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.......:-)
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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 22n=n.
And, probability of getting 2 from all even integers can be expressed as nlog2, 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
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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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2^{34}
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\sum_{i=1}^3
\sin \theta
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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.....
Should not the answer be nlog(2)?
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Since number of positive even integers for first 2n numbers is simply 22n=n. And, probability of getting 2 from all even integers can be expressed as nlog2, 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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I am getting the answer as n∗log(2)+log(n!)log(2)
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mine is the same.....then, you are right absolutely........could you prove your result.......??......:)