Probability....!!!!!!

[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.

[Ninja]

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

Take care.

[coolrahine]

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.


rahine

[killerguy]

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.

[doofus3]

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

anywho, is the answer 1/6?

any explanations?

[killerguy]

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.

[rajsher]

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 (good question on probability..........)

[khanna_sumit]

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

[suprovo2000]

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.

[killerguy]

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 !!!