Random Tree

Calculus Level 5

f ( x , y ) f(x, y) is defined as a random positive integer with equal chances of being chosen between x x and y y inclusive. For example, f ( 1 , 3 ) f(1, 3) is either 1, 2, or 3, each with the same probability 1 3 . \frac 13.

What is the probability that f ( 1 , f ( 2 , f ( 3 , f ( 4 , f ( 5 , . . . ) ) ) ) ) = 1 ? f\big(1, f(2, f(3, f(4, f(5, ...))))\big)=1?

Bonus: For any positive integer n , n, what is the probability that f ( 1 , f ( 2 , f ( 3 , f ( 4 , f ( 5 , . . . ) ) ) ) ) = n ? f\big(1, f(2, f(3, f(4, f(5, ...))))\big)=n?


The answer is 0.36787944117.

Liam Lutton by Liam Lutton , 20, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Liam Lutton
Jun 19, 2018

Bonus answer: for any n, the probability that n = f(1, f(2, f(3, f(4, f(5, ...))))) is 1 ( e ∗ ( n − 1 ) ! ) \frac{1}{(e*(n-1)!)}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Give a more detailed solution rather than a answer please.

rajdeep brahma - 2 years, 11 months ago

Please give a detailed solution otherwise don't write solutions.Its just waste of your time.

D K - 2 years, 9 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...