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Sadman is correct. The way I tackled it is by thinking of just as ways I can arrange A, B, C and D without having any letters match the column that they're in (A, B, etc.). If you just write on your paper ABCD, then you can simply list them all out. Here they are:
BADC
BDAC
BCDA
CDBA
CADB
CDAB
DCBA
DCAB
DABC
You'll notice that each letter is used 3 times in each column. That is because that's how many wrong addresses there are, multiplied by 3 for the number of total ways. The numerator is 9, therefore, and the denominator is 432*1 which equals 24. The answer is 9/24 or 3/8.

Three winning tickets are drawn from an urn of - 100 tickets . what is the probability of winning for a person who buys - (i) 4 tickets ? (ii) only one tickets?

for first letter , since there are 4 envelopes, it can go to right address in only one case.... and wrong in three cases,
so the probability of going to wrong address is 3/4

for 2nd letter , since there are 3 envelopes left now,it can go to right address in only one case.... and wrong in two cases,
so for 2nd letter the probability of going to wrong address is 2/3

for 3rd letter , since there are 2 envelopes left now,it can go to right address in only one case.... and wrong in one case,
so the probability of going to wrong address is 1/2

for 4th letter , since there is 1 envelope left and we have been assuming that others have gone to wrong addresses , so it must go to the wrong address.
so its probability of going to wrong address is 1.

multiplying all we get, 3/4 * 2/3 * 1/2 * 1 = 1/4

(here multiplication is done, since we consider all of them are going to wrong addresses i.e
simplifying we can write....

probability of (first letter will go to wrong address and 2nd letter will go to wrong address and 3rd letter will go to wrong address and 4th letter will go to wrong address and )

i.e probability of (first letter will go to wrong address * 2nd letter will go to wrong address *
3rd letter will go to wrong address * 4th letter will go to wrong address and )

Answer is 9/24.You are making a big mistake in the second step,Pavel. Consider a situation; letter no 1 is put in the envelope of any other letter, say in the envelope of letter 2. Now what if letter 2 is considered next. This case letter 2 has 3 choices again.As letter1 is occupying its envelope it can enter in any remaining envelope.Got it?
Sadman is correct. This can only be solved using derangement formula .You can see it in any combinatorics books.It can be proved using logic of inclusion-exclusion of set theory.

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## Comments

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TopNewestThe answer is $\displaystyle \frac{9}{24}$ or $\displaystyle \frac38$

I am not going details. Just note that this is called Derangement .Try the link and reply if anything seems difficult.

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Thanks, please give me details calculation

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Sadman is correct. The way I tackled it is by thinking of just as ways I can arrange A, B, C and D without having any letters match the column that they're in (A, B, etc.). If you just write on your paper ABCD, then you can simply list them all out. Here they are: BADC BDAC BCDA CDBA CADB CDAB DCBA DCAB DABC You'll notice that each letter is used 3 times in each column. That is because that's how many wrong addresses there are, multiplied by 3 for the number of total ways. The numerator is 9, therefore, and the denominator is 4

32*1 which equals 24. The answer is 9/24 or 3/8.Log in to reply

A person writes 4 letters & addresses 4 envelops randomly. What is the probability that every letters goes to the wrong address?

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1/4

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Three winning tickets are drawn from an urn of - 100 tickets . what is the probability of winning for a person who buys - (i) 4 tickets ? (ii) only one tickets?

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to me the answer is 1/4 ......

Calculation details:

for first letter , since there are 4 envelopes, it can go to right address in only one case.... and wrong in three cases, so the probability of going to wrong address is 3/4

for 2nd letter , since there are 3 envelopes left now,it can go to right address in only one case.... and wrong in two cases, so for 2nd letter the probability of going to wrong address is 2/3

for 3rd letter , since there are 2 envelopes left now,it can go to right address in only one case.... and wrong in one case, so the probability of going to wrong address is 1/2

for 4th letter , since there is 1 envelope left and we have been assuming that others have gone to wrong addresses , so it must go to the wrong address. so its probability of going to wrong address is 1.

multiplying all we get, 3/4 * 2/3 * 1/2 * 1 = 1/4

(here multiplication is done, since we consider all of them are going to wrong addresses i.e simplifying we can write....

probability of (first letter will go to wrong address and 2nd letter will go to wrong address and 3rd letter will go to wrong address and 4th letter will go to wrong address and )

i.e probability of (first letter will go to wrong address * 2nd letter will go to wrong address *

3rd letter will go to wrong address * 4th letter will go to wrong address and )

therefore, P= ( 3/4 * 2/3 * 1/2 * 1 ) = 1/4.

so, the answer is 1/4 . )

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Answer is 9/24.You are making a big mistake in the second step,Pavel. Consider a situation; letter no 1 is put in the envelope of any other letter, say in the envelope of letter 2. Now what if letter 2 is considered next. This case letter 2 has 3 choices again.As letter1 is occupying its envelope it can enter in any remaining envelope.Got it? Sadman is correct. This can only be solved using derangement formula .You can see it in any combinatorics books.It can be proved using logic of inclusion-exclusion of set theory.

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yeah i got it..... thnqu :)

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