Question

If the probability of success is 0.09, how many trials are needed to have a probability of atleast one success as 1/3 or more?

Answer 1

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,

`""^"n""C"_0 (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`. 

Thus we need minimum 5 trial or more.