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Probabilty Question Help >

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by ramkarpa » Sun Aug 13, 2006 6:53 pm
Tanya prepared 4 different letters to be sent to 4 different addresses,For each letter she prepared an envelope with correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability that only 1 will be put in the correct address ?

The answer is 1/3. Any ideas any one ?

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solution

by viksbhat » Tue Aug 15, 2006 6:08 am
4 letters .. 4 envelopes ..
probability that first envelop gets correct letter . 1/4
probability that 2nd envelop gets wrong letter 2/3
probability that 3rd envelop gets wrong letter 1/2
probability that 3rd envelop gets wrong letter 1/1

total no of cases can happen 4 .

the probability of condition to happen is 4*1/4*2/3*1/2*1/1 = 1/3

Hope this helps.

Please feel free to comment
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by ajith » Thu Oct 19, 2006 12:47 am
Basically This is a Derangement problem..

Total no Ways in which u can put letters = 4! =24

according to the question we have to put 1 letter in right place

we can select this letter in 4c1 ways ie 4 ways..

We have now to put all the remaining letters in wrong envelopes this can be done 2 ways (1st letter goes to 2nd 2 to 3rd 3rd to 1st or 1st to 3rd 2nd to 1st and 3rd to 2nd , these are the only ways

so total no of ways in this condition is satisfied is 4*2=8 ways

so the probability is 8/24= 1/3

you can read more about derangements in https://en.wikipedia.org/wiki/Derangement or
https://mathworld.wolfram.com/PartialDerangement.html
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