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Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelop with its correct addresses. If the 4 leters are to be put into the 4 envelops at random , what is the probability that only 1 letter will be put into the envelop with its correct address.
a) 1/24
b) 1/8
c) 1/4
d) 1/3
e) 3/8
==> probability = number of specific event / total number of cases
Total number of cases = letters(A,B,C,D), envelops(a,b,c,d) thus becomes 4*3*2*1.(A has 1 to choose from a,b,c,d, making it 4, while B has 3, C has 2 and D have 1)
The specific event has the cases (A,a) with (B,d) for (A,a), (C,b), (D,c) or (B,c), (C,d), (D,b), thus there is only 2 case (Since there should be only one letter in the envelope with the exact address)
Since the case applies to (B,b), (C,c), (D,d) as well, the number of cases that only 1 letter will be in the envelope with the exact address =2*4=8.
Thus, probability = 8/4*3*2*1=1/3. Therefore the answer is D.
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