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Probability....!!!!!!
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suprovo2000 suprovo2000 is offline
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Probability....!!!!!! - 12-05-2005, 01:14 AM

Five different letters are put at a random inside five addressed envelopes.Find the probability of putting exactly 2 letters in the correct envelopes.
Please tell me this answer in details properly.
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12-05-2005, 01:21 AM

Please post your question on the already existing threads for Probabilty.

Take care.


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hi - 12-05-2005, 01:22 AM

suppose the letters are a, b , c, d, e ... so elaborating further we can tell that let ab be put correctly or bc , or de, nand it goes on so what i feel it shud be of the order 5c2 ways that is 10 ways........ is this correct this is what i feel.


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12-05-2005, 01:31 AM

i think the answer to this problem should be 20. here goes the explaination....

for puttin the letters in the right envelope, u choose 2 out of 5, ie u can do it in 5C2 ways = 10
now for the 3 wrong envelopes (suppose a,b,c are wrong)... the arrangement can be...

in envelope a b c
letters ----- b c a
OR --------- c a b

so 2 ways.

thus the total no. of ways = 10*2 =20.

Last edited by killerguy; 12-05-2005 at 01:33 AM.
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1/6?? - 12-05-2005, 09:30 AM

people are giving the no of ways when the probability is asked.

anywho, is the answer 1/6?

any explanations?


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12-05-2005, 09:46 AM

yes boss..the answer is definately 1/6. n we are giving the no of ways bcoz its the funda using which u solve the problem. definately if the no. of ways u can arrange is 20, then probability is 20/(5!) = 1/6.

note : 5! bcoz no of ways in which u can put 5 letters in 5 envelops.
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12-05-2005, 10:34 AM

Kindly re post ur query in this thread and get the answer from more ppl thr.......


http://www.pagalguy.com/cat/showthread.php?t=8833


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12-05-2005, 01:34 PM

Quote:
Originally Posted by suprovo2000
Five different letters are put at a random inside five addressed envelopes.Find the probability of putting exactly 2 letters in the correct envelopes.
Please tell me this answer in details properly.
we will choose which 2 letters to be put into the correct envelops which can be done in
C(5,2) = 10 ways....further 3 letters are left which are to be dearranged such that no letter goes to the correct envelop which can be done in 3!/2! - 3!/3! = 3 - 1 = 2 ways....hence total possible cases are 10 * 2 = 20 out of 5! = 120 cases....so probability is 20/120 = 1/6

Last edited by khanna_sumit; 12-05-2005 at 01:38 PM. Reason: error
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13-05-2005, 12:46 AM

Quote:
Originally Posted by khanna_sumit
we will choose which 2 letters to be put into the correct envelops which can be done in
C(5,2) = 10 ways....further 3 letters are left which are to be dearranged such that no letter goes to the correct envelop which can be done in 3!/2! - 3!/3! = 3 - 1 = 2 ways....hence total possible cases are 10 * 2 = 20 out of 5! = 120 cases....so probability is 20/120 = 1/6
I do not understand ....3!/2! -3!/3!....Please explain this properly.......

Suprovo.
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13-05-2005, 09:47 AM

Hasnt it been too much pondering on one question...we have the answer and the explaination...so PG's, lets move on... post more questions !!!
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