gmatmillenium wrote:Tanya prepared 4 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 into 4 envelopes at random, what is the probability that only 1 letter will be put its envelope with correct address?
Probability is defined as:
(total number of good outcomes)/(total number of possible outcomes)
Sometimes the best approach is to determine each part of the fraction separately.
Let's say the 4 letters are
ABCD, and to place each in the correct envelope, they have to be placed in alphabetical order:
ABCD.
To determine the total number of possible outcomes, we need to determine how many ways we can arrange the 4 letters:
4 * 3 * 2 * 1 =
24 total possible outcomes.
A good outcome is when 1 letter is in the correct position and the other 3 are not.
Let's say we want only letter A to be in the right position. We'd have only 2 good arrangements:
ACDB and ADBC
2 good outcomes so far.
Extending this logic, we know that:
If we want only B to be in the right position, we'd get
another 2 good outcomes.
If we want only C to be in the right position, we'd get
another 2 good outcomes.
If we want only D to be in the right position, we'd get
another 2 good outcomes.
So altogether we have 2 + 2 + 2 + 2 =
8 good outcomes.
So (good outcomes)/(total outcomes) = 8/24 = 1/3.Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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