Given p = 0.09 (success)
q = 0.91 (failure)
We have to find number of trials ‘n.’
According to the problem,
P(X ≥ 1 ) > 1/3
(We must have atleast one success)
1 – P(X < 1) > 1/3
1 – P(X = 0) > 1/3
(or) P(X = 0) < 2/3
Using p.m.f, we have,
nC0(0.09) 0 (0.91) n < 2/3
(0.91)n < 2/3
we can use log tables to calculate (or) by trial method try for n = 1, 2,…… using calculator.
We observe that (0.91)5 < 2/3 but (0.91)4 = 0.6857 > 2/3. Thus we need a minimum of 5 trials or more.