2 More generally show that P T1 x1 T2 x2 Tr xr Nn r

2.) More generally , show that P ( T1 = x1, T2 = x2, …, Tr = xr | Nn = r) = 1/ (n choose r) ,

0 < x1 < x2 < ...< xr n

This shows that given there are r successes during the first n trials, the trials at which these successes occur a random sample of size r ( without replacement)

This is the second part of a question that a person answered below and I thank the person for showing me the method. The above is a more general version of question 1 below.

1.) Let Nn be the number of successes in the first n trials. Show that P(T1 = x : Nn = 1) = 1/n , x = 1, 2, ..., n.

Solution

p=prob.of success

P ( T1 = x1, T2 = x2, …, Tr = xr | Nn = r) = P( T1 = x1, T2 = x2, …, Tr = xr AND Nn = r) / P(Nn =r)....

= [p^x1 * q^(1-x1) ] [ p^x2 * q^(1-x2) ] .....[ p^xr * q^(1- xr) ] / [ ( n choose r) p^r * q^(1-r) ]

=p^(x1+x2+..+xr) * q^ (1-x1-x2-....-xr) / [ ( n choose r) p^r * q^(1-r) ]

= p^r * q^ (1-r) / [ ( n choose r) p^r * q^(1-r) ]

= 1/ ( n choose r).......