I SAID,I WOULD POST A QUESTION DAILY FOR DISCUSSING AMONG YOURSELVES..... SO HERE IS A NEW PROBLEM FOR TODAY.....THANKS TO ALL WHO JOINS THIS DISCUSSIONS .....WISHING ALL THE BEST.......:-)

I AM HERE GIVING THE SOLUTION......TODAY I WOULD POST A NEW QUESTION....:)....
If the integer is 1 its probability is log(1),if the integer is 2 its probability is log(2),if the integer is 3 its probability is log(3)……likewise if the integer is n then its probability is log(n)……..let A denote the statement that the chosen number is even…..and B denote the statement that the chosen number is 2…. So (A intersection B)=2……hence the required probability is
P(B/A)=[P(A intersection B)]/P(A)=P(B)/P(A)=log(2)/[log(2)+log(4)+log(6)+log(8)+…….log(2n)]
=log(2)/log[(2^n)n!]=log(2)/[nlog(2)+log(n!)]………….thanx for joining....:).....try my new problem.....

Since number of positive even integers for first \(2n\) numbers is simply \(\frac{2n}{2}=n\).
And, probability of getting 2 from all even integers can be expressed as \(\frac{log 2}{n}\), according to the question

sorry,your understanding about the question is wrong....please go through the question minutely.....you have not understood it......try it...best of luck....:)

@Aditya Parson
–
i have already solved it.....:)...here you must think of conditional probability......that's my last hint.....try to think of this statement.......when we pick a number it is 2 given that the number is even........now i have almost said you what to do.......best of luck.....:)

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TopNewestI AM HERE GIVING THE SOLUTION......TODAY I WOULD POST A NEW QUESTION....:).... If the integer is 1 its probability is log(1),if the integer is 2 its probability is log(2),if the integer is 3 its probability is log(3)……likewise if the integer is n then its probability is log(n)……..let A denote the statement that the chosen number is even…..and B denote the statement that the chosen number is 2…. So (A intersection B)=2……hence the required probability is P(B/A)=[P(A intersection B)]/P(A)=P(B)/P(A)=log(2)/[log(2)+log(4)+log(6)+log(8)+…….log(2n)] =log(2)/log[(2^n)

n!]=log(2)/[nlog(2)+log(n!)]………….thanx for joining....:).....try my new problem.....Log in to reply

I am getting the answer as \(\frac{log(2)}{n*log(2) + log(n!)}\)

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mine is the same.....then, you are right absolutely........could you prove your result.......??......:)

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Should not the answer be \( \frac {log(2)}{n} \)?

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Since number of positive even integers for first \(2n\) numbers is simply \(\frac{2n}{2}=n\). And, probability of getting 2 from all even integers can be expressed as \(\frac{log 2}{n}\), according to the question

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sorry,your understanding about the question is wrong....please go through the question minutely.....you have not understood it......try it...best of luck....:)

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