Sanya prepared 4 different letters to 4 different addresses.
Sanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?
Answer/Solution
1/3
Steps/Work
Total Sanya# of ways - 4! = 24.
Desired:
A-Mapped to the only correct address----------x 1 way only
B-Mapped to other two incorrect addresses - x 2 ways
C-Mapped to other two incorrect addresses - x 2 ways
D-Mapped to other two incorrect addresses - x 2 ways
Therefore, 1*2*2*2/24 = 1/3.?
Desired:
A-Mapped to the only correct address----------x 1 way only
B-Mapped to other two incorrect addresses - x 2 ways
C-Mapped to other two incorrect addresses - x 2 ways
D-Mapped to other two incorrect addresses - x 2 ways
Therefore, 1*2*2*2/24 = 1/3.?