I think, Total ways of selecting the envolpes and letters are 4*4 ways.

Not getting the right letter into the envolope is 4+3+2+1 ways into fowur letters.

Total probability of not getting the letter into rt envolope is 10/16 ways.

Probability of getting rt letter into the rt envolope is 1 - 10/16 = 6/16, 3/8

Correct me if i am wrong.

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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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rakaisraka wrote: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 envelope with its correct address.
a) 1/24
b) 1/8
c) 1/4
d) 1/3
e) 3/8

Let the 4 letters be A, B, C and D.
Total ways to arrange the 4 letters = 4! = 24.
Let the correct ordering of the 4 letters be ABCD.

Write out the ways that ONLY A can be put in the correct position:
ACDB
ADBC
Total ways = 2.

Using the same reasoning, there will be 2 ways that ONLY B can be put in the correct position, 2 ways that ONLY C can be put in the correct position, and 2 ways that ONLY D can be put in the correct position.
Thus, the total number of ways to put EXACTLY 1 letter in the correct position = 2+2+2+2 = 8.

Thus:
P(exactly 1 letter is put in the correct position) = 8/24 = 1/3.