1 (1 letter left, and it must be wrong since the correct letter was placed in the 3rd envelope)
Since we need all of these events to happen, we multiply the fractions:
1/4 * 1/3 * 1/2 * 1/1 = 1/24.
Since, of the 4 envelopes, any combination of 2 could get the correct corresponding letters, the result above needs to be multiplied by 4C2 = 6:
6*24 = 1/4.
P(all 4 correct):
P(1st envelope gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(next envelope gets the correct letter) = 1/3 (3 letters left, 1 of them correct)
P(next envelope gets the correct letter) = 1/2 (2 letters left, 1 of them correct)
P(last envelope gets the correct letter) = 1/1 (1 letter left, and it must be correct)
Since we need all of these events to happen, we multiply the fractions:
1/4 * 1/3 * 1/2 * 1/1 = 1/24.
Thus, P(none) = 1 - 1/3 - 1/4 - 1/24 = 1 - 15/24 = 9/24 = 3/8.
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[where 24 is the total number of arrangements]
