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# 4TH day: SOLVE THIS INTERESTING PROBLEM

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.......:-)

Note by Raja Metronetizen
3 years, 8 months ago

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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..... · 3 years, 8 months ago

I am getting the answer as $$\frac{log(2)}{n*log(2) + log(n!)}$$ · 3 years, 8 months ago

mine is the same.....then, you are right absolutely........could you prove your result.......??......:) · 3 years, 8 months ago

Should not the answer be $$\frac {log(2)}{n}$$? · 3 years, 8 months ago

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 · 3 years, 8 months ago

sorry,your understanding about the question is wrong....please go through the question minutely.....you have not understood it......try it...best of luck....:) · 3 years, 8 months ago

Can you point out my mistake, given that you have already solved it. · 3 years, 8 months ago