00:01
Hi guys in this problem referring to example 5d so we have the desired probability okay p is equal to summation over i starts from 0 to k for i over k power r times i over k k power 1 minus i over k okay power n minus r over summation over j from 0 to k for j over k power r times 1 minus g power n minus r okay, so if k is large, we can use integral approximation.
00:59
Okay, so to find this, we need to prove that c of n and m, which is equal to integration from 0 to 1 for y power n times 1 minus y power m, d y, okay, and this is equal to n factorial, m factorial over n plus m plus 1, factorial okay so this is equal to y power n plus 1 over n plus 1 times 1 minus y power m we have the limits 1 and 0 minus integration from 0 to 1 for m times 1 minus y power m times 1 minus y power m minus 1 times negative 1 times y power n plus 1 over n plus 1 d y okay okay so this is d y so this is equal to m over n plus 1 times integration from 0 to 1 for y power n plus 1 times 1 minus y power m minus 1 d y this is m over n plus 1 times c of n plus 1 and m minus 1.
02:45
Okay, so the induction proof, c of n and 0, this is integration from 0 to 1 for y power n d y, this is equal to 1 over n plus 1.
03:02
Okay, so if m is equal to 1, so we have c of n and 1, which is integration from 0 to 1...